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.
\(\textbf{Problem 1. }\) Suppose that \( A \) and \( B \) are \( n \times n \) matrices over the complex numbers.
\(\textbf{Proof(1). }\)Suppose that \(A\) and \(B\) are two similar \(n\times n\) matrices. Then we have \(A = PBP^{-1}\) for some invertible \(n\times n\) matrices over \(\mathbb{C}\). Suppose the characteristic polynomial of \(A\) is \(\chi_A(x)\). Then we have Hence, we have \[ \begin{align} \chi_A(x) &= \text{det}(xI - A)\\ &= \text{det}(xI - PBP^{-1})\\ &= \text{det}(PxIP^{-1} - PBP^{-1})\\ &= \text{det}(P(xI - B)P^{-1})\\ &= \text{det}(P)\text{det}(xI - B)\text{det}(P^{-1})\\ &= \text{det}{P}\text{det}{P^{-1}}\text{det}(xI - B)\\ &= \text{det}(PP^{-1})\text{det}(xI - B)\\ &= \text{det}(I)\text{det}(xI - B)\\ &= \text{det}(xI - B)\\ &= \chi_B(x). \end{align} \] Thus, we show that the characteristic polynomial of \(A\) and \(B\) are the same. Now we need to show that the minimal polynomial of \(A\) and \(B\) are the same. Suppose that the minimal polynomial of \(A\) is \(\mu_A(x):=(x-\lambda_1)^{e_1}\cdots(x - \lambda_m)^{e_m}\) and the minimal polynomial of \(B\) is \(\mu_B(x):=(x-\lambda_1)^{f_1}\cdots(x - \lambda_m)^{f_m}\). Then we have \[ \begin{align} \mu_A(A) &= 0\\ \mu_A(PBP^{-1}) &= (\lambda_1I - PBP^{-1})^{e_1}\cdots(\lambda_mI - PBP^{-1})^{e_m}\\ &= (P\lambda_1IP^{-1} - PBP^{-1})^{e_1}\cdots(P\lambda_mIP^{-1} - PBP^{-1})^{e_m}\\ &= P(\lambda_1I - B)^{e_1}P^{-1}\cdots P(\lambda_mI - B)^{e_m}P^{-1}\\ &= P(\lambda_1I - B)^{e_1}\cdots(\lambda_mI - B)^{e_m}P^{-1}\\ P(\lambda_1I - B)^{e_1}\cdots(\lambda_mI - B)^{e_m}P^{-1} &= 0\\ (\lambda_1I - B)^{e_1}\cdots(\lambda_mI - B)^{e_m} &= P^{-1}0P = 0\\ \end{align} \] Thus, we can know that \(e_i\geq f_i\) for \(i = 1, \ldots, m\). Similarly, we can know that \[ \begin{align} \mu_B(B) &= 0\\ \mu_B(P^{-1}AP) &= (\lambda_1I - P^{-1}AP)^{f_1}\cdots(\lambda_mI - P^{-1}AP)^{f_m}\\ &= (P^{-1}\lambda_1IP - P^{-1}AP)^{f_1}\cdots(P^{-1}\lambda_mIP - P^{-1}AP)^{f_m}\\ &= P^{-1}(\lambda_1I - A)^{f_1}P\cdots P^{-1}(\lambda_mI - A)^{f_m}P\\ &= P^{-1}(\lambda_1I - A)^{f_1}\cdots(\lambda_mI - A)^{f_m}P\\ P^{-1}(\lambda_1I - A)^{f_1}\cdots(\lambda_mI - A)^{f_m}P &= 0\\ (\lambda_1I - A)^{f_1}\cdots(\lambda_mI - A)^{f_m} &= P0P^{-1} = 0\\ \end{align} \] Thus, we can know that \(e_i\leq f_i\) for \(i = 1, \ldots, m\). Hence, we have \(e_i = f_i\) for \(i = 1, \ldots, m\). Therefore, we show that they share the same minimal polynomial.\(\blacksquare\)
\(\textbf{Solution 2.}\) The converse of (i) does hold when \(n = 3\). The converse of (i) does not hold when \(n = 4\).
Problem 4.Recall that a group \(G\) is said to be solvable if there exists a sequence of subgroups \[ \{e\} = G_0\subset G_2\subset \dots \subset G_{n-1}\subset G_n = G, \] such that for each \(i\), \(G_i\) is a normal subgroup of \(G_{i+1}\) and the factor group \(G_{i+1}/G_i\) is abelian. Prove
Problem 5. Let \( R \) be an integral domain with quotient field (i.e., field of fractions) \( K \). Let \( \mathcal{J} \) denote the collection of ideals \( (R : \alpha) \) such that \( \alpha \in K \), where \( (R : \alpha) := \{ r \in R \mid r\alpha \in R \} \). Show that \( R \) is a UFD if the following conditions hold (15 points):
Proof. Let \(\mathcal{P}\) be the set of all principal ideals of \(R\). We will prove that \(\mathcal{P} = \mathcal{J}\cup\{0\}\). Since \(\mathcal{J}\) is a set of principal ideals of \(R\) by assumption, we can know that \(\mathcal{J}\cup\{0\}\subset \mathcal{P}\). Now, let \(r\in R\) and \(r\) is not a unit or zero. Hence, we can know that \(r^{-1}\notin R\). However, we have \(r^{-1}\in K\) since \(K\) is the field of fractions of \(R\). We have \(r\in (R: r^{-1})\) since \(r\cdot r^{-1} = 1\in R\), which implies that \(\langle r\rangle \subset (R: r^{-1})\). If \(x\in (R: r^{-1})\), we know that \(xr^{-1} = s\in R\) for some \(s\in R\). Hence, we have \(x = sr\in \langle r\rangle\), which implies that \((R: r^{-1})\subset \langle r\rangle\). Now, we have \(\langle r\rangle = (R: r^{-1})\) when \(r\) is not a unit or zero. It is not hard to see that \((R: \alpha)=R\) when \(\alpha\in R\). Thus, we can know that \(\mathcal{P} = \mathcal{J}\cup\{0\}\). Since \(\mathcal{J}\) satisfies the ascending chain condition, we can know that \(\mathcal{J}\cup \{0\}\) satisfies the ascending chain condition as well (i.e. add \(\{0\}\) to the chain as the least element). Hence, we can know that \(\mathcal{P}\) satisfies the ascending chain condition. Once the ascending chain condition holds on the principal ideals of \(R\), we can know that every element of \(R\) is a product of irreducible elements. Now, we need to show that every irreducible element is prime. Let \(p\in R\) be an irreducible element. Suppose that \(p\mid ab\) where \(a, b\in R\) and, without loss of generality, we can assume that \(p\nmid b\). If \(p\nmid b\), we can know that \(b/p\notin R\). But, we know that \((R: b/p)\) is a principal ideal of \(R\) since \(b/p\in K\) according to the assumption. So we will show that \((R: b/p) = \langle p\rangle\). If \(s\in\langle p\rangle\), we have \(s = pr\) for some \(r\in R\). Hence, \(s\cdot b/p = pr\cdot b/p = rb\in R\), which implies that \(s\in (R: b/p)\) and \(\langle p\rangle\subset (R: b/p)\). Based on the assumption, let \((R: b/p)= \langle k\rangle \) for some \(k\in R\). Then, we have \(\langle p\rangle \subset \langle k\rangle\). Thus, we can know that \(k\mid p\). Since \(p\) is irreducible, we have \(pu = k\) for some unit \(u\in R\). It means that \(\langle p\rangle = \langle k\rangle = (R: b/p)\). Since \(p\mid ab\), we have \(ab/p\in R\). Thus, we know that \(a\in (R: b/p)\). It shows that \(a\in \langle p\rangle\), which means that \(p\mid a\). Hence, we can know that every irreducible element is prime. Once we know that every irreducible is prime, we want to show that the uniqueness of factorization holds in \(R\). Since every element is a product of irreducible elements, we assume that \[a = p_1p_2\cdots p_n = q_1q_2\cdots q_m\] where \(p_i, q_j\) are irreducible elements of \(R\). We also know that \(p_i, q_j\) are prime elements of \(R\). Hence, we can know that \(p_1\mid q_1q_2\cdots q_m\). If \(p_1\not\mid q_1\), we can know that \(p_1\mid q_2\cdots q_m\). By repeating this process, we can know that \(p_1\mid q_i\) for some \(i\). And \(q_i\) is irreducible, which implies that \(u_1p_1 = q_i\) for some unit \(u_1\in R\). In that case, we rearrange the expression of \(q_1q_2\cdots q_m\) and have \(u_1p_1q_1\cdots q_m = p_1p_2\cdots p_n\). Similarly, we can know that \(p_2\mid q_j\) for some \(j\), which implies that \(q_j = u_2p_2\) for some unit \(u_2\in R\). Since it is a finite product, we can know that, by repeating this process, it will eventually reach the end. If \(n\lt m\), we have \(p_1\cdot p_n = up_1\cdot p_nq_{n+1}\cdots q_m\) for some unit \(u\in R\). We can have \(p_1\cdot p_n - up_1\cdot p_nq_{n+1}\cdots q_m = p_1\cdot p_n(1-uq_{n+1}\cdots q_m) = 0\). Thus, we have \(1-uq_{n+1}\cdots q_m = 0\), which implies that \(uq_{n+1}\cdots q_m = 1\). It is a contradiction since \(q_{n+1}\cdots q_m\) is a product of irreducible elements. For the other direction, we can have the same result. Thus, we can know that the uniqueness of factorization holds in \(R\). \(\blacksquare\)
\(\textbf{Problem 6.}\) Show that \(\mathbb{Q}(\sqrt[3]{2})\) is not contained in \(\mathbb{Q}(\epsilon)\) for any root of unity \(\epsilon \in \mathbb{C}\).
\(\textbf{Proof. }\) We will prove it with mathematical contradiction. Suppose that it is true that \(\mathbb{Q}(\sqrt[3]{2})\) is contained in \(\mathbb{Q}(\epsilon)\) for some \(n\)th root of unity \(\epsilon \in \mathbb{C}\). Firstly, we know that \(\mathbb{Q}(\epsilon)\cong \mathbb{Q}/\Phi_n(x)\), where \(\Phi_n(x)\) is the \(n\)th cyclotomic polynomial. Since \(\Phi_n(x)\) is irreducible over \(\mathbb{Q}\), we can know that \(\Phi_n(x)\) is separable over \(\mathbb{Q}\) since any irreducible polynomial over a characteristic zero field is separable. Then, we can know that \(\mathbb{Q}(\epsilon)\) is a galois extension over \(\mathbb{Q}\). Since \(\mathbb{Q}(\sqrt[3]{2})\) is a subfield of \(\mathbb{Q}(\epsilon)\), we can know that \(\mathbb{Q}(\sqrt[3]{2})\) is a galois extension over \(\mathbb{Q}\). The minimal polynomial of \(\sqrt[3]{2}\) is \(x^3 - 2\), which is irreducible over \(\mathbb{Q}\). However, we can know that \(x^3 - 2\) is not separable over \(\mathbb{Q}\) since it does not contain the complex roots in \(\mathbb{Q}(\sqrt[3]{2})\). Hence, we have a contradiction. Therefore, we can know that \(\mathbb{Q}(\sqrt[3]{2})\) is not contained in \(\mathbb{Q}(\epsilon)\) for any root of unity \(\epsilon \in \mathbb{C}\).\(\blacksquare\)