KU 2021 (August) ALGEBRA QUALIFYING EXAM

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.

Firstly, we write \(\langle\begin{bmatrix} 4\\ 1\end{bmatrix}, \begin{bmatrix}6\\3\end{bmatrix}\rangle \) as a column space generated by \(A\) such that \[ A = \begin{bmatrix} 4 & 6\\ 1 & 3 \end{bmatrix} \] We first want to calculate the characteristic polynomial of the matrix. We have \[ \begin{align} \begin{vmatrix} 4 - \lambda & 6\\ 1 & 3 - \lambda \end{vmatrix} &= (4 - \lambda)(3 - \lambda) - 6\\ &= \lambda^2 - 7\lambda + 6\\ &= (\lambda - 1)(\lambda - 6). \end{align} \] Hence, we can see that the matrix has two distinct eigenvalues, which implies that the matrix is diagonalizable: there exists \(P\) such that \(P^{-1}AP = D\), where \(D\) is \[ \begin{bmatrix} 1 & 0\\ 0 & 6 \end{bmatrix}. \] Hence, we can know that the column space of \(A\) is isomorphic to the column space of \(D\). Thus, we can see that \(\langle (4, 1), (6, 3)\rangle \cong \langle (1, 0), (0, 6)\rangle \cong \mathbb{Z}\oplus \mathbb{Z}_6\). Hence, \[ \mathbb{Z}^2/\langle (4, 1), (6, 3)\rangle \cong \mathbb{Z}^2/\mathbb{Z}\oplus \mathbb{Z}_6 \cong \mathbb{Z}_6. \] Since \(\gcd(2, 3) = 1\) and \(2\cdot 3 = 6\). We have \[ \mathbb{Z}_6 \cong \mathbb{Z}_2\times \mathbb{Z}_3. \]

Proof (a). Given that \(A\) and \(B\) are two commutative subrings of \(R\) with unity and \(AB = \{ \sum a_i b_i \mid a_i \in A, b_i \in B \}\). We can know that \(0\in AB\) since \(0 = 0\cdot 0\) and \(1\in AB\) since \(1\cdot 1 = 1\). Then, we want to show that \(AB\) is commutative group under addition. Suppose that \(\alpha = \sum a_i b_i\) and \(\beta = \sum a_i' b_i'\). Given that \(a_i\in A\), we can know that \(-a_i\in A\) since \(A\) is a ring, thus we have \(-\alpha = -\sum a_ib_i = \sum (-a_i)b_i\in AB\). Since \(AB\subset R\), we can see that \(\alpha + \beta = \beta + \alpha\). For the similar reason (i.e. \(AB\subset R\)\), we can have \(\alpha + (\beta + \gamma) = (\alpha + \beta) + \gamma\) for any \(\alpha, \beta, \gamma\in AB\). Then, we want to show that \(AB\) is closed under multiplication. Suppose that \(\alpha = \sum a_i b_i\) and \(\beta = \sum a_i' b_i'\). Then, we have \(\alpha\beta = (\sum a_i b_i)\cdot( \sum a_i' b_i' )\) After that,

Proof (b). We will prove it with mathematical contradiction. Suppose that neither \(A\) nor \(B\) contains a polynomial whose lowest term has degree one and \(AB = \mathbb{C}[x]\). Suppose that \(a = a_nx^{n} + a_{n-1}x^{n-1} + \dots + a_2x^2 + a_0\) and \(b = b_mx^{m} + b_{m-1}x^{m-1} + \dots + b_2x^2 + b_0\). Then we can know that \[ a\cdot b = a_nb_mx^{n+m} + (a_nb_{m-1} + a_{n-1}b_m)x^{n+m-1} + \dots + a_2b_2x^4 + (a_2b_0 + a_0b_2)x^2 + a_0b_0. \] Given that \(AB = R\) for any \(f(x)\in \mathbb{C}[x]\) there exists \(g\in A, h\in B\) such that \(gh = f(x)\). In that case, if \(f(x)\) is a polynomial whose lowest term has degree one, then there exists \(g\in A\) or \(h\in B\) such that \[ gh = f(x) = a_nx^{n} + a_{n-1}x^{n-1} + \dots + a_2x^2 + a_0, \] which does not contain a polynomial where one of its terms has degree one. Therefore, it is a contradiction, and we can know that at least one of \(A\) or \(B\) contains a polynomial whose lowest term has degree one. \(\blacksquare\)

Solution (c). \

Solution. Firstly, we can know that \(x^4 + 1 = (x^2 - i)(x^2 + i)\). Since \(i = \cos(\pi/2) + \sin(\pi/2)i = e^{\pi i/2}\) and \(-i = \cos(-\pi/2) + \sin(-\pi/2)i = \cos(-\pi/2) + \sin(3\pi/2)i = e^{3\pi i/2}\) Hence, we can know that \(\pm(e^{3\pi i/2})^{1/2} =\pm e^{3\pi i/4} = \pm (-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i) \) and \(\pm(e^{3\pi i/2})^{1/2} =\pm e^{3\pi i/4} = \pm (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i)\) are the roots of \(x^2 + i\) and \(x^2 - i\), respectively. Then, we can know that \(\mathbb{Q}(-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i, \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i)\) is the splitting field of \(x^4 + 1\) over \(\mathbb{Q}\). And we want to show that \(\mathbb{Q}(-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i, \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i) = \mathbb{Q}(i, \sqrt{2})\). It is not hard to see that \(\mathbb{Q}(i, \sqrt{2})\subset \mathbb{Q}(-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i, \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i)\). For the other direction, we can know that \[ \begin{align} \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i + \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i &= \sqrt{2}\\ \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i &= \sqrt{2}i.\\ \dfrac{\sqrt{2}i}{\sqrt{2}} &= i. \end{align} \] Thus, we get \(\mathbb{Q}(-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i, \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i) \subset \mathbb{Q}(i, \sqrt{2})\). Hence, we can know that \(\mathbb{Q}(-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i, \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i) = \mathbb{Q}(i, \sqrt{2})\). Then, we need to know the degree of \(\mathbb{Q}(\sqrt{2}, i)\) over \(\mathbb{Q}\). Since \(\sqrt{2}\) is a root of \(x^2 - 2\), which is irreducible over \(\mathbb{Q}\). Thus, we have \(\mathbb{Q}(\sqrt{2})\cong \mathbb{Q}[x]/(x^2 - 2)\), which is a degree 2 extension of \(\mathbb{Q}\). Then, we can know that \(x^2 + 1\) is the irreducible polynomial of \(i\) over \(\mathbb{Q}(\sqrt{2})\). Thus, we have \(\mathbb{Q}(i, \sqrt{2})\cong \mathbb{Q}(\sqrt{2})[x]/(x^2 + 1)\), which is a degree 2 extension of \(\mathbb{Q}(\sqrt{2})\). Therefore, \[ [\mathbb{Q}(\sqrt{2}, i):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2}, i):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2 \times 2 = 4. \] Hence, the degree of the splitting field of \(x^4 + 1\) over \(\mathbb{Q}\) is 4. \(\blacksquare\)

Solution. However, it is not the case for any positive integer \(a\). For example, let \(a = 4\). Then, we can know that \(x^4 + 4\) is irreducible over \(\mathbb{Q}\). Then, we can know that \(x^4 + 4 = (x^2 - 2i)(x^2 + 2i)\). Since \(2i = 2\cos(\pi/2) + 2\sin(\pi/2)i = 2e^{\pi i/2}\) and \(-2i = 2\cos(-\pi/2) + 2\sin(-\pi/2)i = 2\sin(3\pi/2)i = 2e^{3\pi i/2}\). Hence, we can get that \(\pm(2e^{\pi i/2})^{1/2} =\pm \sqrt{2}e^{\pi i/4} = \pm \sqrt{2}(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i) = \pm (1 + i) \) and \(\pm(\sqrt{2}e^{3\pi i/2})^{1/2} =\pm \sqrt{2}e^{3\pi i/4} = \pm \sqrt{2}(-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i) =\pm( -1 + i)\) are the roots of \(x^2 + 2i\) and \(x^2 - 2i\), respectively. Then, we can know that \(\mathbb{Q}(1 + i, -1 + i)\) is the splitting field of \(x^4 + 4\) over \(\mathbb{Q}\). After that, we can show that \(\mathbb{Q}(1 + i, -1 + i) = \mathbb{Q}(i)\). Firstly, we have \(\mathbb{Q}(i)\subset \mathbb{Q}(1 + i, -1 + i)\). For the other direction, we can know that \[ \begin{align} -1 + i + 1 + i &= 2i\\ \frac{1}{2}\cdot 2i &= i. \\ \end{align} \] Hence, we can know that \(\mathbb{Q}(1 + i, -1 + i) = \mathbb{Q}(i)\). Then, we need to know the degree of \(\mathbb{Q}(i)\) over \(\mathbb{Q}\). Since \(i\) is a root of \(x^2 + 1\), which is irreducible over \(\mathbb{Q}\). Thus, \(x^2+1\) is the minimal polynomial of \(i\) over \(\mathbb{Q}\). Hence, we have \(\mathbb{Q}(i)\cong \mathbb{Q}[x]/(x^2 + 1)\), which is a degree 2 extension of \(\mathbb{Q}\). And it is not \(4\).

Proof (a). Suppose that \(A\) is diagonalizable, then there exists an invertible matrix \(P\) such that \(P^{-1}AP = D\), where \(D\) is a diagonal matrix. Then we have \[ \begin{align} p(A) &= p(PDP^{-1})\\ &= Pp(D)P^{-1}. \end{align} \] Since \(D\) is a diagonal matrix, we can know that \(p(D)\) is also a diagonal matrix. Hence, we have \(p(A)\) is diagonalizable.\(\blacksquare\)

Proof (b). Firstly, we want to show that \(x^r - x - 1\) is separable over the complex numbers. We took the derivative of \(x^r - x - 1\), we have \(rx^{r-1} - 1\). Then, we want to find the greatest common divisor of \(x^r - x - 1\) and \(rx^{r-1} - 1\) by using the division algorithm. \[ \begin{align} x^r - x - 1 &= (rx^{r-1} - 1)\left(\frac1r x\right) + \left(\frac{1-r}{r}\right)x - 1 \\ rx^{r-1} - 1 &= \left(\frac{1-r}{r}x - 1\right)\left(\frac{r^2}{1-r}x^{r-2} + \frac{r^3}{(1 - r)^2}x^{r-3} + \cdots + \frac{r^r}{(1 - r)^{r-1}}\right) + \frac{r^r}{1 - r} - 1 \end{align} \] Then, we can see that \(\gcd(x^r - x - 1, rx^{r-1} - 1) = 1\). Hence, we can know that \(x^r - x - 1\) is separable over the complex numbers. In other words, we can know that \(x^r - x - 1\) has \(r\) distinct roots in \(\mathbb{C}\). Now, Since \(f(A) = 0\), we denote that \(\mu_A(x)\) as the minimal polynomial of \(A\). Then, we can know that \(\mu_A(x)\) divides \(x^r - x - 1\), which implies that \(\mu_A(x)\) has distinct roots in \(\mathbb{C}\) as well. Hence, we can write \(\mu_A(x) = (x - \lambda_1)\cdots(x - \lambda_m)\), where each \(\lambda_i\) is the distinct eigenvalues of \(A\). Then, we will have \[ \mathbb{C}^n = \ker(A - \lambda_1I) \oplus \cdots \oplus \ker(A - \lambda_mI). \] In that case, we can know that the geometric multiplicity of each eigenvalue is equal to the algebraic multiplicity of each eigenvalue and \(A\) is diagonalizable. \(\blacksquare\)