My name is Hanzhang Yin, and I have developed this website as a resource to facilitate the review of key concepts in abstract algebra, and it is not fully completed. Feel free to email me at hanyin@ku.edu if there are any errors or suggestions for improvement.
Question 1. Let \( V \) be the space of \( 2 \times 2 \) matrices with entries in \( \mathbb{C} \), the complex numbers. Let \( T \) be the operator on \( V \) that takes \[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] and outputs \[ \begin{bmatrix} a + c & a + b \\ c + d & b + d \end{bmatrix}. \] Prove that \( T \) is diagonalizable.
Firstly, we know that \(V\) has dimension \(4\) and it has basis \[ \left\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \right\}. \] Now, we define a map \(\varphi: V\to \mathbb{C}^4\) such that \[ \begin{align} \varphi\left(\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\right) &= \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix},\\ \varphi\left(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\right) &= \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix},\\ \varphi\left(\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}\right) &= \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix},\\ \varphi\left(\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\right) &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}. \end{align} \]
\(\textbf{Question 2.}\) For \(a, b \in \mathbb{C}\). Let \(M(a, b)\) be the matrix \( \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & a & b \\ \end{pmatrix}. \) Find the Jordan canonical form of \(M(a, b)\) (depending on the values \(a, b\)).
\(\textbf{Solution. }\)We can get the characteristic polynomial of \(M(a, b)\) is \(\chi_{M(a, b)}(x)=x^3-bx^2-ax = x(x^2 - bx - a)\).
Question 3. Let \( M_n(\mathbb{R}) \) be the set of \( n \times n \) matrices with entries in \( \mathbb{R} \). For \( X, Y \in M_n(\mathbb{R}) \), \( Y \) is called \( X \)-consistent if \( X + \alpha Y \) is invertible for all \( \alpha \in \mathbb{R} \).
1. Assume \( n > 1 \). Prove that \( X \in M_n(\mathbb{R}) \) is invertible if and only if there is a nonzero \( X \)-consistent matrix.
2. If \( X \in M_n(\mathbb{R}) \) is an invertible skew-symmetric matrix (i.e., the transpose of \( X \) is equal to \( -X \)), then \( I_n \) (the identity matrix in \( M_n(\mathbb{R}) \)) is \( X \)-consistent.
Proof (1). Firstly, we want to show that if \(X + \alpha Y\) is invertible for all \(\alpha \in \mathbb{R}\), then \(X\) is invertible. We can set \(\alpha = 0\), then we know that \(X + 0\cdot Y = X\) is invertible. Hence, we have \(X\) is invertible. For the other direction, let \(N\) to be a nilpotent \(n\times n\) real matrix, such that \(N^m = 0\) for some \(m > 0\). Hence, we can see that \[ (I + \alpha N)\cdot (I - \alpha N + \alpha^2 N^2 - \cdots + (-1)^{m-1}\alpha^{m-1}N^{m-1}) = I - \alpha^m N^m = I, \] for any \(\alpha\in \mathbb{R}\). Thus, we know that \(I + \alpha N\) has an inverse, and it is invertible. Hence, we can know that \(\det(I + \alpha N)\neq 0\) for all \(\alpha\in \mathbb{R}\). Given that \(X\) is invertible, we can know that \(\det(X)\neq 0\). Then, \[ \det(X + \alpha X\cdot N) = \det(X)\det(I + \alpha N) \neq 0. \] Thus, we have \(X + \alpha X\cdot N\) is invertible for all \(\alpha\in \mathbb{R}\), and \(X\cdot N\) is \(X\)-consistent. \(\blacksquare\)
Proof (2). Suppose that \(X\) is an invertible skew-symmetric matrix. Then we have \(X^T = -X\). Suppose that \(\lambda\) is an eigenvalue of \(X\) and \(v\) is the corresponding eigenvector. Then we have \(Xv = \lambda v\) and \[ \begin{align} \lambda\langle v, v\rangle = \langle \lambda v, v \rangle = \langle Xv, v \rangle = \langle v, X^T v \rangle &= \langle v, -Xv \rangle = \langle v, -\lambda v \rangle = -\overline{\lambda}\langle v, v \rangle.\\ (\lambda + \overline{\lambda})\langle v, v \rangle &= 0. \end{align} \] Since \(v\) is an eigenvalue, we have \(\lambda\neq 0\) and \(\langle v, v \rangle > 0\). Now, we can know that \(\lambda + \overline{\lambda} = 0\). Assume that \(\lambda\) is real. Then we have \(\lambda + \overline{\lambda} = 2\lambda = 0\), which implies that \(\lambda = 0\). However, we know that \(\lambda\neq 0\) since \(X\) is invertible. Hence, we can know that \(\lambda\) is not real. Then, we can know that \(\lambda\) has to be complex. Since \(\det(X + \alpha I) = \det(X - (-\alpha)I)\), we know \(\det(X + \alpha I) = 0\) if and only if \(-\alpha\) is an eigenvalue of \(X\). Again, we showed that all eigenvalues of \(X\) are complex. Hence, we can know that \(\det(X + \alpha I) \neq 0\) for all \(\alpha\in \mathbb{R}\). Therefore, we showed that \(I\) is \(X\)-consistent. \(\blacksquare\)
Question 4. Let \( f(x) = x^4 + 5 \in \mathbb{Q}[x] \) and \( K \) denote the splitting field of \( f(x) \) over \( \mathbb{Q} \).
Proof (i). By Eisenstein's criterion, we choose \(p = 5\), then we have \(f(x) = x^4 + 5\) is irreducible over \(\mathbb{Z}\). If \(f(x)\) is irreducible over \( \mathbb{Z} \), then it is irreducible over \( \mathbb{Q} \). \(\blacksquare\)
Proof (ii). Firstly, we can know that \(x^4 + 5 = (x^2 - \sqrt{5}i)(x^2 + \sqrt{5}i)\). We know that \(i = 0 + 1\cdot i = \cos(\pi/2) + \sin(\pi/2)i = e^{\pi i/2}\), then \((\sqrt{5}i)^{1/2} = \sqrt[4]{5}e^{\pi i/4} = \frac{\sqrt[4]{5}\sqrt{2}}{2} + \frac{\sqrt[4]{5}\sqrt{2}}{2}i\). Similarly, \(-i = 0 + (-1)\cdot i = \cos(-\pi/2) + \sin(-\pi/2)i = e^{-\pi i/2}\), then \((-\sqrt{5}i)^{1/2} = \sqrt[4]{5}e^{-\pi i/4} = \frac{\sqrt[4]{5}\sqrt{2}}{2} - \frac{\sqrt[4]{5}\sqrt{2}}{2}i\). Thus, we have \[ x^4 + 5 = \left(x + \frac{\sqrt[4]{5}\sqrt{2}}{2} + \frac{\sqrt[4]{5}\sqrt{2}}{2}i\right)\left(x - \frac{\sqrt[4]{5}\sqrt{2}}{2} - \frac{\sqrt[4]{5}\sqrt{2}}{2}i\right)\left(x + \frac{\sqrt[4]{5}\sqrt{2}}{2} - \frac{\sqrt[4]{5}\sqrt{2}}{2}i\right)\left(x - \frac{\sqrt[4]{5}\sqrt{2}}{2} + \frac{\sqrt[4]{5}\sqrt{2}}{2}i\right). \] Thus, we can know that the splitting field of \(x^4 + 5\) is \(\mathbb{Q}\left(\frac{\sqrt[4]{5}\sqrt{2}}{2} + \frac{\sqrt[4]{5}\sqrt{2}}{2}i, \frac{\sqrt[4]{5}\sqrt{2}}{2} - \frac{\sqrt[4]{5}\sqrt{2}}{2}i\right)\). Then, we want to show that \(\mathbb{Q}\left(\frac{\sqrt[4]{5}\sqrt{2}}{2} + \frac{\sqrt[4]{5}\sqrt{2}}{2}i, \frac{\sqrt[4]{5}\sqrt{2}}{2} - \frac{\sqrt[4]{5}\sqrt{2}}{2}i\right) = \mathbb{Q}\left(\sqrt[4]{5}\sqrt{2}, i\right)\). Since \[ \begin{align} \sqrt[4]{5}\sqrt{2} &= \frac{\sqrt[4]{5}\sqrt{2}}{2} + \frac{\sqrt[4]{5}\sqrt{2}}{2}i + \frac{\sqrt[4]{5}\sqrt{2}}{2} - \frac{\sqrt[4]{5}\sqrt{2}}{2}i,\\ \sqrt[4]{5}\sqrt{2}i &= \frac{\sqrt[4]{5}\sqrt{2}}{2} + \frac{\sqrt[4]{5}\sqrt{2}}{2}i - \frac{\sqrt[4]{5}\sqrt{2}}{2} - \frac{\sqrt[4]{5}\sqrt{2}}{2}i,\\ i &= \dfrac{\sqrt[4]{5}\sqrt{2}i}{\sqrt[4]{5}\sqrt{2}}, \end{align} \] we have \(\mathbb{Q}\left(\sqrt[4]{5}\sqrt{2}, i\right)\subset \mathbb{Q}\left(\frac{\sqrt[4]{5}\sqrt{2}}{2} + \frac{\sqrt[4]{5}\sqrt{2}}{2}i, \frac{\sqrt[4]{5}\sqrt{2}}{2} - \frac{\sqrt[4]{5}\sqrt{2}}{2}i\right)\). It is not hard to see that \(\mathbb{Q}\left(\frac{\sqrt[4]{5}\sqrt{2}}{2} + \frac{\sqrt[4]{5}\sqrt{2}}{2}i, \frac{\sqrt[4]{5}\sqrt{2}}{2} - \frac{\sqrt[4]{5}\sqrt{2}}{2}i\right) \subset \mathbb{Q}\left(\sqrt[4]{5}\sqrt{2}, i\right)\). Therefore, we have \(\mathbb{Q}\left(\frac{\sqrt[4]{5}\sqrt{2}}{2} + \frac{\sqrt[4]{5}\sqrt{2}}{2}i, \frac{\sqrt[4]{5}\sqrt{2}}{2} - \frac{\sqrt[4]{5}\sqrt{2}}{2}i\right) = \mathbb{Q}\left(\sqrt[4]{5}\sqrt{2}, i\right)\). After that, we want to show that \(x^4 - 20\) is the minimal polynomial of \(\sqrt[4]{5}\sqrt{2}\) over \(\mathbb{Q}\). It is not hard to see that \(\sqrt[4]{5}\sqrt{2}\) is a root of \(x^4 - 20\). Then, we can know that there does not exists a root of \(x^4 - 20\) in \(\mathbb{Q}\), which implies that \(x^2 - 20\) cannot be factored as \((x + a)(x^3 + bx + cx + d)\) where \(a, b, c, d\in \mathbb{Q}\). We suppose that \(x^4 - 20 = (x^2 + ax + b)(x^2 + cx + d)\). Then we have \[ (x^2 + ax + b)(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd = x^4 - 20. \] Hence, we have \[ \begin{align} a + c &= 0, \\ b + d + ac &= 0, \\ ad + bc &= 0, \\ bd &= 20. \end{align} \] Then, we can know that \(a = -c\) and \(ad + bc = ad - ab = a(d - b) = 0\). In that case, we have either \(a = 0\) or \(d - b = 0\). Suppose that (a = 0\). Then, we have \(c = 0\) as well. And \(b + d + ac = b + d = 0\), which implies that \(b = -d\). And \(bd = -b^2 = 20\), which is impossible if \(b\in \mathbb{Q}\). Then, if \(d - b = 0\), which means that \(d = b\). We can get \(bd = b^2 = 20\). However, \(20\) is not a perfect square in \(\mathbb{Q}\), which is a contradiction. Thus, we can know that \(x^2 - 20\) is irreducible in \(\mathbb{Q}\) and it is a minimal polynomial of \(\sqrt[4]{5}\sqrt{2}\). Then, we will have \[ \mathbb{Q}(\sqrt[4]{5}\sqrt{2})\cong \mathbb{Q}[x]/(x^4 - 20), \] Hence, \([\mathbb{Q}(\sqrt[4]{5}\sqrt{2}): Q] = 4\). For the similar reason, we can know that \(x^2 + 1\) has no root in \(\mathbb{R}\), and \(\mathbb{Q}(\sqrt[4]{5}\sqrt{2})\subset \mathbb{R}\). Hence, \(x^2 + 1\) is irreducible over \(\mathbb{Q}(\sqrt[4]{5}\sqrt{2})\), which implies that it is the minimal polynomial of \(i\) over \(\mathbb{Q}(\sqrt[4]{5}\sqrt{2})\). Then, we can have \[ \mathbb{Q}(\sqrt[4]{5}\sqrt{2})[x]/(x^2 + 1)\cong \mathbb{Q}(\sqrt[4]{5}\sqrt{2})(i) = \mathbb{Q}(\sqrt[4]{5}\sqrt{2}, i) \] And we can see that \([\mathbb{Q}(\sqrt[4]{5}\sqrt{2}, i): \mathbb{Q}(\sqrt[4]{5}\sqrt{2})] = 2\). In that case, \[ [\mathbb{Q}(\sqrt[4]{5}\sqrt{2}, i): \mathbb{Q}] = [\mathbb{Q}(\sqrt[4]{5}\sqrt{2}, i): \mathbb{Q}(\sqrt[4]{5}\sqrt{2})][\mathbb{Q}(\sqrt[4]{5}\sqrt{2}):\mathbb{Q}] = 2\cdot 4 = 8. \] Thus, we can know that there are \(8\) elements in the basis of the splitting field. Then, \[ \begin{align} (\sqrt[4]{5}\sqrt{2})^2 &= 2\sqrt{5} \\ (\sqrt[4]{5}\sqrt{2})^3 &= 2\sqrt{5}\sqrt[4]{5}\sqrt{2} = 2\sqrt[4]{5^3}\sqrt{2} \\ (\sqrt[4]{5}\sqrt{2})^4 &= 2\sqrt[4]{5^3}\sqrt{2}\sqrt[4]{5}\sqrt{2} = 20 \\ \end{align} \] Hence, we can know that the basis of \(\mathbb{Q}(\sqrt[4]{5}\sqrt{2})\) is \(\{1, \sqrt[4]{5}\sqrt{2}, \sqrt{5}, \sqrt[4]{5^3}\sqrt{2}\}\). Then, we know the elements in \(\mathbb{Q}(\sqrt[4]{5}\sqrt{2}, i)\) is in the form of \(a + b\cdot i\) where \(a, b\in mathbb{Q}(\sqrt[4]{5}\sqrt{2})\). Hence, we have the basis of \(\mathbb{Q}(\sqrt[4]{5}\sqrt{2}, i)\) as is \(\{1, \sqrt[4]{5}\sqrt{2}, \sqrt{5}, \sqrt[4]{5^3}\sqrt{2}, i, \sqrt[4]{5}\sqrt{2}i, \sqrt{5}i, \sqrt[4]{5^3}\sqrt{2}i\}\).
Proof (iii). Firstly, we can know that \(x^4 + 5\) has \(4\) distinct root in the splitting field. Hence, the order of its galois group should be the same as the degree fo the splitting field, which is \(8\). Besides that, the element of \(\text{Gal}(\mathbb{Q}(\sqrt[4]{5}\sqrt{2}, i)/\mathbb{Q})\) send the roots of \(x^4 + 5\) to the roots. Hence, we have several options for the automorphisms: \[ \begin{align} \text{id}: \text{id}(\sqrt[4]{5}\sqrt{2}) &= \sqrt[4]{5}\sqrt{2}, \text{id}(i) = i\\ \sigma_1: \sigma_1(\sqrt[4]{5}\sqrt{2}) &= -\sqrt[4]{5}\sqrt{2}, \sigma_1(i) = -i\\ \sigma_2: \sigma_2(\sqrt[4]{5}\sqrt{2}) &= \sqrt[4]{5}\sqrt{2}, \sigma_2(i) = -i\\ \sigma_3: \sigma_3(\sqrt[4]{5}\sqrt{2}) &= -\sqrt[4]{5}\sqrt{2}, \sigma_3(i) = i\\ \sigma_4: \sigma_4(\sqrt[4]{5}\sqrt{2}) &= \sqrt[4]{5}\sqrt{2}i, \sigma_4(i) = i\\ \sigma_5: \sigma_5(\sqrt[4]{5}\sqrt{2}) &= -\sqrt[4]{5}\sqrt{2}i, \sigma_5(i) = -i\\ \sigma_6: \sigma_6(\sqrt[4]{5}\sqrt{2}) &= -\sqrt[4]{5}\sqrt{2}i, \sigma_6(i) = i\\ \sigma_7: \sigma_7(\sqrt[4]{5}\sqrt{2}) &= \sqrt[4]{5}\sqrt{2}i, \sigma_7(i) = -i\\ \end{align} \] Then, we want to show that it is not an abelian group. \[ \begin{align} \sigma_6(\sigma_7(\sqrt[4]{5}\sqrt{2})) &= \sigma_6(\sqrt[4]{5}\sqrt{2}i) = \sigma_6(\sqrt[4]{5}\sqrt{2})\sigma_6(i) = -\sqrt[4]{5}\sqrt{2}i\cdot i = \sqrt[4]{5}\sqrt{2}\\ \sigma_7(\sigma_6(\sqrt[4]{5}\sqrt{2})) &= \sigma_7(-\sqrt[4]{5}\sqrt{2}i) = -\sigma_7(\sqrt[4]{5}\sqrt{2})\sigma_6(i) = -\sqrt[4]{5}\sqrt{2}i\cdot (-i) = -\sqrt[4]{5}\sqrt{2} \end{align} \] Thus, we can see that \(\sigma_6\circ \sigma_7\neq \sigma_7\circ\sigma_6\), which implies that \(\text{Gal}(\mathbb{Q}(\sqrt[4]{5}\sqrt{2}, i)/\mathbb{Q})\) is not abelian. \(\blacksquare\)
Question 5. Let \( G \) be a finite group, and set \( Z := Z(G) \), the center of \( G \). Recall that if \( H \) is a subgroup of \( G \), the normalizer of \( H \) is the subgroup \( N_G(H) = \{ g \in G \mid gHg^{-1} = H \} \). Suppose \( G/Z \) is abelian.
(i) Prove that if \( H \) is a proper subgroup of \( G \), then \( N_G(H) \) properly contains \( H \).
(ii) Use the fact that, if \( P \) is a Sylow subgroup of \( G \) then \( N_G(N_G(P)) = N_G(P) \), to prove that \( G \) is the (internal) direct product of its Sylow subgroups.
Proof (i). For any \(h\in H\), we have \(h\cdot H\cdot h^{-1} = H\), which implies that \(h\in N_G(H)\). Thus, we know that \(H\subset N_G(H)\). Now we want to show that \(N_G(H)\) properly contains \(H\) by contradiction. suppose that \(N_G(H) = H\). For any \(z\in Z\), we have \(zh = hz\) for all \(h\in H\) by the definition of the center. It shows that \(zHz^{-1} = H\) for any \(z\in Z\). Hence, we can get \(Z\subset N_G(H)\). Since \(H = N_G(H)\), we have \(Z\subset H\). Since \( H \) is a proper subgroup of \( G \), we know there exists \(g\in G\) and \(g\notin H\). Fix an element \(g\in G\setminus H\) and an element \(h\in H\). Given \(G/Z\) is abelian, we know that \(gh Z = hgZ\). In other words, for any \(z_1\in Z\), there exists a \(z_2\in Z\) such that \(ghz_1 = hgz_2\). Hence, \[ \begin{align} ghz_1 H &= hgz_2 H\\ gh H &= hg H\text{ for }Z\subset H,\\ gH &= hg H. \end{align} \] Now, we know there exists \(h_1, h_2\in H\) such that \[ \begin{align} gh_1 &= hgh_2, \\ gh_1h_2^{-1} &= hg \\ g^{-1}hg &= h_1h_2^{-1}\in H. \end{align} \] Since \(h\) is a random element of \(H\), we have \(g^{-1}Hg = H\), where \(g\notin H\). Thus we have \(g\in N_G(H)\) and \(g\notin H\), which is a contradiction. Thus, we can know that \(N_G(H)\) properly contains \(H\). \(\blacksquare\)
Proof (ii). Given that \(g\) is a finite group, we suppose that \(|G| = n = p_1^{e_1}\cdot p_2^{e_2}\cdots p_r^{e_r}\), where each \(p_i\) is a distinct prime. Since \(p_i^{e_i}\not\mid p_1^{e_1}\cdot p_{i-1}^{e_{i-1}}p_{i+1}^{e_{i+1}}\cdots p_r^{e_r}\), according to \(\textbf{First Sylow's Theorem}\), we can know that the Sylow \(p_i\)-subgroup, \(P_i\), exists and \(|P_i| = p_i^{e_i}\). Now, we want to show that \(P_i\cap P_j = \{1\}\), where \(i\neq j\). Firstly, we know that \(P_i\cap P_j\) is a subgroup of both \(P_i\) and \(P_j\). Hence, we have \(|P_i\cap P_j|\mid |P_i|\) and \(|P_i\cap P_j|\mid |P_j|\). Hence, \(|P_i\cap P_j|\) is a common divisor of \(p_i^{e_i}\) and \(p_j^{e_j}\). Since \(p_i\) and \(p_j\) are prime and \(p_i\neq p_j\), we know that \(\gcd(p_i^{e_i}, p_j^{e_j}) = 1\). Thus, we have \(|P_i\cap P_j| = 1\), which implies that \(P_i\cap P_j = \{e\}\). Given that \(P_i\) is a Sylow subgroup of \(G\) and we know that \(N_G(N_G(P_i)) = N_G(P_i)\), we can know that \(N_G(P_i) = G\) by contrapositive. Hence, we have that each \(P_i\) is a normal subgroup of \(G\). (to be continued...)
\(\textbf{Question 6. }\)Let \( R \) be a unique factorization domain satisfying the following property: If \( p \in R \) is a prime element, then there are no proper ideals properly containing \( pR \). Show that:
\(\textbf{Proof (a). }\) Suppose that \(I\) is a prime ideal in \(R\). Suppose that \(r\in I\) and \(r = p_1\cdots p_n\) for some prime elements \(p_1, \ldots, p_n\) since \(R\) is a unique factorization domain. Then we have \(p_1\in I\) or \(p_2\in I\) or \(\cdots\) or \(p_n\in I\). In general, there exists a prime \(p\in I\). Hence, we have \(pR \subset I\). Since \(I\) is a prime ideal, it is proper and there are no proper ideals properly containing \(pR\). Thus, it forces \(I = pR\). Hence, we have every prime ideal in \(R\) is a principal ideal. \(\blacksquare\)
\(\textbf{Proof (b). }\) Suppose that \(a, b\in R\) are non-unit and non-zero elements in \(R\). Suppose that \(d\) is the greatest common divisor of \(a\) and \(b\). Then we have \(d\mid a\) and \(d\mid b\). It shows that \(a\in dR\) and \(b\in dR\). Hence, we have \(\langle a, b \rangle \subset dR\). For the other direction, suppose that \(c\in dR\). Then we have \(c = dr\) for some \(r\in R\).