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2. Let \( R \) be a unique factorization domain with quotient field \( K \). Write \( R[x] \) and \( K[x] \) respectively, for the polynomial rings over \( R \) and \( K \). Fix \( 0 \ne f(x) \in R[x] \). Let \( I \) denote the principal ideal \( f(x)R[x] \) and \( J \) denote the ideal \( f(x)K[x] \cap R[x] \). Prove that there exists \( 0 \ne a \in R \) such that \( I = aJ \), where \( aJ := \{ aj \mid j \in J \} \).
Proof. Firstly, we want to show that \(I\subset J\). Suppose that \(a\in I\), we can know that \(a\in f(x)K[x]\) since \(a\in f(x)R[x]\) and \(R[x]\subset K[x]\) for \(K\) is the quotient field of \(R\). Since \(I\) is an ideal of \(R[x]\), we can know that \(a\in R[x]\). Thus, we have \(a\in f(x)K[x]\) and \(a\in R[x]\), which implies that \(a\in J\). Hence, we have \(I\subset J\). Since \(K\) is a field, we can know that \(K[x]\) is a PID. And \(R[x]\) is UFD, for \(R\) is a UFD. Thus, we can know that \(J=(g(x))\) for some \(g(x)\in K[x]\). Since \(J\subset R[x]\), we can know that \(g(x)\in R[x]\) by contradiction. Since \(J\) is also a ideal of \(R[x]\), we can know that \(f(x)\mid g(x)\). Thus, we have \(g(x)=f(x)h(x)\) for some \(h(x)\in R[x]\). Since \(R[x]\) is a UFD, there exists only one \(h(x)\in R[x]\) such that \(g(x)h(x) = f(x)\). Since \[ I = \{f(x)r(x)\,|\, r(x)\in R[x]\} = \{g(x)h(x)r(x)\,|\, r(x)\in R[x]\} = \{h(x)(g(x)r(x))\,|\, r(x)\in R[x]\}, \] and \(g(x)r(x)\in J\), we have \(I\subset h(x)J\). For the other direction, we know that every \(j(x)\in J\), we have \(j(x) = g(x)r(x)\) for some \(r(x)\in R[x]\). Thus, we have \(h(x)\cdot g(x)r(x) = (g(x)h(x))r(x) = f(x)r(x) \in I\), which implies that \(h(x)J\subset I\). Hence, we can know that \(I = h(x)J\). Since we know that \(f(x)\neq 0\), we can know that \(h(x)\neq 0\) given that \(g(x)h(x) = f(x)\). Thus, we have \(I = h(x)J\neq 0\). Hence, we can know that there exists \(0\neq a\in R\) such that \(I = aJ\). \(\blacksquare\)
\(\textbf{Problem 6. }\) Consider the matrix \( A = \begin{bmatrix} 0 & 1 & 0 & 1 \\ -1 & 1 & 0 & 0 \\ -2 & 0 & -1 & -2 \\ 1 & -1 & 0 & 0 \\ \end{bmatrix} \), with entries in \( \mathbb{C} \). Find \( J \), the Jordan canonical form for \( A \) and an invertible matrix \( P \) such that \( J = P^{-1}AP \).
\(\textbf{Solution. }\)We firstly find the characteristic polynomial of \(A\). \[ \chi_A(x) = x^4 -x^2 = x^2(x+1)(x-1) = (x - 0)^2(x - 1)(x + 1). \] Now we try to determine the minimal polynomial of \(A\), which leaves us two possibilities: \(x^2(x-1)(x+1)\) and \(x(x-1)(x+1)\). We plug in \(A\) to \(x(x-1)(x+1)\) and find that it is not the zero matrix. Thus, we have the minimal polynomial is the same as the characteristic polynomial. Hence, we have the Jordan canonical form of \(A\) is \[ J = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix}. \] Now we calculate the eigenvectors of \(A\). Then we have the eigenvectors of \(A\) are \(v_1 = (0, 1, 1, -1)^T\) for eigenvalue \(1\), \(v_2 = (0, 0, 1, 0)^T\) for eigenvalue \(1\), \(v_3 = (1, 1, 0, -1)^T\) for eigenvalue \(0\). We need to find another column vector of \(P\). Suppose that it is \((a, b, c, d)^T\) such that \[ A \begin{bmatrix} a \\ b \\ c \\ d \\ \end{bmatrix} = v_3 = \begin{bmatrix} 1 \\ 1 \\ 0 \\ -1 \\ \end{bmatrix}. \] Then we have \((a, b, c, d)^T = (1, 2, 0, -1)^T\). Hence, we have a matrix such that \[ B = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 2 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ -1 & -1 & 0 & -1 \\ \end{bmatrix}. \] Then, we calculate the determinant of \(B\) and find that it is not zero. Hence, we have \(P = B\), where \(A = P^{-1}JP\).