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.
2. Let \(F\) be a field and \(f(x)\) an irreducible, separable polynomial over \(F\). Write \(E\) for the splitting field of \(f(x)\) over \(F\), and suppose that \(\alpha\) and \(\alpha + 1\) are roots of \(f(x)\).
(a) Prove that the characteristic of \(F\) is not zero.
(b) Prove that there exists a field \(L\) between \(F\) and \(E\) such that \([E : L]\) equals the characteristic of \(F\) .
Proof (a). We will prove it with contradiction. Suppose that \(F\) has characteristic \(0\). Since \(E\) ist he splitting field of \(f(x)\), which is irreducible, we can know that there exists an isomorphism \(\sigma:E\to E\) such that \(\sigma(\alpha) = \alpha + 1\) given that \(\alpha\) and \(\alpha + 1\) are two roots of \(f(x)\) and \(\sigma(a) = a\) for any \(a\in F\). Since, we can know that the \(\sigma\) sends the roots of \(f(x)\) to the roots of \(f(x)\), then we have \[ \begin{align} \sigma(\alpha + 1) &= \sigma(\alpha) + \alpha(1) = \alpha + 1 + 1 = \alpha + 2\\ \sigma(\alpha + 2) &= \sigma(\alpha) + \alpha(2) = \alpha + 1 + 2 = \alpha + 3\\ \vdots\\ \sigma(\alpha + n) &= \sigma(\alpha) + \alpha(n) = \alpha + 1 + n = \alpha + n+1\\ \vdots \\ \end{align} \] Given that the characteristic of \(F\) is \(0\), we can know that \(\alpha + n\) is a root of \(f(x)\) for any \(n\in \mathbb{N}\). In that case, \(f(x)\) has infinitely many roots, which is a contradiction. Hence, we can know that the characteristic of \(F\) is not \(0\). \(\blacksquare\)
Proof (b). Given that \(F\) does not have characteristic \(0\), we can know that the characteristic of \(F\) is \(p\), where \(p\) is a prime number. According to part (a), we can know that \(f(x)\) contains the following roots: \[ \alpha, \alpha + 1, \alpha + 2, \ldots, \alpha + p - 1. \] Fix the following the isomorphism we found in part (a), which is \(\sigma:E\to E\). Then we have \[ \sigma(\alpha) = \alpha + 1, \sigma(\alpha + 1) = \alpha + 2, \ldots, \sigma(\alpha + p - 1) = \alpha. \] In other words, we have \(\sigma^p(\alpha) = \alpha\). Hence, we can know that \(\sigma\in \text{Gal}(E/F)\) has order \(p\). Thus, we can know that \(\langle \sigma \rangle\) is a subgroup of \(\text{Gal}(E/F)\) with order \(p\). According to the \(\textbf{Galois Correspondence}\), we can know that there exists a field \(L\) between \(F\) and \(E\) such that \([E:L] = p\). \(\blacksquare\)
3. Let \(K\) be a field and R a commutative integral domain containing \(K\). Assume that \(R\) is a finite dimensional vector space over \(K\). Prove that \(R\) is a field.
Let \(r\) be any nonzero element in \(R\). And define the map \(\phi_r: R \to R\) by \(\phi_r(x) = rx\). Since \(\phi_r(x + y) = r(x + y) = rx + ry = \phi_r(x) + \phi_r(y)\) and \(\phi_r(k\cdot x) = rk\cdot x = k\cdot rx = k\phi_r(x)\), we can know that \(\phi_r\) is a \(K\)-linear transformation. According to \(\textbf{Fundamental Theorem of Linear Algebra}\), we have \(\dim V = \dim \text{Im}(\phi_r) + \dim \text{Ker}(\phi_r)\). Given that \(R\) is an integral domain, we can know that \(\ker(\phi_r) = \{0\}\) since \(r\neq 0\). Thus, we can know that \(\dim \ker(\phi_r) = 0\). Hence, we have \(\dim \text{Im}(\phi_r) = \dim R\). Given that \(\text{Im}(\phi_r)\) is a subspace of \(R\), we can know that \(\text{Im}(\phi_r) = R\). Hence, we can know that \(1\in \text{Im}(\phi_r)\). In other words, there exists an element \(s\in R\) such that \(rs = 1\). Thus, we can know that \(r\) is a unit in \(R\) if \(r\neq 0\). In other words, we can know that \(R\) is a field. \(\blacksquare\)
Problem 5. Let \( V \) be a finite dimensional vector space over the field \( F \) and \( T : V \to V \) is a linear transformation. Suppose that \( \chi_T(x) \), the characteristic polynomial of \( T \), factors as \( \chi_T(x) = p_1(x)^{f_1} \cdots p_r(x)^{f_r} \), where each \( p_i(x) \) is irreducible over \( F \). Prove that the minimal polynomial of \( T \) equals \( p_1(x)^{e_1} \cdots p_r(x)^{e_r} \), where for each \( 1 \leq i \leq r \), \( e_i \) is the first exponent for which the kernel of \( p_i(T)^{e_i} \) equals the kernel of \( p_i(T)^{e_i+1} \).
Proof. We will prove it with mathematical contradiction. Given that the characteristic polynomial of \( T \) is \( \chi_T(x) = p_1(x)^{f_1} \cdots p_r(x)^{f_r} \), where each \( p_i(x) \) is irreducible over \( F \). Then we have the minimal polynomial of \( T \) is \( p_1(x)^{e_1} \cdots p_r(x)^{e_r} \), where \( 1 \leq e_i \leq f_i \) for each \(i\). By the following theorem:
Theorem. Let \( V \) be a finite-dimensional vector space, \( T \) an operator on \( V \), and assume the minimal polynomial of \( T \) is \( \mu_T(x) = p_1(x)^{e_1} \cdots p_t(x)^{e_t} \), where the polynomials \( p_i(x) \) are irreducible and distinct. For each \( i \), let \[ V_i = V(p_i) = \{ v \in V \mid p_i(T)^{e_i}(v) = 0 \} = \text{Ker}(p_i(T)^{e_i}). \] Then each of the spaces \( V_i \) is \( T \)-invariant and \[ V = V_1 \oplus V_2 \oplus \cdots \oplus V_t. \]
Now, suppose that \(e_i\) is not the first exponent for which the kernel of \(p_i(T)^{e_i}\) equals the kernel of \(p_i(T)^{e_i+1}\). In other words, \(\ker(p_i(T)^{e_i-1}) = \ker(p_i(T)^{e_i})\) (to be continued...)6. Let \( V \) be a finite dimensional inner product space over \( \mathbb{C} \) and \( T: V \to V \) a linear transformation. Write \( T^* \) for the adjoint of \( T \) and let \( B \) denote an orthonormal basis for \( V \).
(a) Prove that the matrix of \( T^* \) with respect to \( B \) is the conjugate transpose of the matrix of \( T \) with respect to \( B \).
(b) Let \( A \) denote the matrix of \( T \) with respect to \( B \). What properties does \( A \) have if \( T \) is: (i) Hermitian or (ii) Unitary.
Proof (a). Let \( B = \{v_1, v_2, \ldots, v_n\} \) be an orthonormal basis for \( V \). Let \( A \) be the matrix of \( T \) with respect to \( B \) and \( A^* \) be the matrix of \( T^* \) with respect to \( B \). Then we have \[ [T(v_i)]_B = \begin{bmatrix} a_{1i} \\ a_{2i} \\ \vdots \\ a_{ni} \end{bmatrix} \quad \text{and} \quad [T^*(v_j)]_B = \begin{bmatrix} b_{1j} \\ b_{2i} \\ \vdots \\ b_{nj} \end{bmatrix} \] where \( a_{ij} \) and \( b_{ij} \) are the entries of \( A \) and \( A^* \) respectively. By the definition of adjoint, we have \[ \begin{align*} \langle T(v_i), v_j \rangle &= \langle v_i, T^*(v_j) \rangle \\ (a_{1i}v_1 + \dots + a_{ni}v_n)^*v_j &= v_i^*(\overline{b_{1j}}v_1 + \dots + \overline{b_{nj}}v_n) \\ a_{1i}v_1^*v_j + \dots + a_{ni}v_n^*v_j &= v_i^*\overline{b_{1j}}v_1 + \dots + v_i^*\overline{b_{nj}}v_n \\ a_{ji}v_j^*v_j &= \overline{b_{ij}}v_i^*v_i \\ a_{ji} &= \overline{b_{ij}} \end{align*} \] Hence, we have \( A^* = \overline{A}^T \). Therefore, the matrix of \( T^* \) with respect to \( B \) is the conjugate transpose of the matrix of \( T \) with respect to \( B \). \(\blacksquare\)
Solution (b). If \(T\) is Hermitian, then we have \(T^* = T\). Hence, the matrix of \(T\) is equal to the matrix of \(T^*\). Then, we can know that \(A = A^*\) and \(A\) is diagonalizable since \(AA^* = A^*A\), which implies that \(A\) is normal. Besides that, we can know that all eigenvalues of \(A\) is real and here is the proof: Let \(\lambda\) be an eigenvalue of \(A\) and \(v\) be an eigenvector of \(A\) corresponding to \(\lambda\). Then we have \[ \begin{align*} \langle Av, v \rangle &= \langle \lambda v, v \rangle = \lambda \langle v, v \rangle, \\ \langle v, Av \rangle &= \langle v, \lambda v \rangle = \overline{\lambda} \langle v, v \rangle, \\ \lambda \langle v, v \rangle &= \overline{\lambda} \langle v, v \rangle, (\lambda - \overline{\lambda})\langle v, v \rangle = 0. \end{align*} \] We know that \(v\neq 0\), which implies that \(\langle v, v\rangle >0\). Hence, we can know that \(\lambda - \overline{\lambda} = 0\), which implies that \(\lambda\) is real.