This document is written in reference to qualifying exams given at the University of Louisville in past years. These solutions are not given from the University, but of my work alone as a way to study for my own qualifying exam. If any tips or recommendations come up and you feel you should share, feel free to raise an issue on GitHub where I have this document saved and open to the public here. To see the qualifying exams for yourself, visit this link. Thank you.
Jacob Townson
Let \(G/Z\) be cyclic with generator \(xZ\). Every element in \(G/Z\) can be written as \(x^k Z\) where \(k \in \mathbb{Z}\) and \(z \in Z\). We want to show first for all \(g \in G\) that \(g\) can be written in the form \(x^k z\) for some \(k \in \mathbb{Z}\) and \(z \in Z\). We know that \(G/Z\) is cyclic, so there exists some \(g \in G\) such that \(gZ = (xZ)^m = x^m Z\) for some \(m\). Well, this is true iff \((x^m)^{-1} g \in Z\). So there exists a \(z \in Z\) such that \((x^m)^{-1} g = z \implies g = x^m z\).
Now let \(g,h \in G\). Then
\[g=x^a z\] and \[h = x^b w\] for \(a,b \in \mathbb{Z}\) and \(z,w \in Z\).
Thus \[gh = x^a z x^b w = x^{a+b} zw = x^{b+a} wz = x^b w x^a z = hg\] because \(z,w \in Z\). THus \(gh = hg\) which implies that G is Abelian. QED
Let \(P\) be a prime ideal in the PID \(R\). So \(ab \in P\) iff \(a\) or \(b \in P\). But \(R\) is a PID, so \(P =(a)\) where \(a \in R\). So we want to show \(P\) is maximal using this information, ie. if \(I\) is an ideal that contains \(P\), then \(I=P\) or \(I=R\). So let \(I\) be an ideal containing \(P\) such that \(P \subset I \subset R\). Let \(I =(b)\) where \(b \in R\) because \(R\) is a PID. Now, \(a \in (a) \subset (b)\) which implies that there exists \(c \in R\) such that \(a =bc\). Since \(a = bc \in P\), we know either \(b\) or \(c\) is in \(P\) because it’s prime.
If \(b \in P\), then \(I = (b) \subset P\) which implies \(P=I\). On the other hand, if \(c \in P =(a)\), then there exists \(d \in R\) such that \(c = ad\). So \[a =bc =bad\]
Because \(R\) is an integral domain (because it’e a PID), we can reduce the above formula to \[1 = bd\]
Thus \(b\) is a unit which implies that \(I = (b) = R\).
Thus either \(P=I\) or \(I=R\) for all \(I\) containing \(P\), so \(P\) is maximal in \(R\). QED
We will prove this using set containment.
Because \(I\) and \(J\) are ideals in \(R\), for all \(x_i \in I\) and all \(y_i \in J\), \(x_i y_i \in I\) and \(x_i y_i \in J\). Moreover, \(\sum_{i=1}^{n} x_i y_i \in I\) and \(\sum_{i=1}^{n} x_i y_i \in J\). Thus \[\sum_{i=1}^{n} x_i y_i \in IJ\] which implies \[\sum_{i=1}^{n} x_i y_i \in I \cap J\] Thus \(IJ \subset I \cap J\)
Now we just need to show that \(IJ \supset I \cap J\). To do this we will use the fact that \(I+J=R\). Let \(a \in I \cap J\). We know \(1 \in R = I + J\). So \(1 = i + j\) for some \(i \in I\) and \(j \in J\). Then since \(R\) is commutative, \[a = a 1 = a (i+j) = ai + aj = ia + aj\] However \(ia \in IJ\) since \(i \in I\) and \(a \in J\) and \(aj \in IJ\) for similar reasons as our previous argument. By the definition of \(IJ\), this shows us \(a \in IJ\). Thus \(I \cap J \subset IJ\).
Thus we’ve proven that \(IJ = I \cap J\) in this circumstance. QED
To do this proof, we will use the broader proof of Cayley’s Theorem. This theorem states that if \(G\) is a group of order \(n < \infty\), then \(G\) is isomorphic to a subgroup of \(S_n\).
Let \(S\) be the set of all elements of \(G\). Now consider the action of \(G\) on \(S\), \[G \times S \rightarrow S, a*b:=ab\] This action defines a homomorphism \(\gamma: G \rightarrow \mathrm{Perm}(S)\). It is fairly easy to see that this homomorphism is one-to-one and onto. Thus, it follows that \(G\) is isomorphic to a subgroup of \(\mathrm{Perm}(S)\). And finally, since \(|S| = n\) we have that \(\mathrm{Perm}(S) \cong S_n\) This tells us that \(G \cong S_n\).
Thus we just need to show that \(S_n \cong A_m\). Let \(\Gamma : S_n \rightarrow A_{n+2}\) such that \(\Gamma : \sigma \mapsto \sigma\) if \(\sigma\) is even, or \(\Gamma : \sigma \mapsto (n+1 \ n+2) \sigma\) if \(\sigma\) is odd where \(\sigma \in S_n\). We just need to show that \(\Gamma\) is an isomorphism. Let \(\sigma_1, \sigma_2 \in S_n\). If \(\sigma_1, \sigma_2\) are even, then \[\Gamma (\sigma_1 \sigma_2) = \sigma_1 \sigma_2 = \Gamma (\sigma_1) \Gamma (\sigma_2)\] If \(\sigma_1, \sigma_2\) are both odd, then \[\Gamma (\sigma_1 \sigma_2) = \sigma_1 \sigma_2 = \Gamma (\sigma_1) \Gamma (\sigma_2)\] This is true because \((n+1 \ n+2)^2\) is the identity permutation, and \(\sigma_1 \sigma_2\) is even. Finally WLOG let \(\sigma_1\) be even and \(\sigma_2\) be odd. Then \[\Gamma(\sigma_1) \Gamma(\sigma_2) = \sigma_1 (n+1 \ n+2) \sigma_2 = (n+1 \ n+2) \sigma_1 \sigma_2 = \Gamma (\sigma_1 \sigma_2)\] Thus \(\Gamma\) is a homomorphism. It is also obvious that for all \(\sigma\) there exists a \(\Gamma(\sigma)\), so we know that \(\Gamma\) is onto. And finally, it is obvious that if \(\Gamma(\sigma_1) = \Gamma(\sigma_2)\), then \(\sigma_1 = \sigma_2\). Thus \(\Gamma\) is an isomorphism.
Using this fact and Cayley’s Theorem, we then know that \(G \cong S_n \cong A_{n+2}\) which proves that any finite group is isomorphic to a subgroup of \(A_n\). QED
In order to prove that \(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}\) is or is not Galois, we will use the fundamental theorem of Galois theory. Thus, we know if we prove that \(\mathbb{Q}(\sqrt[3]{2})\) is a splitting field over some function \(t(x)\) in \(\mathbb{Q}\), then we know the extension is Galois. Consider the polynomial \(t(x) = x^3 - 2\). This polynomial has no roots in \(\mathbb{Q}\), but has root \(\sqrt[3]{2} \in \mathbb{Q}(\sqrt[3]{2})\). Unfortunately, this is not enough for us to factor \(t(x)\) into linear factors, as we will need some \(\omega \in \mathbb{C}\) to further factor the polynomial. Thus we know that \(\mathbb{Q}(\sqrt[3]{2})\) is not a splitting field, and thus it must not be Galois by the fundamental theorem of Galois theory. QED
First off, to save time let \([G,G] =C\). So we want to show that for all \(aC, bC \in G/C\) that \[abC = aCbC = bCaC = baC\] Well, \(aba^{-1} b^{-1}\) generates \(C\). So, \(C = aba^{-1} b^{-1}C\), which implies that \(abC = baC\) through cancellation. Thus \(G/C\) is Abelian. QED
First we must show that \(C \subseteq K\). Well, because \(G/K\) is Abelian, we know that \(abK = baK\) which implies that \(a^{-1}b^{-1}abK = K\). Thus \(C \subseteq K\). Then, because \(C\) is a subgroup of \(G\), we know that \(C\) must then be a subgroup of \(K\) since \(K\) is a group containing \(C\). Lastly, we must show that \(C \unlhd K\). Well, because \(C \unlhd G\), \(K \unlhd G\), and \(C \leq K\), we can easily see that these things imply that \(C\) is normal in \(K\). Thus \(C \unlhd K\). QED
(\(a \implies b\))
Suppose \(I\) is a nonzero ideal of a field \(R\). So \(I\) contains some \(0 \neq a\). Then since \(R\) is a field, \(a^{-1}\) exists and \(1 = a^{-1} a \in I\), so \(I\) contains \(1\) and hence \(I = R\). The trivial ideal also exists, and is trivial to show.
(\(b \implies a\))
Let \(R\) have only two ideals, the trivial and \(R\). Suppose \(R\) is not a field. Then some nonzero element \(a\) does not have an inverse. If we set \(aR = \{ar | r \in R \}\). This is a proper nonzero ideal, which gives us a contradiction to our original statement. Thus \(R\) must be a field for \(R\) to only contain 2 ideals, the trivial and \(R\) itself.
Thus the two statements are equivalent, \(R\) is a field and \(R\) contains only two ideals, those being the trivial ideal and \(R\) itself. QED
Here we can use what I like to call the double extension lemma to find the degree of the extension. So we know that \([F: \mathbb{Q}] = [\mathbb{Q}(i, \sqrt{3}): \mathbb{Q}(\sqrt{3})][\mathbb{Q}(\sqrt{3}): \mathbb{Q}]\). Thus because the minimum polynomials for the extensions are \(x^2 +1\) and \(x^2 - 3\) respectively, we know that the degree of the extension is \(2*2=4\). A basis for this extension is \([1, i, \sqrt{3} , i \sqrt{3}]\). QED
The primitive element of this extension is \(i+\sqrt{3}\) as this element makes the simple extension \(\mathbb{Q}(i+\sqrt{3}) = F\). This can be shown through the following containment arguments:
First let \(\mathbb{Q}(i+\sqrt{3}) = K\). It is obvious that \(K \subset F\), so we’re good there. Thus we just need to show that \(F \subset K\). So we need to show that \(i, \sqrt{3} \in K\). Well, let \(a = i+\sqrt{3}\). We know \(a, a^2 \in K\). \(a^2 = -1 +2i \sqrt{3} +3\), which implies that \(2i \sqrt{3} \in K\). Then we can look at \(2i \sqrt{3} * (i + \sqrt{3}) \in K\) which implies that \(3i - \sqrt{3} \in K\), which implies that \((3i-\sqrt{3})+(i +\sqrt{3}) \in K \implies i \in K\). Thus \(i+ \sqrt{3} -i \in K \implies \sqrt{3} \in K\). So finally we have proven both directions of containment, implying that the primitive element \(i+\sqrt{3}\) makes \(F\) into a simple extension of \(\mathbb{Q}\) which we call \(K\). QED
To begin, we must understand what it means for a group to be simple. A simple group is a group of whose only normal subgroups are the trivial group and the entire group itself. Thus we want to show there are no proper normal subgroups other than the trivial group.
To do this we will use Sylow’s Theorem and a proof by contradiction.
Let \(G\) be a group of order \(300\) such that \(G\) is simple. Well, \(300 = 3 \times 2^2 \times 5^2\). We will denote the number of Sylow-p subgroups as \(n_p\). Notice then that \(n_5 | 12\) and \(n_5 = 1 + 5k\). This implies that \(n_5 = 1\) or \(6\). However, because \(G\) is simple there cannot only be one Sylow-5 subgroup, otherwise we would have a proper nontrivial normal subgroup. Thus \(n_5 = 6\). Because of this, and because \(G\) is simple, we can say that \(G\) embeds into \(S_6\) if we let \(G\) act on the Sylow-5 subgroups by conjugation. But then by Lagrange’s Theorem, \(300\) should divide \(6!\), which it does not. Thus we have a contradiction, which implies that \(G\) is not simple, thus a group of order \(300\) cannot be simple. QED
Here we can use what I like to call the double extension lemma to find the degree of the extension. So we know that \([F: \mathbb{Q}] = [\mathbb{Q}(i, \sqrt{3}): \mathbb{Q}(\sqrt{3})][\mathbb{Q}(\sqrt{3}): \mathbb{Q}]\). Thus because the minimum polynomials for the extensions are \(x^2 +1\) and \(x^2 - 3\) respectively, we know that the degree of the extension is \(2*2=4\). A basis for this extension is \([1, i, \sqrt{3} , i \sqrt{3}]\). QED
Recall for this problem the definition of an ideal for rings. \(I\) is an ideal of the ring \(R\) if for all \(x \in R\) and \(y \in I\), then \(xy, yx \in I\).
We will use this definition to show \(A/B\) is an ideal of \(R/B\). Notice \(R/B = \{ r + B|r \in R \}\). So let’s consider two elements, \(r+B \in R/B\) and \(a + B \in A/B\) such that \(r \in R\) and \(a \in A\). We want to show that \((r+B)(a+B) \in A/B\) and \((a+B)(r+B) \in A/B\). Well, \[(r+B)(a+B) = ra+aB+rB+B = ra+B\] because \(B\) is an ideal of \(R\). But \(A\) is also an ideal of \(R\) so \(ra \in A\), which implies that \(ra + B \in A/B\). Looking at the other direction, we find that \[(a+B)(r+B) = ar+aB+rB+B = ar+B\] and for similar reasons, we then know that \(ar+B \in A/B\). Thus \(A/B\) is an ideal of \(R/B\). QED
We will need to separate this problem into three parts.
First we must prove that \(f\) is well-defined. So we want to show that if \(x=y \in R\), then \(f(x+B)=f(y+B)\). Well, if \(x=y\), then \[f(x+B) = x+A = y+A = f(y+B)\] Thus \(f\) is well-defined by definition.
Next we will show that \(f\) is an onto function. So we need to show that for all \(x \in R\), and \(x+B \in R/B\), then \(f(x+B) \in R/A\). Well, by definition this is obvious, because for all \(x \in R\) and \(x+B \in R/B\), \(f(x+B) = x+A \in R/A\). Thus the function must be onto.
Finally we must show that \(f\) is a homomorphism. First note that \(f(B)=A\) by definition, so the identity maps to the respective identity. Now we must prove that addition and multiplication are preserved. First, let’s look at \(f(x+B+y+B)\) where \(x,y \in R\). So, \[f(x+B+y+B) = f(x+y+B) = x+y+A = (x+A) + (y+A) = f(x+B)+f(y+B)\] Thus addition is preserved. So now we must only prove that multiplication is preserved. Assume a similar situation as above, but let’s look at \(f((x+B)(y+B))\). Well, \[f((x+B)(y+B)) = f(xy+xB+yB+B) = f(xy+B)\] because \(B\) is an ideal. Now we can further this manipulation using that \(A\) is an ideal to see that \[f(xy+B)=xy+A=xy+xA+yA+A=(x+A)(y+A)=f(x+B)f(y+B)\] Thus because identities, addition, and multiplication are preserved, we know then that \(f\) is a homomorphism.
Thus \(f\) is a well-defined, onto, homomorphism. QED
This statement is easily shown by the first isomorphism theorem for rings, where the kernel of \(f\) is simply \(A/B\). A more formal proof is given later in the document.
First note that \(t(x) = x^6 -4 = (x^3 - 2)(x^3 + 2)\). The roots of this polynomial are \(\sqrt[3]{2}, -\sqrt[3]{2}, \omega_1\sqrt[3]{2},\omega_2\sqrt[3]{2},\alpha_1\sqrt[3]{2},\alpha_2\sqrt[3]{2}\) where \(\omega_1, \omega_2, \alpha_1, \alpha_2 \in \mathbb{C}\). We can then see that the splitting field of \(t(x)\) over \(\mathbb{Q}\) is \(\mathbb{Q}(\sqrt[3]{2},i)\). This is the splitting field because in this extension \(t(x)\) can be factored into linear factors. QED
First, recall if \(I \subset R\), \(I\) is a left ideal if for all \(r \in R\) and \(a \in I\), \(ra \in I\).
Assume \(R\) has no proper left ideal, thus the only left ideals are the trivial and the entire ring. So if \(s \in R\) is nonzero and \(R\) has no proper left ideals, then we can define \(I := \{rs|r \in R\} =R\) and there exists \(r \in R\) such that \(rs = 1_R\). Also note \[(rs)r=r(sr)=r \implies r(1_R -sr)=0 \implies sr =1_R\] Because \(rs = 1_R = sr\), \(R\) must be a division ring.
Now we must prove the other direction. Assume \(R\) is a division ring and \(I \neq \{0\}\) is a left ideal of \(R\). Then for all \(s \in I\), \[1_R = s^{-1}s \in I \implies I=R\] Thus \(R\) has no proper left ideals. QED.
Note: This is the proof of the 2nd Isomorphism Theorem for Groups
Recall that \(HK\) can only be a subgroup if \(HK = KH\). Thus if \(H \unlhd G\), and we let \(h \in H\) and \(k \in K\), by assumption \(khk^{-1} \in H\). Thus \[kh = (khk^{-1})k \in HK\] This proves that \(KH \subseteq HK\). Similarly, \(hk = k(k^{-1}hk) \in KH\) which proves the reverse containment. Thus \(HK \leq G\). QED
Since \(K\) normalizes \(H\) and \(H\) normalizes itself trivially, this implies that \(HK\) normalizes \(H\). Thus \(H \unlhd HK\). Since \(H \unlhd HK\), the quotient group \(\frac{HK}{H}\) is well defined. Define the map \(\Gamma : K \rightarrow \frac{HK}{H}\) by \(\Gamma (k) = kH\). Since the group operation in \(\frac{HK}{H}\) is well defined, it’s easy to see that \(\Gamma\) is a homomorphism: \[\Gamma(k_1 k_2) = (k_1 k_2)H = k_1 H *k_2 H = \Gamma(k_1) \Gamma(k_2)\] Alternatively, the map \(\Gamma\) is just the restriction to the subgroup \(K\) of the natural progression homomorphism \(\pi : HK \rightarrow \frac{HK}{H}\), so it’s also a homomorphism. It is clear then from the definition of \(HK\) that \(\Gamma\) is onto. The identity of \(\frac{HK}{H}\) is simply \(H\), so the kernel of \(\Gamma\) consists of the elements \(k \in K\) with \(kH = H\), or where \(k \in H\). Thus \(\mathrm{ker}\Gamma = H \cap K\). So by the first isomorphism theorem, \(H \cap K \unlhd K\). QED
Following the proof of part (b), by the First Isomorphism Theorem, \(\frac{HK}{H} \cong \frac{K}{H \cap K}\) QED
The extension field \(K\) of \(F\) is a splitting field for the polynomial \(f(x) \in F[x]\) if \(f(x)\) factors completely into linear factors in \(K[x]\) and \(f(x)\) does not factor completely into linear factors over any proper subfield of \(K\) containing \(F\).
Let \(f(x) = c_1x^n+c_{n-1}x^{n-1}+...+c_1x+c_0\). Let \(K\) be the splitting field of \(f(x)\) over \(F\). And finally, this all means that \(f(x+a) = c_1(x+a)^n+c_{n-1}(x+a)^{n-1}+...+c_1(x+a)+c_0\). Since \(K\) is the splitting field of \(f(x)\) over \(F\), we then know that \(f(x)\) has roots \(r_1,r_2,...,r_n \in K\). Also note that some of these \(r_i\) values could be duplicates or even zeroes. So by definition, in \(K\), \(f(x) = (x-r_1)(x-r_2)...(x-r_n)\). This means that in \(K\), \(f(x+a) = (x+a-r_1)(x+a-r_2)...(x+a-r_n)\). From here it is easy to see that the roots of \(f(x+a)\) are simply \(r_1-a,r_2-a,...,r_n-a\). All of these values are in \(K\) as well because for all \(i\), \(r_i \in K\) and \(a \in F \subset K\). So because we can factor \(f(x+a)\) into linear factors in \(K\), \(K\) must also be the splitting field of \(f(x+a)\). Note, the situation in which we switch assumptions such that \(K\) is the splitting field over \(f(x+a)\) to prove that \(K\) must also be the splitting field of \(f(x)\) over \(F\) is similar, so we leave this out. QED
Recall that \(HK\) can only be a subgroup if \(HK = KH\). Thus if \(H \unlhd G\), and we let \(h \in H\) and \(k \in K\), by assumption \(khk^{-1} \in H\). Thus \[kh = (khk^{-1})k \in HK\] This proves that \(KH \subseteq HK\). Similarly, \(hk = k(k^{-1}hk) \in KH\) which proves the reverse containment. Thus \(HK \leq G\). QED
Let \(G\) be a group of order \(56 = 2^3 * 7\). We will denote the number of Sylow-\(p\) subgroups by \(n_p\). So \(n_2 = 2k + 1\) where \(k \in \mathbb{Z^+}\) and \(n_2 | 7\). Also \(n_7 = 7j+1\) where \(j \in \mathbb{Z^+}\) and \(n_7 | 2^3 = 8\). So \(n_2 \in \{1,7\}\) and \(n_7 \in \{1,8\}\).
If either \(n_7\) or \(n_2\) are \(1\), then we’re done as this means that there exists a normal proper nontrivial Sylow-\(p\) subgroup in our group \(G\). Consider \(n_7 = 8\). Because \(7\) is prime and the subgroups are cyclic, and because we know the subgroups intersect trivially, \(n_7 = 8 \implies\) there exists \(8*6 = 48\) non identity elements in all of the Sylow \(7\)-subgroups. Notice, \(56=48+8\). But if \(n_2 = 7\), then there are \(8\) distinct elements in each of the Sylow-\(2\) subgroups. But we run into a problem, because \(56=48+8\), so then there must be only 1 Sylow-\(2\) subgroup in order to have \(56\) elements. Thus either \(n_2 =1\) or \(n_7 = 1\), thus \(G\) could not be simple as it has a normal proper nontrivial subgroup. QED
Recall that a ring is commutative if the multiplication operation in the ring is commutative. Thus we want to show that for every Boolean ring, if \(a,b \in R\) then \(ab = ba\).
To begin this problem, consider \(a+b = (a+b)^2\). Well, \[(a+b)^2 = (a+b)(a+b)=a(a+b)+b(a+b)\] \[ = a^2 +ab+ba+b^2=a+ab+ba+b\] When cancelling \(a\) and \(b\) from both sides, we find that \(0 = ab+ba \implies ab = -ba\), but because \(R\) is Boolean, we can easily see that \(-c = c\) for all \(c \in R\). Thus we have proven that all Boolean rings are indeed commutative. QED.
Recall an integral domain is a commutative ring with no zero divisors. Also an element in an integral domain is irreducible if it is not equal to the product of two non-units. Also recall that if \(R\) is our integral domain, \(u \in R\) is a unit if for some \(v \in R\), \(uv = e\) where \(e\) is the identity in \(R\). Lastly, recall that \(p \in R\) is prime if \(p|ab \implies p|a\) or \(p|b\).
Let \(p \in R\) be a prime element where \(R\) is an integral domain. Assume \(p = ab\) where \(a,b \in R\). Obviously \(p|ab\), so because \(p\) is prime, \(p|a\) or \(p|b\). WLOG, assume \(p|a\) such that \(px = a\) for some \(x \in R\). If \(1\) is the identity of \(R\), then \[1*a=a=px=(ab)x=a(bx)\] Because \(R\) is an integral domain, \(a\) can be cancelled out implying that \(1=bx\), thus \(b\) is a unit. This proves that \(p\) is also irreducible in an integral domain. QED.
This statement is false! Consider \(L = \mathbb{Q}(\sqrt[4]{2})\), \(K = \mathbb{Q}(\sqrt{2})\), and \(F = \mathbb{Q}\). In this case, \(K/F\) is Galois because \(K\) is a splitting field over \(F\), and it is similar for \(L/K\). However, \(L/F\) is not Galois, as \(L\) does not make a splitting field over \(F\). This can be seen from the polynomial \(f(x)=x^4-2 \in \mathbb{Q}[x]\). Not all of the roots of \(f(x)\) are contained in \(L\), as \(L\) does not contain any imaginary numbers. QED
Assume \(E/F\) is finite. Let \(\beta \in E/F\). We have then that \(F \subseteq F(\beta) \subseteq E\). We know \[[E:F] =n \in \mathbb{N} \implies [F(\beta):F] \leq n\] Consider \(\{1,\beta,\beta ^2, ..., \beta ^n\} \subseteq E\). So there exists \(c_0, ..., c_n \in F\) such that \(c_0+ c_1 \beta + ... + c_n \beta ^n =0\), where at least one \(c_i \neq 0\). Let \(v(x) = c_n x^n +...+c_0\), so \(v(\beta) =0\). Thus if \(E/F\) is finite, then \(E\) is algebraic over \(F\). QED
Here we can use what I like to call the double extension lemma to find the degree of the extension. So we know that \([F: \mathbb{Q}] = [\mathbb{Q}(i, \sqrt{3}): \mathbb{Q}(\sqrt{3})][\mathbb{Q}(\sqrt{3}): \mathbb{Q}]\). Thus because the minimum polynomials for the extensions are \(x^2 +1\) and \(x^2 - 3\) respectively, we know that the degree of the extension is \(2*2=4\). A basis for this extension is \([1, i, \sqrt{3} , i \sqrt{3}]\). QED
The primitive element of this extension is \(i+\sqrt{3}\) as this element makes the simple extension \(\mathbb{Q}(i+\sqrt{3}) = F\). This can be shown through the following containment arguments:
First let \(\mathbb{Q}(i+\sqrt{3}) = K\). It is obvious that \(K \subset F\), so we’re good there. Thus we just need to show that \(F \subset K\). So we need to show that \(i, \sqrt{3} \in K\). Well, let \(a = i+\sqrt{3}\). We know \(a, a^2 \in K\). \(a^2 = -1 +2i \sqrt{3} +3\), which implies that \(2i \sqrt{3} \in K\). Then we can look at \(2i \sqrt{3} * (i + \sqrt{3}) \in K\) which implies that \(3i - \sqrt{3} \in K\), which implies that \((3i-\sqrt{3})+(i +\sqrt{3}) \in K \implies i \in K\). Thus \(i+ \sqrt{3} -i \in K \implies \sqrt{3} \in K\). So finally we have proven both directions of containment, implying that the primitive element \(i+\sqrt{3}\) makes \(F\) into a simple extension of \(\mathbb{Q}\) which we call \(K\). QED
Consider the map \(\theta : G \rightarrow \mathrm{Aut}(G)\) defined as \(\theta(g) = \gamma_g\), where \(\gamma_g\) is the automorphism of \(G\) defined by \(\gamma_g(h) = ghg^{-1}\).
Lemma 1: \(\theta\) is a homomorphism.
Proof: We have \(\theta(g_1 g_2) = \gamma_{g_{1}g_{2}}\), and \[\gamma_{g_{1}g_{2}}(h)=(g_1 g_2)h(g_1 g_2)^{-1} =g_1(g_2 h g_2 ^{-1})g_1 ^{-1} = \gamma_{g_1}(\gamma_{g_2}(h)).\]
Lemma 2: \(\mathrm{ker}(\theta) = Z(G)\).
Proof: We have \[ker(\theta) = \{g: \theta (g) = e \} = \{g : \gamma_g = e\}\] \[= \{g: \gamma_g (h) = h\} = \{g: gh = hg\} = Z(G)\]
Finally, by the first isomorphism theorem, we have \(G/ker(\theta) = G/Z(G) \cong im (\theta) = \mathrm{Inn}(G)\), as desired. QED.
Since \(P \cap H\) is a \(p\)-group, as it is contained \(P\), we only need to show that \(P \cap H\) has maximal \(p\)-power order in \(N\). By the Sylow Theorem, \(P \cap H\) is contained in a conjugate of \(P\), which we’ll call \(K\) such that \(K \subset gPg^{-1}\). Thus \(g^{-1}Kg \subset P\). Also \(g^{-1}Kg \subset g^{-1}Hg = H\), so \(g^{-1}Kg \subset P \cap H \subset K\). Since \(|K| = |g^{-1}Kg|\), we get that \(|P \cap H| = |K|\), so \(P \cap H = K\) is a \(p\)-Sylow subgroup of \(H\). QED
Let \(IJ \subseteq P\) and suppose WLOG that \(J \nsubseteq P\). Then there is some \(b \in J\) such that \(b \notin P\). But then \(aRb \subseteq P\) for all \(a \in I\), so by (i), \(a \in P\) for all \(a \in I\).
Let \(IJ \subseteq P\) so that \(I \subseteq P\) or \(J \subseteq P\). Then if we let \(a \in I\) and \(b \in J\), if \(aRb \subseteq P\), then \(RaRb \subseteq P\), so by (ii), \(a \in P\) or \(b \in P\) which implies by definition that \(P\) is prime.
Recall that the Galois group of a separable polynomial \(f(x) \in F[x]\) is defined to be the Galois group of the splitting field of \(f(x)\) over \(F\).
Well, \(f(x) = x^4 - 5x^2 +6 = (x^2-2)(x^2-3)\). We cannot factor \(f(x)\) any futher in \(\mathbb{Q}\) but in \(\mathbb{Q}(\sqrt{2}, \sqrt{3})\), \(f(x)\) can be factored into linear factors. So the splitting field over \(\mathbb{Q}\) is \(\mathbb{Q}(\sqrt{2}, \sqrt{3})\). So the automorphisms in the Galois group map \(\sqrt{2} \rightarrow -\sqrt{2}\), \(\sqrt{3} \rightarrow -\sqrt{3}\), the identity map, and interchanging the \(\sqrt{2}\) and \(\sqrt{3}\). Thus the Galois group has order \(4\) and is not cyclic, implying that the Galois group must be \(\mathbb{Z}_2 \times \mathbb{Z}_2\). QED
For \(g \in N_G(H)\), it is clear that \(\sigma_g:h \rightarrow ghg^{-1}\) defines an automorphism of \(H\). Define a map: \[ \sigma : N_G(H) \rightarrow Aut(H)\] where \(g \rightarrow \sigma_g\). Since for all \(a,b \in N_G(H), h \in H\), we have that \(\sigma_a \sigma_b (h) = abhb^{-1} a^{-1} = \sigma_{ab}(h)\). Then \(\sigma\) is a group homomorphism from \(N_G(H)\) to \(Aut(H)\). Note that \[\mathrm{Ker}(\sigma) = \{g \in N_G(H) | ghg^{-1} = h, \forall h \in H\}\] \[ = C_G(H) \cap N_G(H) = C_G(H)\] From the first isomorphism theorem we then have that \(N_G(H)/C_G(H)\) is isomorphic to a subgroup of \(Aut(H)\). QED
Assume \(D\) is Artinian, and has a unit \(1_D\) that is a multiplicative identity element. For any \(0 \neq a \in D\), consider the principle ideal generated by \(a^i\), \((a^i)\). Since \(a^{n+1}d=a^n (ad)\) for any \(d \in D\), we have that \((a^{n+1}) \subseteq (a^n)\). Thus we have our descending chain of ideals \[(a) \supseteq (a^2) \supseteq ... \supseteq (a^n) \supseteq ...\] which will stabilize because \(D\) is Artinian. Thus for some positive integer \(m\), \[(a^m)=(a^{m+1})\] Then since \(a^m = a^m 1_D \in (a^m)\), there exists \(s \in D\) such that \(a^{m+1} s = a^m\) or \(a^m (1_D - a s)=0\). Using the properties of an integral domain, this implies that \(as=1_D\). Thus we have shown that every \(0 \neq a \in D\) has a multiplicative inverse thus \(D\) is also a field.
Assume \(D\) is an integral domain and a field. Then \(D\) has only two ideals, the trivial and \(D\) itself. Thus every strictly descending chain of ideals is of finite length, implying that \(D\) is Artinian. QED
Let \(p \in R\) be a prime element where \(R\) is an integral domain. Assume \(p = ab\) where \(a,b \in R\). Obviously \(p|ab\), so because \(p\) is prime, \(p|a\) or \(p|b\). WLOG, assume \(p|a\) such that \(px = a\) for some \(x \in R\). If \(1\) is the identity of \(R\), then \[1*a=a=px=(ab)x=a(bx)\] Because \(R\) is an integral domain, \(a\) can be cancelled out implying that \(1=bx\), thus \(b\) is a unit. This proves that \(p\) is also irreducible in an integral domain.
Now assume that \(p\) is irrecucible. To complete the proof, we want to show that \(p\) is prime. If \(p\) is irreducible, then \((p)\) is maximal in the set of all proper principal ideals of \(R\). Since \(R\) is a principal ideal domain then every ideal is principal, so in fact \((p)\) is maximal in \(R\) itself. Since \(R\) is an integral domain (because it’s a PID), then \(R\) has an identity. So then \((p)\) must be prime which implies \(p\) must then be prime. QED
Let \(G\) be a group of order \(56 = 2^3 * 7\). We will denote the number of Sylow-\(p\) subgroups by \(n_p\). So \(n_2 = 2k + 1\) where \(k \in \mathbb{Z^+}\) and \(n_2 | 7\). Also \(n_7 = 7j+1\) where \(j \in \mathbb{Z^+}\) and \(n_7 | 2^3 = 8\). So \(n_2 \in \{1,7\}\) and \(n_7 \in \{1,8\}\).
If either \(n_7\) or \(n_2\) are \(1\), then we’re done as this means that there exists a normal proper nontrivial Sylow-\(p\) subgroup in our group \(G\). Consider \(n_7 = 8\). Because \(&\) is prime and the subgroups are cyclic, and because we know the subgroups intersect trivially, \(n_7 = 8 \implies\) there exists \(8*6 = 48\) non identity elements in all of the Sylow \(7\)-subgroups. Notice, \(56=48+8\). But if \(n_2 = 7\), then there are \(8\) distinct elements in each of the Sylow-\(2\) subgroups. But we run into a problem, because \(56=48+8\), so then there must be only 1 Sylow-\(2\) subgroup in order to have \(56\) elements. Thus either \(n_2 =1\) or \(n_7 = 1\), thus \(G\) could not be simple as it has a normal proper nontrivial subgroup. QED
Assume \(E/F\) is finite. Let \(\beta \in E/F\). We have then that \(F \subseteq F(\beta) \subseteq E\). We know \[[E:F] =n \in \mathbb{N} \implies [F(\beta):F] \leq n\] Consider \(\{1,\beta,\beta ^2, ..., \beta ^n\} \subseteq E\). So there exists \(c_0, ..., c_n \in F\) such that \(c_0+ c_1 \beta + ... + c_n \beta ^n =0\), where at least one \(c_i \neq 0\). Let \(v(x) = c_n x^n +...+c_0\), so \(v(\beta) =0\). Thus if \(E/F\) is finite, then \(E\) is algebraic over \(F\). QED
We will denote the minimal polynomial here as \(m(x)\). Consider \(m(x) = x^4 - 2x^2 -1\). First we can see that \(m(a)=0\). We can also see from Eisenstein’s Criteria that \(m(x)\) is irreducible in \(\mathbb{Q}\). Thus \(m(x)\) is indeed the minimal polynomial of \(a\) over \(\mathbb{Q}\).
We know that \([E:\mathbb{Q}] = \mathrm{deg}(m(x))\). Thus \([E:\mathbb{Q}]=4\).
Well, \(Gal(E/\mathbb{Q})\) fixes all \(q \in \mathbb{Q}\), so to find the automorphisms we just need to observe the roots. Well, there are \(4\) possible automorphism maps for the roots: \[\sqrt{1+\sqrt{2}} \rightarrow \sqrt{1+\sqrt{2}}\] \[\sqrt{1+\sqrt{2}} \rightarrow -\sqrt{1+\sqrt{2}}\] \[\sqrt{1+\sqrt{2}} \rightarrow \sqrt{1-\sqrt{2}}\] \[\sqrt{1+\sqrt{2}} \rightarrow -\sqrt{1-\sqrt{2}}\] Notice first off that these automorphisms do not make a cyclic group, so our Galois group is not \(\mathbb{Z}_4\), and our Galois group turns out to be \(\mathbb{Z}_2 \times \mathbb{Z}_2\). QED
(Note, this answer depends on how you define the Galois group. Is it the group of Automorphisms only, or does the extension need to be Galois?)
This is irreducible because it is a cyclotomic polynomial.
Let \(\psi\) be the primitive \(8^{th}\) root of unity. Then \(\psi, \psi^2, \psi^3, \psi^4\) are the roots of \(q(x)\).
Let \(S = \mathbb{Q}(\psi)\). Since \(q(x)\) is irreducible in \(\mathbb{Q}\), and \(q(x)\) can be reduced into linear factors in \(\mathbb{Q}(\psi)\), \(S\) must be the splitting field of \(q(x)\) over \(\mathbb{Q}\). Thus, with \(q(x)\) as the minimal polynomial, \([S:\mathbb{Q}] = \mathrm{deg}(q(x))=4\).
We know by the fundamental theorem of Galois Theory that \(|\mathrm{Gal}(S/\mathbb{Q})| = 4 = [S:\mathbb{Q}]\). Thus \(\mathrm{Gal}(S/\mathbb{Q}) = \mathbb{Z}_4\) where \(\psi\) maps to all possible powers of \(\psi\).
Assume that \(A_8\) does in fact have a subgroup of index \(7\). We’ll call this subgroup \(H\). Then \([A_8:H] = 7\). Then there exists a group action of \(A_8\) on the cosets of \(H\) by conjugation. Define \(f\) to be a homomorphism such that \(f\) embeds \(A_8\) into \(S_7\). Thus by Lagrange’s Theorem, \(7!\) should divide the order of \(A_8\), \(8!/2\), which is not true. Thus we have a contradiction, so \(A_8\) has no subgroup with index \(7\). QED
For \(a \in X\), let \(O(a) = \{g(a)| g\in G\}\). This is the orbit of \(a\) under the action of \(G\) on \(X\). Also for \(a \in X\), let \(G_a = \{g \in G | g(a) = a\}\). This is the stablilizer of \(a\) in \(G\). Let \(\mathbb{O} = \{O(a)|a \in X\}\) be the set of orbits, and let \(G/G_a = \{hG_a | h \in G\}\) be the set of all left cosets of \(G_a\) in \(G\).
Let \(a \in X\). We want to show that for all \(g,h \in G_a\), \(gh^{-1} \in G_a\). Well, let \(g,h \in G_a\). Then \[gh(a) = g(h(a)) = g(a) = a\] by associativity, so \(gh \in G_a\). Also note that \(e \in G_a\) where \(e\) is the identity element of \(G\), so, \[e(a) = a = h^{-1}h(a) = h^{-1}(a)\] which implies that \(h^{-1} \in G_a\). Thus from here we can tell that \(gh^{-1} \in G_a\) which implies that \(G_a \leq G\). QED
Define \(\phi : G/G_a \rightarrow O(a)\). So for any \(gG_a \in G/G_a\), \(\phi(gG_a) = g(a)\) where if \(h \in G_a\), \[h(a) = \phi(hG_a) = \phi(G_a) = a\] To show that \(\phi\) is well-defined, let \(g = h \in G\). Then \[\phi(gG_a) = g(a) = h(a) = \phi(hG_a)\] So \(\phi\) is indeed well-defined. In order to show \(\phi\) is onto, notice that for all \(g(a) \in O(a)\), there exists sone \(gG_a \in G/G_a\) such that \(\phi(gG_a) = g(a)\). Thus \(\phi\) is indeed onto. Finally, to show that \(\phi\) is one-to-one, let \[\phi(gG_a) = g(a) = h(a) = \phi(hG_a)\] Because \(g(a) = h(a)\), we can easily see that either \(g = h\) or \(g,h \in G_a\), which implies that \(gG_a = hG_a\). Thus \(\phi\) is one-to-one. QED
The orbits are the conjugacy classes of \(G\). \(x\) belongs to an orbit of size \(1\) iff \(x \in Z(G)\) since \(gxg^{-1} = x\) for all \(g \in G\) iff \(gx = xg\) for all \(g \in G\) iff \(x \in Z(G)\). By the orbit stabilizer theorem, an orbit size that is greater than \(1\) must divide \(|G|\) and therefore must be a positive power of \(p\). If \(Z(G) = \{1\}\), then we have one orbit of size \(1\) with all other orbit sizes being defined as \(0 \mathrm{mod}p\). Thus \(|G| = 1 \mathrm{mod} p\), contradicting the assumption that \(G\) is a non-trivial \(p\)-group. QED
The groups are \(D_{12}, \mathbb{Z}_{12}, A_4, \mathbb{Z}_6 \times \mathbb{Z}_2\), and the dicyclic group of order \(12\). The dicyclic group is not isomorphic to any of these because it is not Abelian but cyclic. The group \(\mathbb{Z}_{12}\) is not isomorphic to any of these because it is cyclic and Abelian. The group \(A_4\) is not isomorphic to these because it is not Abelian, and has a normal Sylow-\(2\) subgroup. The group \(D_{12}\) is not isomorphic to any of these because it is not Abelian, and doesn’t have any normal Sylow-\(2\) subgroups. And finally \(\mathbb{Z_6} \times \mathbb{Z_2}\) is not isomorphic to any of these because it is Abelian and has a normal Sylow-\(2\) subgroup. QED
Let \(IJ \subseteq P\) where \(P\) is prime. Suppose WLOG that \(J \nsubseteq P\). Then there is some \(b \in J\) such that \(b \notin P\). But then \(aRb \subseteq P\) for all \(a \in I\), so because \(P\) is prime, \(a \in P\) for all \(a \in I\).
We will prove this using set containment.
Because \(I\) and \(J\) are ideals in \(R\), for all \(x_i \in I\) and all \(y_i \in J\), \(x_i y_i \in I\) and \(x_i y_i \in J\). Moreover, \(\sum_{i=1}^{n} x_i y_i \in I\) and \(\sum_{i=1}^{n} x_i y_i \in J\). Thus \[\sum_{i=1}^{n} x_i y_i \in IJ\] which implies \[\sum_{i=1}^{n} x_i y_i \in I \cap J\] Thus \(IJ \subset I \cap J\)
Now we just need to show that \(IJ \supset I \cap J\). To do this we will use the fact that \(I+J=R\). Let \(a \in I \cap J\). We know \(1 \in R = I + J\). So \(1 = i + j\) for some \(i \in I\) and \(j \in J\). Then since \(R\) is commutative, \[a = a 1 = a (i+j) = ai + aj = ia + aj\] However \(ia \in IJ\) since \(i \in I\) and \(a \in J\) and \(aj \in IJ\) for similar reasons as our previous argument. By the definition of \(IJ\), this shows us \(a \in IJ\). Thus \(I \cap J \subset IJ\).
Thus we’ve proven that \(IJ = I \cap J\) in this circumstance. QED
Consider \(2 \in D\). Note, the units of \(D\) are \(\{1,-1\}\). Assume \(2 = xy\) where \(x,y \in D\). Then if we define finding the norm as the map \(N\), then \(N(2) = 4\). Then if \(x = a+b\sqrt{-5}\) and \(y = c+d\sqrt{-5}\), then \[N(xy) = N(x)N(y) = (a^2+5b^2)(c^2+5d^2)\] Assume \(a^2+5b^2 = 2\). Then \(b=0\) and \(a = \sqrt{2}\) which isn’t an integer. Thus \(x\) must be a unit, which shows us that \(2\) is irreducible. QED
In order for \(2\) to be prime, if \(2\) divides \(ab\), then \(2\) divides \(a\) or \(2\) divides \(b\). Well, \(2\) divides \(6\), and \[6 = 3*2 = (1+\sqrt{-5})(1-\sqrt{-5})\] however \(2\) does not divide \((1 \pm \sqrt{-5})\) which implies that \(2\) is not prime.
\(D\) is not a UFD because \[6 = 3*2 = (1+\sqrt{-5})(1-\sqrt{-5})\] where \(2,3,(1 \pm \sqrt{-5})\) are irreducible and not associates. QED
First, we will prove that \(G^2 \unlhd G\). Let \(h \in G^2\) and \(g \in G\). So \[g^{-1}hg = g^{-1}g^{-1}ghg = (g^{-1})^2h^{-1}(hg)^2 \in G^2\] because \((g^{-1})^2, g^2, h \in G^2\). Thus, \(G^2 \unlhd G\).
Next we just need to prove that \(\frac{G}{G^2}\) is Abelian. Well, for all \(yG^2, xG^2 \in \frac{G}{G^2}\), we want to show that \(xG^2 yG^2 = yG^2 xG^2\). First, note that \(xG^2 yG^2 = xyG^2\) and \(yG^2 xG^2 = yxG^2\). And \(xyG^2 = yxG^2\) iff \(xy(yx^{-1}) \in G^2\). So if we can show that \(xyx^{-1}y^{-1} \in G^2\), then we know that \(\frac{G}{G^2}\) is Abelian. Well, \[xyx^{-1}y^{-1} = x^{-1}xxyyy^{-1}x^{-1}y^{-1} = x^{-1}(xy)^2y^{-1}x^{-1}y^{-1} = h(x^{-1}y^{-1})^2\] Where \(h \in G^2\) because \(G^2 \unlhd G\). Thus we have \(h(x^{-1}y^{-1})^2 \in G^2\), implying that \(xyx^{-1}y^{-1} \in G^2\), which shows us that \(\frac{G}{G^2}\) must be Abelian. QED
Well, \(NH \leq G\) iff \(NH = HN\). So for all \(n \in N\) and \(h \in H\), \(nh \in NH\) and \(hn \in HN\). Also because \(N \unlhd G\), for all \(h \in H\), \(hnh^{-1} \in N\). So \(hnh^{-1}*h \in NH \implies hn \in NH \implies NH \subseteq HN\). And \(h^{-1}hnh^{-1} \in HN \implies nh^{-1} \in HN \implies HN \subseteq NH\). Thus \(NH = HN\) implying that \(NH \leq G\).
Now assume that \(H \unlhd G\). We want to show that \(NH \unlhd G\). Well, then we want to show that for all \(nh \in NH\), \(gnhg^{-1} \in NH\). Well, \(gn = n'g\) and \(hg^{-1} = g^{-1}h'\) where \(n' \in N\) and \(h' \in H\). So \[gnhg^{-1} = n'gg^{-1}h' = n'h' \in NH\] Thus \(NH \unlhd G\). QED
First note that \(105 = 3*5*7\). Let’s denote the number of Sylow-\(p\) subgroups of \(G\) by \(n_p\). Then \(n_3 = 3k+1\) and \(n_3|35\); \(n_5 = 5k+1\) where \(n_5|21\); and finally \(n_7 = 7k+1\) where \(n_7|15\), where in all of these cases \(k\) is some integer. We know that if \(n_p=1\) for any \(p\), then \(G\) cannot be simple. Notice \(n_5 = \{1,21\}\) and \(n_7 = \{1,15\}\).
Assume \(G\) is simple. Then \(n_5 = 21\), \(n_7 = 15\). This implies that there are \(21*4=44\) non-identity elements in Sylow-\(5\) subgroups and \(15*6=90\) non-identity elements in Sylow-\(7\) subgroups, all of which intersect trivially. But \(44+90=134\) which gives us too many elements, thus either \(n_5=1\) or \(n_7 =1\) which implies that \(G\) must not be simple. QED
Let \(I \subseteq F[x]\). If \(I = (0)\) or \(I=(1)=F\), then the case is trivial and we’re done. Suppose \(p \in I\) is a nonzero polynomial of smallest possible degree. So, if \(f \in I\), \(\mathrm{deg}(f) \geq \mathrm{deg}(p)\). Also there exists a unique \(q,r \in F\) such that \(f=pq+r\) and \(0 \leq \mathrm{deg}(r) \leq \mathrm{deg}(p)\). Because \(I\) is an ideal containing \(p\) and \(f\), \(r = f-pq \in I\) which implies that \(r=0\) and \(f=pq\). Thus for all \(f \in I\), \(f\) is generated by \(p\). Thus \(I\) is principal. QED
Note, this trait is not true necessarily if \(F\) is not a field. This is because a field is a Euclidean Domain, meaning that we can use the division algorithm. However, not all commutative rings are Euclidean Domains.
Let \(a+bi\sqrt{3},c+di\sqrt{3} \in R\). Then \[\phi(a+bi\sqrt{3})\phi(c+di\sqrt{3}) = (a^2+3b^2)(c^2+3d^2) = a^2c^2+a^23d^2+3b^2c^2+9b^2d^2\] Also, \[\phi((a+bi\sqrt{3})(c+di\sqrt{3})) = (a^2+3b^2)(c^2+3d^2) = a^2c^2+a^23d^2+3b^2c^2+9b^2d^2\] Thus \(\phi(uv)=\phi(u)\phi(v)\). Now, because \(a,b \in \mathbb{Z}\), and \(a^2,b^2 \in \mathbb{Z^+}\), we can easily see that \(\phi(u) > 3\) whenever \(u \notin \{0,1,-1\}\). If \(u \in \{0,1,-1\}\), then \(\phi(u) \in \{0,1\}\) where \(\phi(0) = 0\) and \(\phi(\pm{1})=1\), both of which cases make \(\phi(u) < 3\).
An element \(u\) of \(R\) is a unit if \(uv=1\). This is only true if \(b=0\) and \(a = \pm 1\), otherwise we will have a complex element. Thus the only units of \(R\) are \(\pm 1\).
Notice that \(\phi\) is simply the norm of our ring \(R\).
Let \(uv = 2\). Then \(\phi(2) = 4 = \phi(u) \phi(v)\). WLOG, assume \(u\) is not a unit. So \(\phi(u)\phi(v) = 4\). Suppose \(u = a+bi \sqrt{3}\). This implies that \(a^2+3b^2 = 2\) or \(4\). But for \(a^2+3b^2 = 2\), \(b=0\) and \(a = \sqrt{2} \notin \mathbb{Z}\). Thus \(\phi(a) = 4\) which implies that \(\phi(v) = 1\) so \(v = \pm 1\). Thus \(v\) is a unit and \(2\) is irreducible.
Let \(uv = 1+i\sqrt{3}\). Then \(\phi(1+i\sqrt{3})=4= \phi(u) \phi(v)\). But as we showed above, this implies that either \(u\) or \(v\) is a unit, thus \(1+i\sqrt{3}\) is irreducible in \(R\).
Let \(uv = 1-i\sqrt{3}\). Then \(\phi(1-i\sqrt{3})=4= \phi(u) \phi(v)\). But as we showed above, this implies that either \(u\) or \(v\) is a unit, thus \(1-i\sqrt{3}\) is irreducible in \(R\).
We can deduce from here that \(R\) is not a UFD. This can be seen because \(2*2 = 4 = (1+\sqrt{-3})(1-\sqrt{-3})\). This shows us that \(R\) is a UFD because \(2, 1+\sqrt{-3}, 1-\sqrt{-3}\) are not associates in \(R\) because none of them differ by a unit (\(\pm 1\)). QED
A subring of \(R\) is a subset of \(R\) that is closed under multiplication. So we want to show that for any \(\frac{r}{s}, \frac{t}{q} \in D\) that \(\frac{rt}{sq} \in D\). Well, we know that \((r,s)=1=(t,q)\) (they are all rel. prime). If \((r,q) = 1 = (t,s)\), then we’re done as the numerator and denominators are both relatively prime, and an odd times an odd makes an odd. But what if \((r,q) \neq 1 \neq (t,s)\)? This is the case we need to consider. Well, WLOG, assume \(t\) and \(s\) are not relatively prime. Then \(t = gh\) and \(s = jh\) such that both \(s\) and \(t\) are divisible by \(h\). Then \[\frac{rt}{sq} = \frac{rgh}{jhq} = \frac{rg}{jq}\] This process can be continued until we find \(\frac{n}{d} = \frac{rt}{sq}\) such that \(n\) and \(d\) are relatively prime and both are contained in \(D\). Also, \(d\) will be odd because if \(a\) and \(b\) are odd such that \(a=cb\) which implies \(c\) is odd, and because you can’t factor an even from an odd. Thus we’ve proven that \(D\) is a subring of \(\mathbb{Q}\). QED
The units of \(D\) are many elements \(\frac{r}{s} \in D\) such that \(r\) and \(s\) are odd and \(r\) and \(s\) are relatively prime. This is because \(\frac{r}{s}*\frac{s}{r} = 1\) which implies that \(\frac{r}{s}\) and \(\frac{s}{r}\) are units in \(D\). It is important to note that both \(r\) and \(s \in \mathbb{Z}\) must be odd for this to be the case. \(1\) and \(-1\) are also units.
A “proof” of this part of the problem can be found here. However, I believe I have a counter example to this in the fraction \(\frac{7}{15} \in D\) as this fraction cannot be generated by \(2^n\) for any \(n \in \mathbb{Z}\).
Consider the polynomial \(f(x) = x^4 - 6x^2 + 2\). In order to be the minimal polynomial of \(\alpha\) over \(\mathbb{Q}\), it must be the polynomial with smallest degree with \(\alpha\) as a root, and be irreducible. Well, it’s fairly obvious that \(f(\alpha)=0\) by simply testing \(x=\alpha\). It is also fairly easy to see that this polynomial has the lowest possible degree that could have \(\alpha\) as a root, as \(\alpha\) is made up of two square roots that can only be cancelled out by a polynomial of degree \(4\). And finally, by checking Eisenstein’s Criteria, we can see that \(f(x)\) is indeed irreducible in \(\mathbb{Q}\). Thus \(f(x)\) is our minimal polynomial for \(\alpha\) over \(\mathbb{Q}\). QED
Well, because \(G\) is the Galois group, we know that \[|G| = [\mathbb{Q}(\sqrt[3]{5},i\sqrt{3}): \mathbb{Q}] = [\mathbb{Q}(\sqrt[3]{5},i\sqrt{3}): \mathbb{Q}(i\sqrt{3})][\mathbb{Q}(i\sqrt{3}):\mathbb{Q}]\] This is equal to \(3*2 = 6\) as the minimal polynomials for each extension listed above are degree \(3\) and \(2\) respectively. Thus the order of \(G\) is \(6\). QED
Here we will use the Fundamental Theorem of Galois Theory. So we know that the group \(G\) fixes the elements of \(\mathbb{Q}\), while the identity group fixes the elements of \(\mathbb{Q}(\sqrt[3]{5},i\sqrt{3})\). So in between these, there is a subgroup \(H\) of \(G\) that fixes a subfield \(K\) of \(\mathbb{Q}(\sqrt[3]{5},i\sqrt{3})\). In our case, since \(G\) is isomorphic to \(\mathbb{Z}_2 \times \mathbb{Z}_3\), we will consider the subgroup \(H = \mathbb{Z}_2\). This subgroup then fixes the elements in the extension \(\mathbb{Q}(\sqrt[3]{5}) = K\), giving us that \([\mathbb{Q}(\sqrt[3]{5},i\sqrt{3}): K] = 2\). QED
We will skip part (b) as it is a repeat of an earlier problem because if \(R\) is Artinian, then it is an integral domain.
So for all descending chains of ideals in \(R\) \[I_1 \supseteq I_2 \supseteq ... \supseteq I_n \supseteq ...\] the descending chain stabilizes such that for some \(I_i\), \(I_i = I_{i+1}\). Well, note that \(R/I = \{rI| \forall r \in R\}\). We will now prove a lemma stating that \(J/I\) is an ideal of \(R/I\) if \(I\) and \(J\) are ideals of \(R\).
For all \(r \in R\), \(rJ = Jr = J\). So then for all \(rI \in R/I\), \(rI(J/I) = r(J/I) = J/I\). So \(J/I\) is an ideal of \(R\).
Now consider a descending chain of ideals \[J_1 \supseteq J_2 \supseteq ... \supseteq J_n \supseteq ...\] where all \(J_i\)’s are ideals in \(R\). Then \[J_1/I \supseteq J_2/I \supseteq ... \supseteq J_n/I \supseteq ...\] is a descending chain of ideals in \(R/I\). Let \(J_m = J_{m+1}\) because \(R\) is Artinian. Then \(J_m/I = J_{m+1}/I\) because \(J_m = J_{m+1}\). So \(R/I\) is Artinian. QED
First off, in order to be a group all elements must be closed under the operation, the operation must be associative, and ther must be an identity and an inverse. Also, for the sake of simplicity, let \[A = \bigl( \begin{smallmatrix}1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{smallmatrix}\bigr) \in H\] \[B = \bigl( \begin{smallmatrix}1 & x & y\\ 0 & 1 & z\\ 0 & 0 & 1 \end{smallmatrix}\bigr) \in H\] \[C = \bigl( \begin{smallmatrix}1 & 1-a & c-b\\ 0 & 1 & -c\\ 0 & 0 & 1 \end{smallmatrix}\bigr) \in H\]
Well, the identity is obviously the identity matrix \(I = \bigl( \begin{smallmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{smallmatrix}\bigr)\). Also, \[A \times B = \bigl( \begin{smallmatrix}1 & x+a & y+az+b\\ 0 & 1 & z+c\\ 0 & 0 & 1 \end{smallmatrix}\bigr) \in H\] So \(H\) is closed under the operation. We can also check that \(A \times C = C \times A = I\), thus we have an inverse. And finally, we know from experience that matrix multiplication is associative, thus \(H\) is indeed a group. So the last thing here to show is just that \(H\) is not an Abelian group. Well, using \(A \times B\) that we found above, we can check \[B \times A = \bigl( \begin{smallmatrix}1 & x+a & cx+y+b\\ 0 & 1 & z+c\\ 0 & 0 & 1 \end{smallmatrix}\bigr) \neq A \times B\] Thus \(H\) is a noncommutative group. QED
For all \(h \in H\) and \(k \in K\) we have \[hkh^{-1}k^{-1} = (hkh^{-1})k^{-1} = h(kh^{-1}k^{-1}) \in H \cap K\] so \(hkh^{-1}k^{-1} = e\) if \(e =1\) is the identity. This implies that \(hk = kh\).
Let \(f : H \times K \rightarrow G\) such that \(f(h,k) = hk\). For any \(h, h' \in H\) and \(k, k' \in K\), \[f(h,k)f(h',k') = (hk)(h'k') = hh'kk'=(hh')(kk') = f(hh',kk') = f((h,k)(h',k'))\] Thus \(f\) is a homomorphism. Since \(HK = G\), any element \(g \in G\) is simply \(g = hk\) for some \(h \in H\) and \(k \in K\), and \(f(h,k) = hk = g\) so \(f\) is obviously onto. Finally if \((h,k) \in \mathrm{ker}(f)\), then \(hk = e\). So \(h = k^{-1} \in H \cap K = \{1\}\) so \((h,k) = (e,e)\) which implies that \(f\) is one-to-one. Thus \(f\) must be an isomorphism. QED
Let \(R\) be an integral domain. Let \(p \in R\) be a prime element in \(R\). Assume \(p=ab\). This implies that \(p|a\) or \(p|b\). WLOG, assume that \(p|a\). Then \(pq = a\) for some \(q \in R\). Thus \[1*a = a = pq = abq \implies 1 = bq\] Thus \(b\) is a unit and \(p\) must be irreducible. QED
In order for \(\mathbb{Z}[i]\) to be an integral domain, it must contain no zero divisors. Let \(a,b \in \mathbb{Z}[i]\) such that \(ab = 0\). Well, if \(a = x+yi\) and \(b = w+zi\) then \[ab = (x+yi)(w+zi) = xw+yi+xzi+yizi\] \[= (xw-yz)+i(yw+xz) = 0\] This implies that \(xw = yz\) and \(yw = -xz\), but this is only the case if either \(x,y = 0\) or \(w,z = 0\). Thus \(\mathbb{Z}[i]\) is an integral domain.
The units of \(\mathbb{Z}[i]\) are \(\{1,-1,i,-i\}\) because \((a+bi)(c+di)=xy = 1\) is only true when these elements are used.
No, because \((2+3i)\) is an associate of \(3-2i\) because \((2+3i)*-i = 3-2i\). Similarly, \(2-3i\) is an associate of \(3+2i\) because \((2-3i)*i = 3+2i\). Thus this does not make a contradiction.
\(5=(1+2i)(1-2i)\)
Let \(R\) be a finite integral domain with \(0 \neq a \in R\). Consider \(a,a^2,a^3,...\), since there are only finitely many elements, we must have \(a^m = a^n\) where \(m < n\). Then \[0 = a^m - a^n = a^m(1-a^{n-m})\] Since there are no zero divisors, we must have \(a^m \neq 0\) and hence \(1-a^{n-m} = 0\) which implies that \(1 = a(a^{n-m-1})\). Thus \(a \in R\) has a multiplicative inverse, thus \(R\) is a field. QED
It is easy to see by Eisenstein’s Criteria that \(f(x)\) is irreducible over \(\mathbb{Q}\).
\[f(x) = x^4 - 2 = (x^2 + \sqrt{2})(x^2 - \sqrt{2})\] \[= (x+i \sqrt[4]{2})(x-i \sqrt[4]{2})(x- \sqrt[4]{2})(x+ \sqrt[4]{2})\] Thus we can easily see that the splitting field of \(f(x)\) over \(\mathbb{Q}\) will be \(F = \mathbb{Q}(\sqrt[4]{2},i)\) as \(f(x)\) can be factored into linear factors in \(F\).
Here we will use what I like to call the double extension lemma. Using it, we find that \[[\mathbb{Q}(\sqrt[4]{2},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{2},i):\mathbb{Q}(i)][\mathbb{Q}(i):\mathbb{Q}]\] Using this we can find that \[[\mathbb{Q}(\sqrt[4]{2},i):\mathbb{Q}] = \mathrm{deg}(x^4 - 2) \times \mathrm{deg}(x^2+1) = 4 \times 2 = 8\] Thus the degree of the extension is \(8\). QED
Well, let \(a,b,c \in G\). Then \[(a \times b) \times c = (a*u*b)*u*c = a*u*(b*u*c) = a \times (b \times c)\] because \(*\) is associative. Thus \(\times\) is associative as well.
Well, \[a \times u^{-1} = a*u*u^{-1} = a = u^{-1}*u*a = u^{-1} \times a\] This implies that \(G\) has an identity under \(\times\) and it’s \(u^{-1}\).
Let \(a \in G\). Then \(u^{-1}*a*u^{-1} \in G\) as well. Thus \[a \times (u^{-1}*a*u^{-1}) = a*u*u^{-1}*a*u^{-1} = u^{-1} \] \[u^{-1}= u^{-1}*a*u^{-1}*u*a = (u^{-1}*a*u^{-1}) \times a\] So for all \(a \in G\) there exists an inverse \(u^{-1}*a*u^{-1} \in G\) under \(\times\).
Let \(\Gamma : (G, \times) \rightarrow (G,*)\). Then for all \(a \in G\), \(\Gamma(a) = a\), but \(\Gamma(a \times b) = a*u*b\). Observe though that \(\Gamma(a) \Gamma(b) = a*b \neq a*u*b\). Thus these groups couldn’t be isomorphic because \(\Gamma\) isn’t even a homomorphism. QED
First, let \(K \unlhd G\) and \(f(g) = h\) for all \(g \in G\) and some \(h \in H\). Let \(k \in K\). This implies that \(gkg^{-1} = k' \in K\). So \(f(gkg^{-1}) = f(k') \in f(K)\). Also note then that \(f(gkg^{-1}) = hf(k)h^{-1}\) because \(f\) is a homomorphism. This implies that \(f(K) \unlhd H\) thus we have proven the first part.
Let \(\Gamma :H \rightarrow G/K\). If we can show that \(\Gamma\) is a homomorphism with a kernel of \(f(K)\), then by the first isomorphism theorem, we’re done. First off, define \(f(g) = h\) and \(f(g') = h'\) for some \(g,g' \in G\) and \(h,h' \in H\). If \(h \in H\) and \(gK \in G/K\), we can define \(\Gamma\) by \(\Gamma(h) = gK\). Then, \[\Gamma(hh') = gg'K = gKg'K = \Gamma(h) \Gamma(h')\] Thus \(\Gamma\) is a homomorphism. Finally, \(\mathrm{ker}(\Gamma) = \{a \in H| \Gamma(a) = K\}\). Well, \(\Gamma(a) = K\) iff \(f(a) \in f(K)\). This implies that \(\mathrm{ker}(\Gamma) = f(K)\). Thus by the first isomorphism theorem, \[\frac{H}{f(K)} \cong \frac{G}{K}\] QED
Let \(f\) and \(g\) be in \(R\). Then \((f+g)(x) = f(x)+g(x) = g(x)+f(x) = (g+f)(x)\). So \((R, +)\) is Abelian. Also for all \(f(x)\) there exists a \(g(x)=-f(x)\) such that \(f(x)+(-f(x)) = 0_R\). Hence because of this and because \(g(x)-f(x)\) is a continuous function from \(0\) to \(1\), so \((R,+)\) is an Abelian group.
Let \(f,g,h \in R\). Consider \[((fg)h)(x) = (fg)h(x) = (fg)(x)h(x) = f(x)g(x)h(x) = f(x)((gh)(x)) = (f(gh))(x)\] Thus multiplication is associative.
Let \(f,g,h \in R\). Consider \[(f(g+h))(x) = f((h+h)(x))= f(g(x) + h(x)) = f(x)g(x) + f(x)h(x) = (fg)(x)+(fh)(x)\] Thus \(R\) follows the distributive property. Thus \(R\) is a ring. It is also easy to see that \((fg)(x) = (gf)(x)\) because all of the functions are continuous on \([0,1]\). Thus \(R\) is a commutative ring with identity \(f(x) = 1\).
The units of \(R\) are just the inverse functions of \(f(x)\). ie. \((fg)(x) = 1\) iff \(g(x) = f^{-1}(x)\).
Let \(f^2 = f\). Then \(f^2(x) = f(x)f(x)\). Then by cancellation, \(f(x) = 1_R\) or \(f(x) = 0\). QED
Assume \(G\) is simple and denote the number of Sylow-\(p\) subgroups as \(n_p\). Then \(n_2 = 2k+1\) where \(k \in \mathbb{Z^+}+\{0\}\) and \(n_2 | 5^4 = 625\). Also, \(n_5 = 5k+1\) where \(k \in \mathbb{Z^+}+\{0\}\) and \(n_5 | 2^4 = 16\). So \(n_5 \in \{1,16\}\). Since \(G\) is not simple by assumption, then \(n_5 = 16\).
Thus because \(G\) is smple and \(n_5 = 16\), there exists \(N \leq G\) such that \([G:N] = 16\). This implies that \(G\) can be embedded in \(S_{16}\) if we let \(G\) act on \(N\) by conjugation. But then by Lagrange’s theorem \(10,000\) should divide \(16!\) which it does not. Thus we have a contradiction, so \(G\) cannot be simple. QED
Consider the ideal generated by \(2\), \((2)\). Then for all \(a \in (2)\), \(ax\) should be in \((2)\) but this is not true. Thus \(\mathbb{Z}[x]\) cannot be a PID.
Now we must decide if it is a UFD. Because \(\mathbb{Z}\) is a UFD, we know a theorem that states that \(\mathbb{Z}[x]\) must be a UFD as well. QED
Let \(r \in R/M\) where there exists an \(x \in R\) such that \(1-rx \in M\). This is true iff \[(1-rx)M = M \iff M - rxM = M \iff -rxM = M\] \[ \iff rM(-xM) = M\] But this is true iff \(rM\) has the inverse \(-xM\) in \(R/M\) which is true iff \(R/M\) is a field. Finally, \(R/M\) is a field iff \(M\) is maximal. Thus using this string of if and only if statements, we know this is true. QED
First we will state the theorem itself:
Let \(\phi : G \rightarrow H\) be a homomorphism of groups, then \(\mathrm{ker} \phi \unlhd G\) and \(\frac{G}{\mathrm{ker}\phi} \cong \phi(G)\).
To shorten things, let \(K = \mathrm{ker}\phi\). Define the homomorphism \(\psi : G/K \rightarrow \phi(G)\) by \(\psi(Kg) = \phi(g)\). We claim that \(\psi\) is an isomorphism. So we need to check that it’s a well-defined homomorphism, and a bijection. First, to check that it’s well-defined, we will check to see that if \(a \in Kg\), then \(\phi(a) = \phi(g)\). If we let \(a = kg\) then for some \(k \in K\), then \[\phi(a) = \phi(kg) = \phi(k)\phi(g) = \phi(g)\] Now we will show that \(\psi\) is a homomorphism. Consider \(Kg, Kh \in G/K\). Then for any \(a,b \in K\) we have that \[\psi(Kg*Kh) = \psi(Kgh) =\phi(gh) = \phi(g) \phi(h) = \psi(Kg) \psi(Kh)\] This proves that \(\psi\) is a well-defined homomorphism. To prove that \(\psi\) is one-to-one, we must prove that \(\mathrm{ker}(\psi) = K\). Well, observe that \[\psi(Kg) = 1 \iff \phi(g) = 1 \iff g \in \mathrm{ker}(\phi) \iff g \in K \iff Kg = K\] So \(\mathrm{ker}(\psi) = K\). Finally we just need that \(\psi\) is onto, which is somewhat obvious from the fact that \(\phi(g) \in \phi(G)\) so that \(\phi(g)\) has preimage \(Kg\). Thus we have proven the First Isomorphism Theorem for Groups. QED
First we must state it. Let \(G\) be an arbitrary group. For any two subgroups \(H\) and \(K\) of \(G\), define \(HK = \{ hk | h \in H, k \in K\}\). Assume \(H \unlhd G\) and \(K \leq G\). Then: (a) \(HK \leq G\). (b) \(H \unlhd HK\) and \((H \cap K) \unlhd K\). (c) \(\frac{HK}{H} \cong \frac{K}{H \cap K}\).
Recall that \(HK\) can only be a subgroup if \(HK = KH\). Thus if \(H \unlhd G\), and we let \(h \in H\) and \(k \in K\), by assumption \(khk^{-1} \in H\). Thus \[kh = (khk^{-1})k \in HK\] This proves that \(KH \subseteq HK\). Similarly, \(hk = k(k^{-1}hk) \in KH\) which proves the reverse containment. Thus \(HK \leq G\). QED
Since \(K\) normalizes \(H\) and \(H\) normalizes itself trivially, this implies that \(HK\) normalizes \(H\). Thus \(H \unlhd HK\). Since \(H \unlhd HK\), the quotient group \(\frac{HK}{H}\) is well defined. Define the map \(\Gamma : K \rightarrow \frac{HK}{H}\) by \(\Gamma (k) = kH\). Since the group operation in \(\frac{HK}{H}\) is well defined, it’s easy to see that \(\Gamma\) is a homomorphism: \[\Gamma(k_1 k_2) = (k_1 k_2)H = k_1 H *k_2 H = \Gamma(k_1) \Gamma(k_2)\] Alternatively, the map \(\Gamma\) is just the restriction to the subgroup \(K\) of the natural progression homomorphism \(\pi : HK \rightarrow \frac{HK}{H}\), so it’s also a homomorphism. It is clear then from the definition of \(HK\) that \(\Gamma\) is onto. The identity of \(\frac{HK}{H}\) is simply \(H\), so the kernel of \(\Gamma\) consists of the elements \(k \in K\) with \(kH = H\), or where \(k \in H\). Thus \(\mathrm{ker}\Gamma = H \cap K\). So by the first isomorphism theorem, \(H \cap K \unlhd K\). QED
Following the proof of part (b), by the First Isomorphism Theorem, \(\frac{HK}{H} \cong \frac{K}{H \cap K}\) QED
First (once again) we must state it first. Let \(K,N \unlhd G\) with \(N \leq K\). Then \(K/N \unlhd G/N\) and \((G/N)/(K/N) \cong G/K\).
The proof that \(K/N \unlhd G/N\) is easy and follows from the 4th Isomorphism Theorem for Groups, so we will leave this aside. Now, define \[\phi: G/N \rightarrow G/K\] by \(\phi(Ng) = Kg\). By the First Isomorphism Theroem, \((G/N)/\mathrm{ker}(\phi) \cong \phi(G)\). It is clear that \(\phi\) is onto, so \(\phi(G) = G/K\). Now \[\mathrm{ker}(\phi) = \{Ng \in G/N | Kg = K\} = \{Ng \in G/N | g \in K\} = K/N\] Thus by the First Isomorphism Theorem, \((G/N)/(K/N) \cong G/K\). QED
If \(\phi: R \rightarrow S\) is a homomorphism of rings, then the kernel of \(\phi\) is an ideal of \(R\), the image of \(\phi\) is a subring of \(S\), and \(R/\mathrm{ker}(\phi) \cong \phi(R)\)
Proof:
If \(I\) is the kernel of \(\phi\), then the cosets of \(I\) are precisely the fibers of \(\phi\). In particular, the cosets \(r+I,s+I,\) and \(rs+I\) are the fibers of \(\phi\) over \(\phi(r), \phi(s),\) and \(\phi(rs)\) respectively. Since \(\phi\) is a ring homomorphism, \(\phi(r) \phi(s) = \phi(rs)\), hence \((r+I)(s+I) = rs+I\). Multiplication of cosets is well defined and so \(I\) is an ideal and \(R/I\) is a ring. The correspondence \(R+I \rightarrow \phi(r)\) is a bijection between rings \(R/I\) and \(\phi(R)\) which respects addition and multiplication, hence is a ring isomorphism. Proving that the image of \(\phi\) is a subring of \(S\) is trivial. QED
Let \(A\) be a subgring and let \(B\) be an ideal of \(R\). Then \(A+B = \{a+b| a \in A, b \in B\}\) is a subring of \(R\), \(A \cap B\) is an ideal of \(A\) and \(\frac{(A+B)}{B} \cong \frac{A}{(A \cap B)}\)
Proof:
\(A\) is a subring and \(B\) is an ideal, so \(1+0 \in A+B\). Let \(a_1 + b_1\) and \(a_2 + b_2\) be elements of \(A+B\). Then \[(a_1+b_1) - (a_2+b_2) = (a_1-a_2)+(b_1-b_2)\] and \[(a_1+b_1)(a_2+b_2) = a_1a_2+a_1b_2+b_1a_2+b_1b_2\] Hence \(A+B\) is a subring of \(R\).
The intersection \(A \cap B\) is nonempty since \(0\) is contained in \(A\) and \(B\). Let \(b_1,b_2 \in A \cap B\) and let \(a \in A\). Then \(b_1+b_2 \in A \cap B\) sinc \(A\) and \(B\) are both closed under addition. Furthermore \(ab_1\) and \(ab_2\) are in \(A \cap B\) since \(B\) is closed under multiplication from \(R \supset A\) and \(A\) is closed under multiplication. Thus \(A \cap B\) is an ideal of \(A\).
Consider the map \(\phi: A \rightarrow (A+B)/B\) which sends an element \(a\) to \(a+B\). THis is a ring homomorphism by definition of addition and multiplication in quotient rings (easy to prove). We claim that it is onto with kernel \(A \cap B\) whcih would complete the proof by the first isomorphism theorem for rings. Consider the elements \(a \in A\) and \(b \in B\). Then \(a + b + B = a +B\) since \(b \in B\), so \(a+b+B \in \mathrm{im} \phi\) and hence \(\phi\) is onto. Let \(a \in A\) be an element of \(\mathrm{ker}(\phi)\). THen \(a+B =B\) which holds iff \(a \in B\) or equivalently if \(a \in A \cap B\). Thus \(\mathrm{ker}(\phi) = A \cap B\) and by the first isomorphism theorem we have our result. QED
Let \(I\) and \(J\) be ideals of \(R\) with \(J \subseteq I\). Then \(\frac{I}{J}\) is an ideal of \(\frac{R}{J}\) and \(\frac{(R/J)}{(I/J)} \cong \frac{R}{I}\).
Proof:
Since \(I\) and \(J\) are ideals, they are nonempty and so \(I/J = \{a+J|a \in I\}\) is also nonempty. Let \(a_1,a_2 \in I\) and let \(r \in R\). By definition of addition and multiplication of cosets, we have \[(a_1+J)+(a_2+J)=(a_1+a_2)+J\] \[(r+J)(a_1+J)=ra_1+J\] and \[(a_1+J)(r+J)=a_1r+J\] Since \(I\) is an ideal, \(a_1+a_2,ra_1,\) and \(a_1r\) are contained in\(I\) so \(I/J\) is an ideal of \(R/J\). Consider the map \(\phi:R/J \rightarrow R/I\) that sends \(r+J\) to \(r+I\). We claim that this is a well-defined onto homomorphism with kernel equal to \(I/J\). This is fairly obvious to see as every input has an image in \(R/I\). As for the kernel, take any \(a+J \in I/J\) where \(a \in I\). Well, \(a+J \in R/J\) as well, and maps to \(a+I = I\) because \(a \in I\). Thus \(I/J\) is the kernel. Thus we have proven the desired result by the first isomorphism theorem for rings. QED
A group is a set of elements \(G\) with a binary operation such that it follows closure under the operation, follows associativity, has an identity, and all elements have an inverse.
Hard to define in words, but remember what the dihedral group is (\(D_2n\), where \(n\) is the number of sides of the geometric representation).
A map \(\Gamma\) is a homomorphism if it maps a group \(G\) to a group \(H\) such that if \(g \in G\) and \(h \in H\) \[\Gamma(gh)=\Gamma(g)\Gamma(h)\]
An isomorphism is a homomorphism that is a bijection
The homomorphism given from \(G\) to \(S_A\) where \(S_A\) is the permutation of the set \(A\) is called the permutation representation associated to the given action.
(One-Step-Subgroup Test) A subset \(H\) of a group \(G\) is a subgroup iff \(H\) is nonempty and for all \(x,y \in H, xy^{-1} \in H\).
Define \(C_G(A) = \{g \in G | gag^{-1} =a, \forall a \in A\}\). This subset of \(G\) is called the centralizer of \(A\) in \(G\).
Define \(Z(G) = \{g \in G | gx = xg, \forall x \in G\}\). This subset of \(G\) is the center
Define \(gAg^{-1} = \{gag^{-1} | a \in A\}\). Define the normalizer of \(A\) in \(G\) to be the set \(N_G(A) = \{g \in G | gAg^{-1}=A\}\).
The kernel of an action of \(G\) on \(S\) is defined as \(\{g \in G | gs = s, \forall s \in S\}\).
A group \(G\) is cyclic if \(G\) can be generated by a single element, in other words, there exists some \(x \in G\) such that \(G = \{x^n | n \in \mathbb{Z}\}\).
If \(\phi\) is a homomorphism of \(G\) to \(H\), the kernel of \(\phi\) is the set \(\{g \in G | \phi(g) = e\}\) where \(e\) is the identity of \(H\).
Let \(N\) be any subgroup of the group \(G\). The set of left cosets of \(N\) in \(G\) form a partition of \(G\). Furthermore, for all \(u,v \in G\), \(uN=vN\) iff \(v^{-1}u \in N\).
The element \(gng^{-1}\) is called the conjugate of \(n \in N\) by \(g\). The set \(gNg^{-1} = \{gng^{-1}| n \in N\}\) is called the conjugate of \(N\) by \(g\). THe element \(g\) is said to normalize \(N\) if \(gNg^{-1} = N\). A subgroup \(N\) of \(G\) is called normal if every element of \(G\) normalizes \(N\).
(Lagrange’s Theorem) If \(G\) is a finite group and \(H\) is a subgroup of \(G\), then the order of \(H\) divides the order of \(G\), and the number of left cosets of \(H\) in \(G\) equals \(\frac{|G|}{|H|}\).
If \(G\) is a group and \(H \leq G\), the number of left cosets of \(H\) in \(G\) is called the index of \(H\) in \(G\), denoted \([G:H]\).
If \(G\) is a group of prime order \(p\), then \(G\) is cyclic, hence \(G \cong Z_p\).
If \(G\) is a finite group and \(x \in G\), then the order of \(x\) divides the order of \(G\). In particular \(x^{|G|} = 1\) for all \(x \in G\).
(Cauchy’s Theorem) If \(G\) is a finite group and \(p\) is a prime dividing \(|G|\), then \(G\) has an element of order \(p\).
If \(H\) and \(K\) are finite subgroupd of a group then \(|HK| = \frac{|H||K|}{|H \cap K|}\)
If \(H\) and \(K\) are subgroups of a group, \(HK\) is a subgroup iff \(HK=KH\).
IF \(G\) is a finite abelian group and \(p\) is a prime dividing \(|G|\), then \(G\) contains an element of order \(p\).
A group \(G\) is called simple if \(|G| > 1\) and the only normal subgroups of \(G\) are \(1\) and \(G\).
In a group \(G\) a sequence of subgroups \(1 = N_0 \leq N_1 \leq N_2 \leq ... \leq N_{k-1} \leq N_k = G\) is called a composition series if \(N_i \unlhd N_{i+1}\) and \(N_{i+1}/N_i\) is a simple group.
If \(G\) is a simple group of odd order, then \(G \cong Z_p\) for some prime \(p\).
A group \(G\) is solvable if there is a chain of subgroups \(1 = G_0 \unlhd G_1 \unlhd G_2 \unlhd ... \unlhd G_s = G\) such that \(G_{i+1}/G_i\) is abelian for all \(i\).
The permutation \(\sigma\) is odd iff the number of cycles of even length in its cycle decomposition is odd.
A group action is faithful if its kernel is the identity.
(Cayley’s Theorem) Every group is isomorphic to a subgroup of some symmetric group. If \(G\) is a group of order \(n\), then \(g\) is isomorphic to a subgroup of \(S_n\).
If \(G\) is a finite group of order \(N\) and \(p\) is the smallest prime dividing \(|G|\), then any subgroup of index \(p\) is normal.
Two elements \(a\) and \(b\) of \(G\) are said to be conjugate in \(G\) is there is some \(g \in G\) such that \(b = gag^{-1}\). The orbits of \(G\) acting on itself by conjugation are called the conjugacy classes of \(G\).
(The Class Equation) \(|G| = |Z(G)| + \sum_{i=1}^{r}|G:C_G(g_i)|\).
Let \(G\) be a group. An isomorphism from \(G\) onto itself is called an automorphism of \(G\).
Let \(G\) be a group and let \(g \in G\). Conjugation by \(g\) is called an inner automorphism of \(G\) and the subgroup of \(Aut(G)\) consisting of all inner automorphisms is denoted by \(Inn(G)\).
A subgroup \(H\) of a group \(G\) is called characteristic in \(G\) if every automorphism of \(G\) maps \(H\) to itself.
(Sylow’s Theorem) Let \(G\) be a group of order \(p^\alpha m\) where \(p\) is a prime not dividing \(m\). Then: (i) Sylow \(p\)-subgroups of \(G\) exist, (ii) If \(P\) is a Sylow \(p\)-subgroup of \(G\) and \(Q\) is any \(p\)-subgroup of \(G\), then there exists \(g \in G\) such that \(Q \leq gPg^{-1}\), in particular two Sylow \(p\)-subgroups of \(G\) are conjugate in \(G\), and (iii) The number of Sylow \(p\)-subgroups of \(G\) is of the form \(1+kp\), or \(n_p \equiv 1(mod (p))\).
If \(P\) is a Sylow \(p\)-subgroup of \(G\), then the following are equivalent: \(P\) is characteristic in \(G\); \(P\) is the unique Sylow \(p\)-subgroup of \(G\) (\(n_p =1\)); \(P\) is normal in \(G\).
A maximal subgroup of a group \(G\) is a proper subgroup \(M\) of \(G\) such that there are no subgroups \(H\) of \(G\) with \(M \leq H \leq G\).
– \((R, +)\) is an abelian group
– \(\times\) is associative
– The distributive law holds for \(R\)
A ring \(R\) is commutative if multiplication is commutative.
The ring \(R\) is said to have an identity if there is an element \(1 \in R\) such that \(1a = a1 = a\) for all \(a \in R\).
A ring \(R\) with \(1\) is called a division ring if every nonzero element \(a \in R\) has a multiplicative inverse. A commutative division ring is called a field.
Let \(R\) be a ring. A nonzero element of \(a \in R\) is called a zero divisor if there is a nonzero element \(b \in R\) such that either \(ab =0\) or \(ba=0\).
Assume \(R\) has an identity \(1 \neq 0\). An element \(u\) of \(R\) is called a unit in \(R\) if there is some \(v\) in \(R\) such that \(uv = vu = 1\).
A commutative ring with identity \(1 \neq 0\) is called an integral domain if it has no zero divisors.
A subring of the ring \(R\) is a subgroup of \(R\) that is closed under multiplication.
Let \(R\) and \(S\) be rings. A ring homomorphism is a map \(\phi : R \rightarrow S\) satisfying:
\(\phi(a+b) = \phi(a) + \phi(b)\) for all \(a,b \in R\)
\(\phi(ab) = \phi(a) \phi(b)\) for all \(a,b \in R\).
The kernel of the ring homomorphism \(\phi\) is the set of elements of \(R\) that map to \(0\) in \(S\).
A bijective ring homomorphism is a ring isomorphism.
Let \(R\) and \(S\) be rings and let \(\phi: R \rightarrow S\) be a homomorphism. Then the image of \(\phi\) is a subgring of \(S\). Also, the kernel of \(\phi\) is a subring of \(R\).
Let \(R\) be a ring and let \(I\) be a subset of \(R\), and let $ r R$. \(rI = \{ra | a \in I\}\) and \(Ir = \{ar | a \in I\}\). A subset \(I\) of \(R\) is a left ideal of \(R\) if \(I\) is a subring and \(I\) is closed under left multiplication by elements from \(R\), ie. \(rI \subseteq I\) for all \(r \in R\). A right ideal is identical to this, just replace left with right. Then an ideal is a subset \(I\) of \(R\) that is both a left and a right ideal.
Let \(a \in R\). \((a)\) is the ideal generated by \(a\).
Let \(I\) be an ideal of \(R\). \(I = R\) iff \(I\) contains a unit. Assume \(R\) is commutative. Then \(R\) is a field iff its only ideals are \(0\) and \(R\).
An ideal \(M\) in a ring \(R\) is a maximal ideal if \(M \neq R\) and the only ideals containing \(M\) are \(M\) and \(R\).
Assume \(R\) is commutative. The ideal \(M\) is a maximal ideal iff the quotient ring \(R/M\) is a field.
Assume \(R\) is commutative. An ideal \(P\) is called a prime ideal if \(P \neq R\) and whenever the product \(ab\) of two elements \(a,b \in R\) is an element of \(P\), then at least one of \(a\) and \(B\) is an element of \(P\).
Any function \(N: R \rightarrow \mathbb{Z}^{+} \cup \{0\}\) with \(N(0) =0\) is called a norm on the integral domain \(R\). If \(N(a) >0\) for \(a \neq 0\) define \(N\) to be a positive norm.
The integral domain \(R\) is said to be a Euclidean Domain if there is a norm \(N\) on \(R\) such that for any two elements \(a\) and \(b\) of \(R\) with \(b \neq 0\) there exists elements \(q\) and \(r\) in \(R\) such that \[a = qb+r | r = 0 \lor N(r) < N(b)\]
A Principle Ideal Domain or PID is an integral domain in which every ideal is principal.
Every nonzero prime ideal in a PID is a maximal ideal.
If \(R\) is any commutative ring such that the polynomial ring \(R[x]\) is a PID (or a Euclidean Domain), then \(R\) is necessarily a field.
Let \(R\) be an integral domain. Suppose \(r \in R\) is nonzero and is not a unit. Then \(r\) is called irreducible in \(R\) if whenever \(r=ab\) with \(a,b \in R\), at least one of \(a\) or \(b\) must be a unit in \(R\). Otherwise \(r\) is *reducible.
Let \(R\) be an integral domain. The nonzero element \(p \in R\) is called prime in \(R\) if the ideal \((p)\) generated by \(p\) is a prime ideal. In other words, a nonzero element \(p\) isa prime if it is not a unit and whenever \(p | ab\) for any \(a,b \in R\), then either \(p|a\) or \(p|b\).
Let \(R\) be an integral domain. Two elements \(a\) and \(b\) of \(R\) differing by a unit are said to be associate in \(R\).
In an integral domain a prime element is always irreducible.
In a PID a nonzero element is a prime iff it is irreducible.
A Unique Factorization Domain or UFD is an integral domain \(R\) in which every nonzero element \(r \in R\) which is not a unit has the following two proporties: (i) \(r\) can be written as a finite product of irreducibles \(p_i\) of \(R\) (not necessarily distinct), so \(r = p_1p_2...p_n\); and (ii) the decomposition in (i) is unique up to associates, namely, if \(r = q_1q_2...q_m\) is another factorization of \(r\) into irreducibles, then \(m =n\) and there is some renumbering of the factors so that \(p_i\) is associate to \(q_i\) for all \(i\).
In a UFD a nonzero element is a prime iff it’s irreducible.
The integers \(\mathbb{Z}\) are a UFD
finite fields \(\subset\) fields \(\subset\) Euclidean domains \(\subset\) PIDs \(\subset\) UFDs \(\subset\) integral domains \(\subset\) commutative rings
If \(F\) is a field, then \(F[x]\) is a PID and a UFD.
(Gauss’ Lemma) Let \(R\) be a UFD with field of fractions \(F\) and let \(p(x) \in R[x]\). If \(p(x)\) is reducible in \(F[x]\) then \(p(x)\) is reducible in \(R[x]\).
\(R\) is a UFD iff \(R[x]\) is a UFD.
Let \(F\) be a field and let \(p(x) \in F[x]\). Then \(p(x)\) has a factor of degree one iff \(p(x)\) has a root in \(F\).
(Eisenstein’s Criterion) Let \(P\) be a prime ideal of the integral domain \(R\) and let \(f(x) = x^n + a_{n-1}x^{n-1}+...+a_1x+a_0\) be a polynomial in \(R[x]\). Suppose \(a_{n-1},...,a_1,a_0\) are all elements of \(P\) and suppose \(a_0\) is not an element of \(P^2\). Then \(f(x)\) is irreducible in \(R[x]\).
Let \(F \subseteq K \subseteq L\) be fields. Then \[[L:F] = [L:K][K:F]\]
The characteristic of a field \(F\) is defined to be the smallest positive integer \(p\) such that \(p*1_F = 0\) if such a \(p\) exists and is defined to be \(0\) otherwise.
The prime subfield of a field \(F\) is the subfield of \(F\) generated by the multiplicative identity \(1_F\) of \(F\).
If \(K\) is a field containing the subfield \(F\), then \(K\) is said to be an extension field of \(F\), denoted \(K/F\).
The degree of a field extension \(K/F\) denoted \([K:F]\) is the dimension of \(K\) as a vector space over \(F\).
Let \(K\) be an extension of the field \(F\) and let \(\alpha, \beta,... \in K\) be a collection of elements in \(K\). Then the smallest subfield of \(K\) containing both \(F\) and the elements \(\alpha, \beta,...\), denoted \(F(\alpha, \beta,...)\) is called the field generated by \(\alpha, \beta,...\) over \(F\).
If the field \(K\) is generated by a single element \(\alpha\) over \(F\), \(K = F(\alpha)\), then \(K\) is said to be a simple extension of \(F\) and the element \(\alpha\) is called a primitive element for the extension.
The element \(\alpha \in K\) is said to be algebraic over \(F\) if \(\alpha\) is a root of some nonzero polynomial \(f(x) \in F[x]\). If \(\alpha\) is not algebraic over \(F\) then \(\alpha\) is said to be transcendental over \(F\). The extension \(K/F\) is said to be algebraic if every element of \(K\) is algebraic over \(F\).
Let \(\alpha\) be algebraic over \(F\). Then there is a unique monic irreducible polynomial \(m_{\alpha, F}(x) \in F[x]\) which has \(\alpha\) as a root. A polynomial \(f(x) \in F[x]\) has \(\alpha\) as a root iff \(m_{\alpha, F}(x)\) divides \(f(x)\) in \(F[x]\).
The element \(\alpha\) is algebraic over \(F\) iff the simple extension \(F(\alpha)/F\) is finite.
If the extension \(K/F\) is finite, then it’s algebraic.
Let \(L/F\) be an arbitrary extension. Then the collection of elements of \(L\) that are algebraic over \(F\) form a subfield \(K\) of \(L\).
If \(K\) is algebraic over \(F\) and \(L\) is algebraic over \(K\), then \(L\) is algebraic over \(F\).
THe extension field \(K\) of \(K\) is called a splitting field for the polynomial \(f(x) \in F[x]\) if \(f(x)\) factors completely into linear factors in \(K[x]\) and \(f(x)\) does not factor completely into linear factors over any proper subfield of \(K\) containing \(F\).
For any field \(F\), if \(f(x) \in F[x]\) then there exists an extension \(K\) of \(F\) which is a splitting field for \(f(x)\).
If \(K\) is an algebraic extension of \(F\) which is the splitting field over \(F\) for a collection of polynomials \(f(x) \in F[x]\) then \(K\) is called a normal extension of \(F\).
(Uniqueness of Splitting Fields) Any two splitting fields for a polynomial \(f(x) \in F[x]\) over a field \(F\) are isomorphic.
A field \(K\) is said to be algebraically closed if every polynomial with coefficients in \(K\) has a root in \(K\).
(Fundamental Theorem of Algebra) The field \(\mathbb{C}\) is algebraically closed.
A polynomial over \(F\) is called separable if it has no multiple roots.
Let \(F\) be a field of characteristic \(p\). Then for any \(a,b \in F\), \[(a+b)^p = a^p + b^p_{}\mathrm{and} (ab)^p = a^p b^p\]
Suppose that \(\mathbb{F}\) is a finite field of characteristic \(p\). Then every element of \(\mathbb{F}\) is a \(p\)th power in \(\mathbb{F}\).
If \(H\) is a subgroup of the group of automorphisms of \(K\), the subfield of \(K\) fixed by all the elements of \(H\) is called the fixed field of \(H\).
Let \(K/F\) be a finite extension. Then \(K\) is said to be Galois over \(F\) and \(K/F\) is a Galois extension if \(|Aut(K/F)| = [K:F]\). If \(K/F\) is Galois the group of automorphisms \(Aut(K/F)\) is called the Galois Group of \(K/F\), denoted \(Gal(K/F)\).
If \(K\) is the splitting field over \(F\) of a separable polynomial \(f(x)\) then \(K/F\) is Galois.
If \(f(x)\) is a separable polynomial over \(F\), then the Galois group of \(f(x)\) over \(F\) is the Galois group of the splitting field of \(f(x)\) over \(F\).
Let \(G= \{\sigma_1 = 1, \sigma_2,...,\sigma_n\}\) be a subgroup of automorphisms of a field \(K\) and let \(F\) be the fixed field. Then \([K:F] = n = |G|\).
The extension \(K/F\) is Galois iff \(K\) is the splitting field of some separable polynomial over \(F\).
(Fundamental Theorem of Galois Theory) Let \(K/F\) be a Galois extension and set \(G=Gal(K/F)\). Then there is an order reversing bijection between the subfields of \(K\) containing \(F\) and the subgroups of \(G\). Theorem continues, but this is biggest idea.