KU 2018 (August) Algebra Qualifying Exam

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Solution (ii). \(I_{1/2} \cap I_{\pi/4} = \{ f \in A \mid f(1/2) = 0 \text{ and } f(\pi/4) = 0 \}\), which is not a prime ideal. Let \[ \begin{align} f(x) &= |x - 1/2|, \\ g(x) &= |x - \pi/4|. \end{align} \] Thus, we can know that both \(f(x)\) and \(g(x)\) are continuous real function and Hence, we can know that \(f(x)\cdot g(x)\in I_{1/2} \cap I_{\pi/4}\), but \(f(x)\) and \(g(x)\) are not in \(I_{1/2}\cap I_{\pi/4}\).

Solution (b). \(\langle 2, x^2 + x + 1 \rangle\) is a maximal ideal.

Proof.Since \(\mathbb{Z}[x]\) has division algorithm, we can know if \(p(x)\in \mathbb{Z}[x]\), then we have \[ p(x) = q(x)(x^2 + x + 1) + r(x), \text{ where } \deg(r(x)) < \deg(x^2 + x + 1). \] In other words, \(r(x)=ax + b\), where \(a, b\in \mathbb{Z}\). Since \(2\in \langle 2, x^2 + x + 1\rangle\), we can know that \(r(x)=ax + b\) only has three possibilities: \(r(x)=0, r(x)=x, r(x)=1, r(x)=x+1\). Thus, we have \[ \mathbb{Z}[x]/\langle 2, x^2 + x + 1\rangle=\{0, 1, x, x + 1\}. \] And \(x(x + 1)=x^2 + x=-1=1\), which implies that \(x\) and \(x + 1\) are units. Thus, we can see that \(\langle 2, x^2 + x + 1\rangle\) is a maximal ideal. \(\blacksquare\)

Solution (c). The ideal generated by the class of \(x\) in the quotient ring \(\mathbb{C}[x, y]/\langle xy\rangle\) is a prime ideal.

Proof. We can know that \(\mathbb{C}[x, y]/\langle xy\rangle = \{a_nx^n + \dots + a_1x + b_my^m + \dots + b_1y + c + \langle xy\rangle \mid a_i, b_j, c\in \mathbb{C}\}\). For \(\langle x\rangle\subset \mathbb{C}[x, y]/\langle xy\rangle\), we have \[ \begin{align} \langle x \rangle + \langle xy \rangle &= \{x(a_nx^n + \dots + a_1x + b_my^m + \dots + b_1y + c) + \langle xy\rangle \mid a_i, b_j, c\in \mathbb{C}\}, \\ &= \{a_nx^{n+1} + \dots + a_1x^2 + b_my^mx + \dots + b_1yx + cx + \langle xy\rangle \mid a_i, b_j, c\in \mathbb{C}\}, \\ &= \{a_nx^{n+1} + \dots + a_1x^2 + cx + \langle xy\rangle \mid a_i, c\in \mathbb{C}\}, \\ &= \{a_nx^{n} + \dots + a_1x + \langle xy\rangle \mid a_i\in \mathbb{C}\}. \end{align} \] Then, we denote \(R = \mathbb{C}[x, y]/\langle xy\rangle\). Thus, we have \[ R/\langle x\rangle = \{b_my^m + b_{m-1}y^{m-1} + \dots b_1y + b_0 + \langle xy\rangle \mid b_i\in \mathbb{C}\} \cong \mathbb{C}[y]. \] Since \(\mathbb{C}[y]\) is a PID, which is an integral domain. Hence, we can know that \(\langle x\rangle\) is a prime ideal.

Proof (a). We will show that \(f(x)\) is irreducible over \(F\) by contradiction. Suppose that \(f(x)\) is reducible over \(F\). Then we have two possibilities: \[ \begin{align} f(x) &= (x + a)(x^4 + bx^3 + cx^2 + dx + e),\\ f(x) &= (x^2 + ax + b)(x^3 + cx^2 + dx + e).\\ \end{align} \] For the first case, we can know that \(f(x)\) has a root in \(\mathbb{Z}/2\mathbb{Z}\). However, \(f(0) = 0^5 + 0^2 + 1 = 1\) and \(f(1) = 1^5 + 1^2 + 1 = 1\), which implies that \(f(x)\) has no root in \(\mathbb{Z}/2\mathbb{Z}\). Hence, it leaves us with the second case. Then we have \[ \begin{align} f(x) &= (x^2 + ax + b)(x^3 + cx^2 + dx + e)\\ &= x^5 + (a + c)x^4 + (b + ac + d)x^3 + (ad + bc + e)x^2 + (ae + bd)x + be. &= x^5 + x^2 + 1. \end{align} \] Then we have \[ \begin{align} be &= 1, \\ ae + bd &= 0, \\ ad + bc + e &= 1, \\ b + ac + d &= 0, \\ a + c &= 0. \end{align} \] Since \(be = 1\), we can have \(b = e = 1\). Then we rewrite the above equations as \[ \begin{align} a + c &= 0, \\ 1 + ac +d &= 0, \\ ad + c + 1 &= 1, \\ a + d &= 0. \end{align} \] Since \(a + c = 0\) and \(a, c\in \mathbb{Z}/2\mathbb{Z}\), we can know that \(a = e\). By the similar reason, since \(a + d = 0\), we have \(a = d\). Now, we can know that \(a = c = d\), and we replace \(c\) and \(d\) with \(a\) and get \[ \begin{align} 1 + a^2 + a &= 0, \\ a^2 + a + 1 &= 1. \\ \end{align} \] Now, we see that \(a^2 + a + 1 \neq a^2 + a + 1\), which is a contradiction. Therefore, we can conclude that \(f(x)\) is irreducible over \(F\). \(\blacksquare\)

Solution (b). Firstly, we know that \(\alpha^5 + \alpha^2 + 1 = 0\). Hence, we have \[ \begin{align} \alpha^5 + \alpha^2 &= -1 = 1, \\ \alpha(\alpha^4 + \alpha) &= 1. \\ \end{align} \] Thus, we can \(\alpha^{-1} = \alpha^4 + \alpha\). Meanwhile, we have \[ \begin{align} \alpha^5 + \alpha^2 + \alpha + 1 &= \alpha \\ \alpha^5 + \alpha^2 + \alpha &= \alpha - 1 = \alpha + 1 \\ \alpha(\alpha^4 + \alpha + 1) &= \alpha + 1 \\ (\alpha^4 + \alpha + 1) &= \dfrac{\alpha + 1}{\alpha} \\ (\alpha^4 + \alpha + 1)^{-1} &= \dfrac{\alpha}{\alpha + 1} = \alpha(\alpha + 1)^{-1} \\ \end{align} \] It shows that we need to find \(\alpha(\alpha + 1)^{-1}\). Given that \[ \begin{align} (\alpha + 1)(\alpha + 1) &= \alpha^2 + 2\alpha + 1 = \alpha^2 + 1 = \alpha^5 \\ (\alpha + 1)\dfrac{\alpha + 1}{\alpha^5} &= (\alpha + 1)(\alpha + 1)(\alpha^5)^{-1} = 1. \end{align} \] Since we know that \(\alpha^{-1} = \alpha^4 + \alpha\), we have \[ \begin{align} (\alpha^5)^{-1} &= (\alpha^{-1})^{5} \\ &= (\alpha^4 + \alpha)^5 = \alpha^5(\alpha^3 + 1)^5 = (\alpha^2 + 1)\cdot (\alpha^3 + 1)^4\cdot (\alpha^3 + 1) \\ &= (\alpha^2 + 1)(\alpha^3 + 1)(\alpha^3 + 1)^2(\alpha^3 + 1)^2 \\ &= (\alpha^5 + \alpha^2 + \alpha^3 + 1)(\alpha^6 + 1)^2 \\ &= (\alpha^3)(\alpha^{12} + 1) \\ &= \alpha^3(\alpha^5\cdot \alpha^5\cdot \alpha^2 + 1) \\ &= \alpha^3((\alpha^2 + 1)^2\cdot \alpha^2 + 1) \\ &= \alpha^3(\alpha^6 + \alpha^2 + 1) \\ &= \alpha^3(\alpha\cdot \alpha^5 + \alpha^2 + 1) \\ &= \alpha^3(\alpha(\alpha^2 + 1) + \alpha^2 + 1) \\ &= \alpha^3(\alpha^3 + \alpha^2 + \alpha + 1) \\ &= \alpha^6 + \alpha^5 + \alpha^4 + \alpha^3 \\ &= \alpha(\alpha^2 + 1) + \alpha^2 + 1 + \alpha^4 + \alpha^3 \\ &= \alpha^3 + \alpha^2 + \alpha + 1 + \alpha^4 + \alpha^3 \\ &= \alpha^4 + \alpha^2 + \alpha + 1 \\ \end{align} \] Thus, we can know that \[ \begin{align} (\alpha + 1)\cdot(\alpha^5)^{-1} &= (\alpha + 1)\cdot (\alpha^4 + \alpha^2 + \alpha + 1) \\ &= \alpha^5 + \alpha^3 + \alpha^2 + \alpha + \alpha^4 + \alpha^2 + \alpha + 1 \\ &= \alpha^4 + \alpha^3 + \alpha^2. \end{align} \] Hence, the inverse of \(\alpha + 1\) is \(\alpha^4 + \alpha^3 + \alpha^2 \). Now, we can know that \[ \begin{align} (\alpha^4 + \alpha + 1)^{-1} &= \alpha(\alpha + 1)^{-1} = \alpha(\alpha^4 + \alpha^3 + \alpha^2) = \alpha^5 + \alpha^4 + \alpha^3 = \alpha^4 + \alpha^3 + \alpha^2 + 1.\\ \alpha(\alpha^4 + \alpha + 1)^{-1} &= \alpha(\alpha^4 + \alpha^3 + \alpha^2 + 1) = \alpha^5 + \alpha^4 + \alpha^3 + \alpha = \alpha^4 + \alpha^3 + \alpha^2 + \alpha + 1. \\ \end{align} \]

Solution. Given that \(A^2 = I\), we have \(A^2 - I = 0\). Hence, we set \(f(x) = x^2 - 1 = (x - 1)(x + 1)\). Then, we have \(f(A) = 0\). Thus, we can know that the minimal polynomial of \(A\), \(\mu_A(x)\), divides \(f(x)\). Hence, we have three possibilities: \(x - 1\) or \(x + 1\) or \(f(x)\). If the minimal polynomial of \(A\) is \(x - 1\), then we can have \(A - I = 0\), which means that \(A = I\). If the minimal polynomial of \(A\) is \(x + 1\), then we can have \(A + I = 0\), which means that \(A = -I = (p - 1)I\). If \(p = 2\), then we can know that \(A = I\) is the solution for both case where \(\mu_A(x) = x - 1\) and \(\mu_A(x) = x + 1\). If \(\mu_A(x) = (x - 1)(x + 1)\), then we can know that \(\gcd(x - 1, x + 1) = 1\), when \(p>2\). Hence, we can have two eigenvalues, which are \(1\) and \(p - 1\). And \(F^2 = \ker(A - I) \oplus \ker(A + I)\). Then, we can know that \(A\) is diagonalizable over \(F\). Thus, we can write \(A = PDP^{-1}\), where \[ D = \begin{bmatrix} 1 & 0 \\ 0 & p - 1 \\ \end{bmatrix}. \] And \(P\) is an invertible matrix with entries in \(F\). Denote \(P = [\vec{v_1}, \vec{v_2}]\), where \(\vec{v_1}\) and \(\vec{v_2}\) are the eigenvectors of \(A\). Besides, that \(\vec{v_1}, \vec{v_2}\) should be independent. Hence, for \(\vec{v_1}\in F^2\), we can know that it cannot be \(0\). Thus, there are \(p^2 - 1\) choices for \(\vec{v_1}\). For \(\vec{v_2}\in F^2\), we can know that it cannot be a multiple of \(\vec{v_1}\). Hence, there are \(p^2 - p\) choices for \(\vec{v_2}\). Then, we will have \((p^2 - 1)(p^2 - p)\) choices for \(P\). Hence, the number of \(2 \times 2\) matrices with entries in \(F\) such that \(A^2 = I\) is \(2 + (p^2 - 1)(p^2 - p)\) when \(p > 2\). When \(p=2\), there is only one matrix, which is the identity one.