Let \(F\) be a field and let \(V\) and \(W\) be vector spaces over \(F\). Make \(V\) and \(W\) into \(F[x]\)-modules via linear operators \(T \in \operatorname{End}_F(V)\) and \(S \in \operatorname{End}_F(W)\) by defining \(x \cdot v = T(v)\) for all \(v \in V\) and \(x \cdot w = S(w)\) for all \(w \in W\). Denote the resulting \(F[x]\)-modules by \(V_T\) and \(W_S\) respectively.
Show that an \(F[x]\)-module homomorphism from \(V_T\) to \(W_S\) consists of an \(F\)-linear transformation \(R: V \to W\) such that \(R T = S R\).
Show that \(V_T \cong W_S\) as \(F[x]\)-modules if and only if there is an \(F\)-linear isomorphism \(P: V \to W\) such that \(T = P^{-1} S P\).
Recall that a module \(M\) is simple (or irreducible) if \(M \ne \{0\}\) and its only submodules are \(\{0\}\) and \(M\). Suppose that \(\dim_F V = 2\). Give an example of \(F\) and \(T\) with \(V_T\) simple.
Assume \(F\) is algebraically closed. Prove that if \(\dim_F V = 2\), then \(V_T\) is never simple.
Solution
Goal: Characterize module homomorphisms and isomorphisms for polynomial modules \(V_T\), and analyze simplicity via invariant subspaces and eigenvalues.
1. Part (a): \(F[x]\)-module homomorphisms correspond to intertwining linear maps (\(R T = S R\)).
1.1. Forward direction (\(\implies\)):- Let \(\varphi: V_T \to W_S\) be an \(F[x]\)-module homomorphism.- Since \(F \subset F[x]\), \(\varphi(c v) = c \varphi(v)\) for all \(c \in F\) and \(\varphi(v_1 + v_2) = \varphi(v_1) + \varphi(v_2)\), so \(\varphi\) is an \(F\)-linear transformation \(R: V \to W\).- Module compatibility on \(x\) gives \(\varphi(x \cdot v) = x \cdot \varphi(v)\).- Since \(x \cdot v = T(v)\) and \(x \cdot \varphi(v) = S(\varphi(v))\), this means \(R(T(v)) = S(R(v))\) for all \(v \in V\).- Thus \(R T = S R\).
1.2. Reverse direction (\(\impliedby\)):- Let \(R: V \to W\) be an \(F\)-linear transformation such that \(R T = S R\).- By induction, \(R T^k = S^k R\) for all \(k \ge 0\).- Let \(p(x) = \sum_{k=0}^d a_k x^k \in F[x]\) be any polynomial.- For any \(v \in V\):\[R(p(x) \cdot v) = R\left( \sum_{k=0}^d a_k T^k(v) \right) = \sum_{k=0}^d a_k R(T^k(v)) = \sum_{k=0}^d a_k S^k(R(v)) = p(x) \cdot R(v).\]- Together with additivity, \(R\) is an \(F[x]\)-module homomorphism.
2. Part (b): \(V_T \cong W_S \iff T = P^{-1} S P\) for an \(F\)-linear isomorphism \(P\).
2.1. An \(F[x]\)-module isomorphism is a bijective \(F[x]\)-module homomorphism.
2.2. By Part (a), a map \(P: V_T \to W_S\) is an \(F[x]\)-module homomorphism if and only if \(P: V \to W\) is \(F\)-linear and \(P T = S P\).
2.3. A map between vector spaces is a bijective module homomorphism if and only if it is an \(F\)-linear isomorphism whose inverse is also compatible.
2.4. If \(P\) is invertible, multiplying \(P T = S P\) on the left by \(P^{-1}\) gives \(T = P^{-1} S P\).
2.5. Conversely, if \(P: V \to W\) is an \(F\)-linear isomorphism with \(T = P^{-1} S P\), then \(P T = S P\), so \(P\) is an \(F[x]\)-module isomorphism.
3. Part (c): Example of a simple \(V_T\) of dimension 2 over \(\mathbb{R}\).
3.1. Let \(F = \mathbb{R}\), \(V = \mathbb{R}^2\), and let \(T: \mathbb{R}^2 \to \mathbb{R}^2\) be given by the matrix\[T = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.\]
3.2. Submodules of \(V_T\) are precisely the \(T\)-invariant \(\mathbb{R}\)-subspaces of \(\mathbb{R}^2\).
3.3. If \(V_T\) had a proper non-trivial submodule \(U\), then \(\dim_{\mathbb{R}} U = 1\).
3.4. A 1-dimensional \(T\)-invariant subspace is spanned by an eigenvector of \(T\).
3.5. The characteristic polynomial of \(T\) is\[\chi_T(x) = \det(x I - T) = \det \begin{pmatrix} x & 1 \\ -1 & x \end{pmatrix} = x^2 + 1.\]
3.6. Since \(x^2 + 1\) has no real roots, \(T\) has no real eigenvalues, and therefore no 1-dimensional \(T\)-invariant subspaces.
3.7. Thus the only \(T\)-invariant subspaces of \(\mathbb{R}^2\) are \(\{0\}\) and \(\mathbb{R}^2\), so \(V_T\) is simple.
4. Part (d): Over an algebraically closed field \(F\), \(V_T\) is never simple when \(\dim_F V = 2\).
4.1. Let \(F\) be algebraically closed and \(\dim_F V = 2\).
4.2. The characteristic polynomial \(\chi_T(x) = \det(x I - T) \in F[x]\) has degree 2.
4.3. Because \(F\) is algebraically closed, \(\chi_T(x)\) has at least one root \(\lambda \in F\).
4.4. The root \(\lambda\) is an eigenvalue of \(T\), so there exists a non-zero eigenvector \(v \in V \setminus \{0\}\) such that \(T(v) = \lambda v\).
4.5. Define the 1-dimensional subspace \(U = \operatorname{span}_F(v) \subset V\).
4.6. For any \(u = c v \in U\) (\(c \in F\)), \(T(u) = c T(v) = c \lambda v = (\lambda c) v \in U\).
4.7. Thus \(U\) is a \(T\)-invariant subspace, hence an \(F[x]\)-submodule of \(V_T\).
4.8. Since \(\dim_F U = 1\) and \(\dim_F V = 2\), \(\{0\} \subsetneq U \subsetneq V_T\), so \(U\) is a proper non-trivial submodule.
4.9. Therefore \(V_T\) is not simple.
5. Conclusion:
$F[x]$-homomorphisms intertwine operators, isomorphisms correspond to similarity, $90^\circ$ rotation on $\mathbb{R}^2$ gives a simple module, and over algebraically closed fields existence of eigenvalues prevents simplicity in dimension 2.