KU 2021 (January) ALGEBRA QUALIFYING EXAM

Problem 1.

Write down all possible Jordan canonical forms for a \( 5 \times 5 \) matrix \( A \) such that \( A^3 = 0 \) (the blocks should have non-increasing size down the diagonal. We work over complex numbers).
Let \(N\) be a nilpotent matrix over any field. Prove that \(I + N\) is diagonalizable if and only if \(N = 0\) (I is the identity matrix of the same size).

\(\textbf{Solution 1.}\)

\(\textbf{Proof 2.}\)Suppose that \(I\) is an identity matrix and \(N\) is a nilpotent matrix. Hence, we know there exists a \(n\in\mathbb{N}\) such that \(N^n = 0\). If \(N = 0\), then we can know that \[ I + N = I + 0 = I. \] Since \(I\) is a diagonal matrix, we can know that \(I + N\) is diagonalizable. To see that, just pick any invertible matrix \(P\). Then, we have \(PIP^{-1} = I\). Hence, we have \(I\) is diagonalizable. Now suppose that \((I+N)\) is diagonalizable. Then, there exists a diagonal matrix \(D\) and an invertible matrix \(P\) such that \(D = P^{-1}(I + N)P\). Then, we have \[ \begin{align} P^{-1}(I + N)P &= D\\ P^{-1}IP + P^{-1}NP &= D\\ I + P^{-1}NP &= D\\ P^{-1}NP &= D - I\\ (P^{-1}NP)^n &= (D - I)^n\\ P^{-1}N^nP &= (D - I)^n\\ P^{-1}0P &= (D - I)^n\\ 0 &= (D - I)^n\\ \end{align} \] Given that \(D\) and \(I\) are diagonal matrices, we can know that \(D - I\) is also a diagonal matrix. Suppose that \[ D - I = \begin{pmatrix} a_{11} & 0 & \cdots & 0\\ 0 & a_{22} & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & a_{mm} \end{pmatrix}. \] Hence, we can know that \[ (D - I)^n = \begin{pmatrix} a_{11}^n & 0 & \cdots & 0\\ 0 & a_{22}^n & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & a_{mm}^n \end{pmatrix}. \] Since \((D - I)^n = 0\), we can know that \(a_{ii}^n = 0\) for all \(i\). Since \(a_{ii}\) is in a field, we can know that \(a_{ii} = 0\) for all \(i\). Hence, we have \(D - I = 0\), which implies that \(D = I\). Hence, we have \(I + N = P^{-1}DP = P^{-1}IP = I\), which implies that \(N = 0\). \(\blacksquare\)

Question 3.

(a) Let \( A \) be a real \( n \times n \) matrix such that \( A^2 = 3A \). Prove that \( A \) is similar (over the real numbers) to a diagonal matrix whose diagonal entries are 3 or 0.

(b) Let \( A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \) (note that \( A^2 = 3A \)). Find a real matrix \( S \) so that \( S^{-1} A S \) has the form as in part (a).

Proof (a). Given that \(A^2 = 3A\), we can know that \(A^2 - 3A = 0\). Set \(f(x) = x^2 - 3x\), then we have \(f(A) = 0\). Hence, we can know that minimal polynomial of \(A\), \(\mu_A(x)\mid f(x)\). Since \(f(x) = x(x - 3)\), we can know that \(\mu_A(x)\) can be \(x(x - 3)\), \(x\) or \(x - 3\). If \(\mu_A(x) = x\), then we can know that \(A = 0\), which is a diagonal matrix with all diagonal entries are \(0\). If \(\mu_A(x) = x - 3\), then we can know that \(A = 3I\), which is a diagonal matrix with all diagonal entries are \(3\). If \(\mu_A(x) = x(x - 3)\), we can see that \(\gcd(x, x - 3) = 1\), which implies that \(\mathbb{R}^n = \ker(A) \oplus \ker(A - 3I)\). Hence, we can know that geometric multiplicity of each eigenvalue is equal to the algebraic multiplicity of each eigenvalue, which implies that \(A\) is diagonalizable. And the diagonal matrix is the form as in the question. \(\blacksquare\)

Solution (b). Given that \(A^2 = 3A\), and \(A\neq 0\) and \(A - 3I\neq 0\), we can know that the minimal polynomial of \(A\) is \(x(x - 3)\). Hence, \(A\) has eigenvalues \(0\) and \(3\). Hence, we need to find the eigenvectors of \(A\). When the eigenvalue is \(0\), let its eigenvector be \(v_1 = [a, b, c]^T\) and we will have \(A\cdot v_1 = 0\). Then we have \[ \begin{align} 2a + b + c &= 0,\\ a + 2b - c &= 0,\\ a - b + 2c &= 0. \end{align} \] Hence, we can find one eigenvector, which is \([1, -1, -1]^T\). When the eigenvalue is \(3\), let its eigenvector be \(v_2 = [d, e, f]^T\) and we will have \((A - 3I)\cdot v_2 = 0\). \[ \begin{align} -d + e + f &= 0,\\ d - e - f &= 0,\\ d - e - f &= 0. \end{align} \] Then, we have two eigenvectors \([1, 1, 0]^T\) and \([0, -1, 1]^T\). Suppose that \[ \begin{align} S &= \begin{bmatrix} 1 & 1 & 0\\ -1 & 1 & -1\\ -1 & 0 & 1 \end{bmatrix},\\ \det(S) &= 1\cdot 1\cdot 1 - 1\cdot ((-1)\cdot 1 - (-1)\cdot(- 1)) = 1 - 1\cdot (-2) = 3\neq 0. \end{align} \] Then, we have \[S^{-1}AS = \begin{bmatrix} 0 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 3 \end{bmatrix}. \]

Question 4. Let \( k \) be a field and \( R = k[X, Y] / I \) where \( I \) is the ideal \( (X^2, Y^2) \). Let \( x, y \) be the image of \( X, Y \) in \( R \).

(a) Prove that \( R \) has only one maximal ideal \(\mathfrak{m} = (x, y) \).

(b) Prove that any ideal of \( R \) strictly contained in \(\mathfrak{m} \) is principal.

Proof (a) (v1). Firstly, we know that if \(a(X, Y)\in (X^2, Y^2)\), we can have \[ a(X, Y) = X^2f(X, Y) + Y^2g(X, Y) = X(Xf(X, Y)) + Y(Yg(X, Y))\in (X, Y). \] Thus, we can know that \((X^2, Y^2)\subset (X, Y)\). Then, we can see that \(k[X, Y]/(X, Y)\) is isomorphic to \(k\), which is a field. It shows that \((X, Y)\) is a maximal ideal. Now, we want to show that any maximal ideal of \(k[x, y]\) containing \((X^2, Y^2)\) contains \(X\) and \(Y\) by contradiction. Suppose that \(M\) is a maximal ideal containing \((X^2, Y^2)\) but it does not contain \(X\) or \(Y\). Hence, we can know that \(M + (X)\) is an ideal strictly containing \(M\) and \(M + (Y)\) is an ideal strictly containing \(M\). Given that \(M\) is a maximal ideal, we can have \(M + (X) = M + (Y) = k[X, Y]\). In that case, we have \(m_1(X, Y)\in M\) and \(f(X, Y)\in k[X, Y]\) such that \[ \begin{align} m_1(X, Y) + f(X, Y)X &= 1, \\ m_1(X, Y) &= 1 - f(X, Y)X. \end{align} \] Since \(m_1(X, Y)\in M\), we have \(X\cdot m_1(X, Y) = X - X^2f(X, Y)\in M\). Given that \(X^2\in M\), we can know that \(X^2f(X, Y)\in M\). Thus, we have \[ X = X - X^2f(X, Y) + X^2f(X, Y)\in M, \] which is a contradiction. Hence, we can have \(x\in M\). For the similar reason, we can also show that \(y\in M\). Thus, for any maximal ideal of \(k[X, Y]\) containing \((X^2, Y^2)\) contains \((X, Y)\). And we already showed that \((X, Y)\) is a maximal ideal, we can know that \((X, Y)\) is the only maximal ideal containing \((X^2, Y^2)\). According to the \(\textbf{Correspondence Theorem of Rings}\), any maximal ideal containing \((X^2, Y^2)\) is in one-to-one correspondence with the maximal ideal of \(k[X, Y]/(X^2, Y^2)\). Hence, we can know that \(R\) has only one maximal ideal \((X, Y) + (X^2, Y^2) = (x, y)\). \(\blacksquare\)

Proof (a) (v2) Firstly, we want to show that \(\mathfrak{m} = (x, y)\) is a maximal ideal. Since \(R = k[X, Y]/(X^2, Y^2)\), if \(r\in R\), then \(r = f(X, Y) + (X^2, Y^2)\), where \(f(X, Y) = aX + bY + cXY + d\) where \(a, b, c, d\in k\). Hence, \(r = f(X, Y) + (X^2, Y^2) = aX + bY + cXY + d+ (X^2, Y^2) = aX + (X^2, Y^2) + bY + (X^2, Y^2) + cXY + (X^2, Y^2) + d + (X^2, Y^2)= ax + by + cxy + d\) since \(x, y\) are the images of \(X, Y\) in \(R\). Since \(\mathfrak{m} = (x, y)\), we have \(r/\mathfrak{m} = ax + by + cxy + d + (x, y) = d + \mathfrak{m}\). Then, we can know that \(R/\mathfrak{m} \cong k\), which is a field. Thus, \(\mathfrak{m}\) is a maximal ideal. Suppose that \(J\) is a proper ideal of \(R\). Then, we can know that \(J\) does not contain a unit of \(R\). If \(j\in J\), we want to show that \(j = ax + by + cxy\), where \(a, b, c\in k\) by contradiction. Suppose that \(j = ax + by + cxy + d\) where \(d\neq 0\). Then we can know that \[ j\cdot xy = dxy \in J. \] Since \(d\neq 0\), we have \(d^{-1}\in k\). Thus, \(dxy\cdot d^{-1} = xy\in J\). If \(xy\in J\) and \(ax + by + cxy + d\in J\), then \[ \begin{align} ax + by + cxy + d - c\cdot xy = ax + by + d\in J, \\ (ax + by + d)\cdot x = ax^2 + bxy + dx = bxy + dx\in J, \\ bxy + dx - b\cdot xy = dx\in J, \\ dx\cdot d^{-1} = x\in J. \end{align} \] Now, we have \(xy, x\in J\). Thus, we can know that \[ \begin{align} ax + by + cxy + d - ax - cxy = by + d\in J, \\ (by + d)\cdot y = by^2 + dy = dy\in J, \\ dy\cdot d^{-1} = y\in J. \end{align} \] Again, we have \(x, y, xy\in J\). Thus, \[ d = ax + by + cxy + d - ax - by - cxy \in J. \] Since \(d\neq 0\), \(d\) is a unit in \(J\), which is a contradiction since \(J\) does not contain a unit. Hence, we can know that if \(J\) is a proper ideal of \(R\), then \(j = ax + by + cxy\in (x, y)\). Hence, we can know that \(J\subset (x, y)\). We know that every proper ideal of \(R\) is contained in a maximal ideal of \(R\). Suppose that \(T\) is a maximal ideal of \(R\), then we can know that \(T\) is a proper ideal of \(R\). Hence, we can know that \(T\subset (x, y)\). Since \(T\) is a maximal ideal, the only proper ideal contains \(T\) is \(T\) itself. Thus, \(T = (x, y)\). Hence, we can know that \(R\) has only one maximal ideal \((x, y)\). \(\blacksquare\)

Question 5. Find the Galois group of \( x^3 + x + 2021 \) over \( \mathbb{Q} \). (2021 = 43 × 47)

Proof. Firstly, we want to show that the polynomial is irreducible over \( \mathbb{Q} \). Given that the degree of the polynomial is degree \(3\). Suppose that it is reducible, then it can only be factored into a polynomial of degree \(2\) and a polynomial of degree \(1\). If it can be factored as a polynomial of degree \(1\), it means it has a root in \( \mathbb{Q} \). Since we know that \(p(x)\) is an irreducible polynomial over \( \mathbb{Q} \) if and only if it is an irreducible polynomial over \( \mathbb{Z} \). For the other direction, we can know that if it is reducible over \( \mathbb{Q} \), then it is reducible over \( \mathbb{Z} \). Then we can know that it has a root in \( \mathbb{Z} \). Now, suppose that \(a\) is a root of the polynomial, then we have \[ \begin{align} a^3 + a + 2021 &= 0\\ a^3 + a &= -2021\\ a(a^2 + 1) &= -2021\\ a(a^2 + 1) &= -43 \times 47\\ \end{align} \] In \(\mathbb{Z}\), we can know that each number of prime factorization. Hence, we can know that \(a^2+1\) is either \(43\) or \(47\). Suppose that \(a^2 + 1 = 43\), then we have \(a^2 = 42\), which is not a perfect square. Hence, we have \(a^2 + 1 = 47\), then we have \(a^2 = 46\), which is not a perfect square. Hence, it is a contradiction. Therefore, we can know that the polynomial is irreducible over \( \mathbb{Z} \), which implies that it is irreducible over \( \mathbb{Q} \). Now, set \(f(x) = x^3 + x + 2021\), we take the derivative of \(f(x)\) and find that it is \(3x^2 + 1\). We try to see what the GCD of \(f(x)\) and \(f'(x)\) is. We will use the Euclidean algorithm to find the GCD. \[ \begin{align} f(x) &= f'(x)\left(\frac13x\right) + \left(\frac23 x + 2021\right) \\ f'(x) &= \left(\frac23 x + 2021\right)\left(\frac92x - \frac{54567}{4}\right) + \frac{110279911}{4}\\ \end{align} \] Then we can know that the GCD of \(f(x)\) and \(f'(x)\) is \(1\). Hence, we can know that the polynomial is separable. Once, we know that \(f(x) = x^3 + x + 2021\) is separable, the galois group of \(f(x)\) over \( \mathbb{Q} \) is the galois group of the splitting field of \(f(x)\) over \( \mathbb{Q} \). Since there are three distinct roots in the splitting field of \(f(x)\) over \( \mathbb{Q} \), and there will be an automorphism that maps one root to another root. Hence, there are \( 6\) automorphisms in the galois group of \(f(x)\) over \( \mathbb{Q} \).