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.
Proof. Firstly, \(\epsilon^6 - 1 = 0\), we know that the minimal polynomial of \(\epsilon\) over \(\mathbb{Q}\), \(\mu_\epsilon(x)\), divides \(x^6 - 1\). Since \[ x^6 - 1 = (x^3 + 1)(x^3 - 1) = (x + 1)(x^2 - x + 1)(x - 1)(x^2 + x + 1), \] We can know that the \(\epsilon\) cannot be
3. Set \( f(x) = x^3 + 2 \in \mathbb{Z}[x] \) and fix \( \alpha \in \mathbb{C} \), a root of \( f(x) \). Set \( K := \mathbb{Q}(\alpha) \). Let \( R \) denote the subset of \( K \) consisting of elements of the form \( a + b\alpha + c\alpha^2 \), with \( a, b, c \in \mathbb{Z} \).
(a) Show that \( R \) is a subring of \( K \).
(b) Show that for every non-zero ideal \( I \subseteq \mathbb{Z} \), \( IR \cap \mathbb{Z} = I \). Here, \( IR \) denotes the ideal of \( R \) generated by \( I \).
(c) Show that for every non-zero ideal \( J \subseteq R \), \( J \cap \mathbb{Z} \ne 0 \).
(d) Show that every non-zero prime ideal in \( R \) is a maximal ideal.
Proof (b). Suppose that \(I\) is a non-zero ideal \(I\subseteq\mathbb{Z}\) and \(IR\) is the ideal of \(R\) generated by \(I\). Since \(\mathbb{Z}\subseteq R\), we can know that \(I\subseteq IR\). And \(I\subseteq\mathbb{Z}\), we can get \(I\subseteq IR\cap\mathbb{Z}\). For any \(c\in \mathbb{Z}\cap R\), we can know that \(c\cdot I \subseteq I\) since \(I\) is an ideal of \(R\). If \(a\in R\setminus I\), we have \(a = a_1 + a_2\alpha + a_3\alpha^2\), where \(a_1, a_2, a_3\in \mathbb{Z}\) and \(a\notin I\). Hence, for any \(c\in \mathbb{I}\), we have \(ca = ca_1 + ca_2\alpha + ca_3\alpha^2\) and \(ca\notin I\). Now, we can know that \(IR\cap\mathbb{Z} = I\).
Proof (c). Suppose that \(J\) is a non-zero ideal \(\{0\}\neq J\subseteq R\). Hence, for any \(j\in J\), we have \(j = a + b\alpha + c\alpha^2\), where \(a, b, c\in \mathbb{Z}\). If we have \(a \neq 0\) and \(b\neq 0\) and \(c\neq 0\), we can know that \(j\neq 0\). Since \(x^3 + 2\) is irreducible over \(\mathbb{Q}\) and \(\alpha\) is a root of \(x^3 + 2\), we can know that \(x^3 + 2\) is the minimal polynomial of \(\alpha\) over \(\mathbb{Q}\). Given that \(\mathbb{Q}\) is a field, we can know that \(\mathbb{Q}(\alpha)\cong \mathbb{Q}[x]/(x^3 + 2)\) is a field. Since \(j\neq 0\), we can know that \(j\) is a unit in \(\mathbb{Q}(\alpha)\). Then, there exists \(j^{-1} = \frac{a_0}{a_1} + \frac{b_0}{b_1}\alpha + \frac{c_0}{c_1}\alpha^2\), where \(a_0, a_1, b_0, b_1, c_0, c_1\in \mathbb{Z}\) and \(a_1, b_1, c_1\neq 0\). Hence, we have \[ \begin{align} 1 = jj^{-1} &= (a + b\alpha + c\alpha^2)\left(\frac{a_0}{a_1} + \frac{b_0}{b_1}\alpha + \frac{c_0}{c_1}\alpha^2\right)\\ 0\neq a_1b_1c_1 &= (a + b\alpha + c\alpha^2)\left(a_0b_1c_1 + b_0a_1c_1\alpha + a_1b_1c_0\alpha^2\right)\in \mathbb{Z} \end{align} \] Thus, we can know that \(J\cap \mathbb{Z}\neq \{0\}\). \(\blacksquare\)
Proof (d). Suppose that \(P\) is a non-zero prime ideal in \(R\). Then, we want to show that \(P\cap \mathbb{Z}\) is an ideal of \(\mathbb{Z}\). Since \(0\in P\) and \(0\in \mathbb{Z}\), we have \(0\in P\cap\mathbb{Z}\). If \(a\in P\cap \mathbb{Z}\), we can know that \(-a\in P\) since \(P\) is an ideal of \(R\), which implies it is an additive subgroup of \(R\). Also, \(-a\in \mathbb{Z}\), we can know that \(-a\in P\cap \mathbb{Z}\). It is clear that \(P\cap \mathbb{Z}\) is abelian, associative, and closed under both addition and multiplication by the definition of an ideal. Hence, we can know that \(P\cap \mathbb{Z}\) is an additive subgroup of \(\mathbb{Z}\). Then if \(p\in P\cap\mathbb{Z}\), we can know that \(p\cdot\mathbb{Z}\subseteq\mathbb{Z}\) and \(pP\subset P\). THus, \(P\cap\mathbb{Z}\) is an ideal of \(\mathbb{Z}\). We know that \(\mathbb{Z}\) is a PID, which implies that \(P\cap\mathbb{Z} = \langle q\rangle\) for some \(q\in \mathbb{Z}\). Now, we want to show that \(P\cap \mathbb{Z}\) is a prime ideal of \(\mathbb{Z}\). If \(ab\in P\), where \(a, b\in \mathbb{Z}\), we have \(a, b\in R\) and \(a\in P\) or \(b\in P) since \(P\) is a prime ideal of \(R\). It shows that we have either \(a\in P\) or \(b\in P) where \(a, b\in \mathbb{Z}\). Hence, we can know that \(P\cap \mathbb{Z}\) is a prime ideal of \(\mathbb{Z}\). Thus, we can know that \(P\cap \mathbb{Z} = p\mathbb{Z}\) for some prime number \(p\in \mathbb{Z}\). Thus, we can know that \(pR\subseteq P\) since \(p\in P\). Since \[ R/pR = \{a + b\alpha + c\alpha^2\mid a, b, c\in \mathbb{Z}/p\mathbb{Z}\}. \] Now, we want to show that \(R/pR\cong \mathbb{Z}/p\mathbb{Z}[x]/(x^3 + 2)\). Given that \(\alpha\) is a root of \(x^3 + 2\) over \(\mathbb{Q}\) and we know that \(\alpha\in\mathbb{C}\). Suppose \(x^3 + 2\) is not the minimal polynomial of \(\alpha\) over \(\mathbb{Z}/p\mathbb{Z}\). Then, we have \(f(x) = x^2 + cx + d\in\mathbb{Z}/p\mathbb{Z}\) such that \(f(\alpha) = 0\). We know there are two roots of \(x^3 + 2\) in \(\mathbb{C}\), which are \(\alpha_1 = \sqrt[3]{2}e^{\pi i/3}\) and \(\alpha_2 = \sqrt[3]{2}e^{-\pi i/3}\). We can know that \(\alpha_1^2\) has a factor of \(3^{2/3}\) and \(\alpha_2^2\) has a factor of \(3^{2/3}\) as well, which implies that \(\alpha^2 + c\alpha\notin \mathbb{Z}/p\mathbb{Z}\) for any \(c\in \mathbb{Z}/p\mathbb{Z}\). Hence, we can know that \(x^3 + 2\) is the minimal polynomial of \(\alpha\) over \(\mathbb{Z}/p\mathbb{Z}\). Thus, we have \(\mathbb{Z}/p\mathbb{Z}[x]/(x^3 + 2)\) is a field and \(\mathbb{Z}/p\mathbb{Z}[x]/(x^3 + 2)=\{a + b\alpha + c\alpha^2\mid a, b, c\in \mathbb{Z}/p\mathbb{Z}\}\). Thus, we can know that \(R/pR\cong \mathbb{Z}/p\mathbb{Z}[x]/(x^3 + 2)\). Hence, we have \(pR\) is a maximal ideal. Given that \(P\) is a prime ideal, which implies that it is a proper ideal, and \(pR\subset P\), we can know that \(P = pR\). Hence, we can know that \(P\) is a maximal ideal. \(\blacksquare\)
5. Let \(A\) and \(B\) be \(n \times n\) commuting matrices over \(\mathbb{C}\).
(a) Prove that \(A\) and \(B\) have a common eigenvector.
(b) Assume that \(B\) has \(n\) distinct eigenvalues. Prove that \(A\) is diagonalizable.
Proof (a). (to be continued...)
Proof (b). Given that \(B\) has \(n\) distinct eigenvalues, and they are \(\lambda_1, \dots, \lambda_n\). Thus, we can know that \(\ker(A-\lambda_1 I)\oplus \dots \oplus \ker(A - \lambda_nI) = \mathbb{C}^n\). Hence, we can know that any two eigenvectors correspondent to different eigenvalues are orthogonal to each other. \ Here is the proof: Suppose that \(v\in \ker(A-\lambda_vI)\) and \(w\in \ker(A-\lambda_wI)\). \[ \lambda Bv, w\lambda \]
Problem 6. Let \( A \) be the \( n \times n \) matrix over \( \mathbb{C} \) all of whose entries equal \(1\).
(a) Find the minimal polynomial of \( A \).
(b) Find the Jordan canonical form of \( A \).
(c) Let \( B \) denote the \( n \times n \) matrix over \( \mathbb{C} \) whose diagonal entries are 0 and all other entries are 1. Find the Jordan canonical form of \( B \).
Solution (a). Let \(A\) be the \(n \times n\) matrix over \(\mathbb{C}\) all of whose entries equal \(1\). Then, we can know that \(A^2 = nA\). Thus, we have \(A^2 - nA = 0\), which implies that \(\mu_A \mid x^2 - nx = x(x - n)\). If \(\mu_A\) is the polynomial of degree \(1\), the \(\mu_A = x\) or \(\mu_A = x - n\). However, we know that \(A\neq 0\) and \(A - n\cdot I\neq 0\), which is the contradiction. Therefore, we can know that \(\mu_A = x(x - n)\).
Solution (b). According to the minimal polynomial, we can know that it only has two eigenvalues, \(0\) and \(n\). Then we try to find the dimension of the eigenspace of \(0\) and \(n\). Since we can see that \(\text{dim}\ker(A-0\cdot I) = n-1\). Besides that, we can know that \(\text{dim}\ker(A - n\cdot I)\geq 1\), which means that \(\text{dim}\ker(A-0\cdot I)+\text{dim}\ker(A-n\cdot I)\geq n\). We also have \(\(\text{dim}\ker(A-0\cdot I)+\text{dim}\ker(A-n\cdot I)\leq n\), which implies that \(\text{dim}\ker(A-0\cdot I)+\text{dim}\ker(A-n\cdot I)= n\). Then, we can know that the geometric Multiplicity equals to the algebraic multiplicity. Hence, the Jordan canonical form of \(A\) is the diagonal matrix with \(0\) and \(n\) on the diagonal.
Solution (c (1)). Let \(A_n\) to denote the \(n\times n\) matrix with all entries equal to 1. Then, we denote \(B_n\) as \(A_n - I_n\) where \(I_n\) is the identity matrix of size \(n\times n\). Since, we know that \(A_n\) has diagonlization in the form of \(A = PDP^{-1}\) where \(D\) is the diagonal matrix with \(0\) and \(n\) on the diagonal entries. Hence, we can know that \[ B_n = A_n - I_n = PDP^{-1} - PI_nP^{-1} = P(D - I_n)P^{-1}. \] It is not hard to see that \(D - I_n\) is a diagonal matrix, which means that \(B_n\) is diagonalizable. Again, since the diagonal entries of \(D\) is either \(0\) or \(n\), we can know that \(D - I_n\) is the diagonal matrix with diagonal entries of \(-1\) or \(n-1\). Therefore, the Jordan Canonical Form of \(B_n\) is the diagonal matrix with diagonal entries of \(-1\) or \(n-1\).
Solution (c (2)). Let \(A_n\) to denote the \(n\times n\) matrix with all entries equal to 1. Then, we denote \(B_n\) as \(A_n - I_n\) where \(I_n\) is the identity matrix of size \(n\times n\). Let \(e_1\) to be the first standard basis vector in \(\mathbb{C}^n\). Then we have \(B\cdot e_1 = [0, 1, 1, \ldots, 1]^T\). Since \(B_n^2 = (A_n - I_n)^2 = A_n^2 - 2A_n + I_n = nA_n - 2A_n + I_n = (n - 2)A_n + I_n\). Hence, we can see that \(B_n^2\) is the \(n\times n\) matrix with all entries equal to \(n-2\) besides the diagonal entries are \(n - 1\). Then, we can get \(B_n^2 \cdot e_1 = [n-1, n-2, n-2, \ldots, n-2]^T\). Thus, we will get the following equations: \[ \begin{align} (n-1)\cdot e_1 + (n-2)\cdot B_n\cdot e_1 &= B_n^2\cdot e_1 \\ B_n^2\cdot e_1 - (n-2)\cdot B_n\cdot e_1 &- (n-1)\cdot e_1 = 0. \end{align} \] Therefore, we can have the minimal polynomial of \(B_n\) is \(x^2 - (n-2)x - (n-1)\). By Quadratic Formula, we can find the two roots of the minimal polynomial, which are \(n-1\) and \(-1\). Hence, we can know that \(\mu_{B_n} = (x - (n-1))(x + 1)\). Now, we try to find the dimension of the eigenspace of \(-1\). Since \(\ker(B_n -(-1)\cdot I) = \ker(A_n) \) and we already knew that \(\text{dim}\ker(A_n) = n-1\), \(\text{dim}\ker(B_n -(-1)\cdot I) = n-1 \). By the similar reason, we can know that \(\text{dim}\ker(B_n - (n-1)\cdot I)\geq 1\) and it shows that \(\text{dim}\ker(B_n - (n-1)\cdot I)+\text{dim}\ker(A_n)\geq n\). With the condition that \(\text{dim}\ker(B_n - (n-1)\cdot I)+\text{dim}\ker(A_n)\leq n\), we have \(\text{dim}\ker(B_n - (n-1)\cdot I)+\text{dim}\ker(A_n)= n\). Again, the geometric Multiplicity equals to the algebraic multiplicity, we the Jordan canonical form of \(B_n\) os the diagonal matrix with \(-1\) and \(n-1\) on the diagonal.