Brian Bi

Problem 3.8.3 I found this problem rather difficult and was not able to solve it by myself. However, the proof of this result (known as Fitting's lemma) is quite intuitive. Although the proof given in the text only covers the case of algebraically closed fields (for which it's possible to write $$W$$ as a direct sum of generalized eigenspaces), the basic idea of the proof carries over to the general case. Instead of considering generalized eigenspaces with eigenvalue zero, we consider the subspace of $$W$$ which is eventually annhilated by $$\theta$$ (these concepts coincide in the case of an algebraically closed field). Instead of considering generalized eigenspaces with nonzero eigenvalue, we consider the subspace of $$W$$ which is eventually reachable under iteration of $$\theta$$ (again they coincide for an algebraically closed field since a Jordan block with $$\lambda \neq 0$$ is invertible).

Thus the proof is as follows. Let $$n = \dim W$$. We claim that $$W$$ is the direct sum $$\ker \theta^n \oplus \im \theta^n$$, from which the desired result will follow.

To see this, first observe that $$\ker \theta \subseteq \ker \theta^2 \subseteq \ldots \subseteq W$$. If an inclusion in this sequence is strict, the dimension must increase by at least 1. If one of the inclusions in this sequence is an equality, then all subsequent inclusions must be equalities as well. So there are at most $$n$$ strict inclusions, implying $$\ker \theta^n = \ker \theta^{n+1} = \ldots$$. Likewise, since $$W \supseteq \im \theta \supseteq \im \theta^2 \supseteq \ldots$$, we can similarly conclude that $$\im \theta^n = \im \theta^{n+1} = \ldots$$.

Now let $$w \in W$$ be given. We want to write $$w$$ in the form $$w = u + \theta^n(v)$$, where $$u \in \ker \theta^n$$. If such $$u, v \in W$$ did in fact exist, then then it would necessarily be the case that $$\theta^n(w) = \theta^{2n}(v)$$. In fact, it suffices to simply choose $$v$$ such that this equation holds (which is possible since $$\im \theta^{2n} = \im \theta^n$$) and put $$u = w - \theta^n(v)$$. For then $$\theta^n(u) = \theta^n(w) - \theta^{2n}(v) = 0$$ by definition of $$v$$. So indeed $$w \in \ker \theta^n + \im \theta^n$$.

Now suppose $$t \in \ker \theta^n \cap \im \theta^n$$. Pick $$s \in W$$ such that $$\theta^n(s) = t$$. Then $$\theta^{2n}(s) = 0$$. But $$\ker \theta^{2n} = \ker \theta^n$$, so $$\theta^n(s) = 0$$, that is, $$t = 0$$. So in fact $$W = \ker \theta^n \oplus \im \theta^n$$ as vector spaces.

It is evident that $$\ker \theta^n$$ is a subrepresentation since for all $$w \in W, a \in A$$, we have $$\theta^n(av) = a\theta^n(v) = 0$$. Similarly $$\im \theta^n$$ is also a subrepresentation, since if $$w = \theta^n(v)$$, then $$aw = a\theta^n(v) = \theta^n(av)$$. Thus the claim that $$W = \ker \theta^n \oplus \im \theta^n$$ as representations is proven.

At this point we finally use the fact that $$W$$ is indecomposable, which implies that either $$\ker \theta^n = 0$$ or $$\im \theta^n = 0$$. If the former, then since $$\ker \theta \subseteq \ker \theta^n$$, it follows that $$\theta$$ is an automorphism. If the latter, then $$\theta$$ is nilpotent.

Problem 3.8.4 The solution is described in a math.SE post. Here are the pieces in order:

1. Let $$V$$ be a vector space over a field $$K$$. Let $$L$$ be an extension field of $$K$$. If $$B$$ is a $$K$$-basis for $$V$$, then $$B' = \{b \otimes_K 1 \mid b \in B\}$$ is an $$L$$-basis for $$V \otimes_K L$$.

Since the $$K$$-span of $$B$$ is already all of $$V$$, the $$L$$-span of $$B'$$ certainly contains all vectors of the form $$v \otimes_K 1$$. But the $$L$$-span of this latter set is all of $$V \otimes L$$, so $$B'$$ is enough to generate $$V \otimes L$$. Now, suppose $$0 = \sum_{i=1}^n c_i (b_i \otimes_K 1)$$ where the $$c_i$$'s are taken from $$L$$ and the $$b_i$$'s are distinct elements of $$B$$. Then we can rewrite this as $$0 = \sum_{i=1}^n b_i \otimes_K c_i$$. But as the $$b_i$$'s are linearly independent over $$K$$, this can only hold if all $$c_i$$'s vanish, so $$B'$$ is also linearly independent over $$L$$. This establishes that $$B'$$ is an $$L$$-basis for $$V \otimes_K L$$.

First, we reduce to the finitely generated case. Let $$V, W, A, K, L$$ be given as in the problem. Let $$\varphi : V \otimes_K L \to W \otimes_K L$$ be an isomorphism of $$A \otimes_K L$$-algebras. As $$L$$ is a subalgebra of $$A \otimes_K L$$, this implies that $$\varphi$$ induces an isomorphism between $$V \otimes_K L$$ and $$W \otimes_K L$$ as $$L$$-modules, or equivalently $$L$$-vector spaces as $$L$$ is a field. According to the Lemma, the domain and codomain both have finite $$L$$-bases $$\{v_1 \otimes_K 1, \ldots, v_n \otimes_K 1\}$$ and $$\{w_1 \otimes_K 1, \ldots, w_n \otimes_K 1\}$$ where $$v_1, \ldots, v_n$$ and $$w_1, \ldots, w_n$$ are respectively $$K$$-bases of $$V$$ and $$W$$. The isomorphism $$\varphi$$ can be written as a matrix $$X$$ with entries in $$L$$ with respect to these two bases, so that $\varphi\left[\sum_{i=1}^n c_i (v_i \otimes_K 1)\right] = \sum_{j=1}^n c_i X_{ij} (w_j \otimes_K 1)$ Let $$M$$ be the subfield of $$L$$ generated by $$K$$ and the elements of $$X$$, that is, $$M = K(X_{11}, X_{12}, \ldots, X_{nn})$$. Denote the restriction of the map $$\varphi$$ to $$V \otimes_K M$$ by $$\varphi'$$. Now

• Whenever $$m \in M, w \in W$$, we have that $$m(w \otimes_K 1) = w \otimes_K m \in W \otimes_K M$$. This implies that the map $$\varphi'$$ is an $$M$$-linear homomorphism to $$W \otimes_K M$$;
• $$V \otimes_K M$$ and $$W \otimes_K M$$ are $$A \otimes_K M$$-modules;
• Since $$A \otimes_K M$$ acts on $$V \otimes_K M$$ and $$W \otimes_K M$$ as the restrictions of the actions of $$A \otimes_K L$$ on $$V \otimes_K L$$ and $$W \otimes_K L$$, respectively, it follows that $$\varphi'$$ is also a homomorphism of representations;
• As $$\varphi'$$ is described by the same matrix $$A$$ as $$\varphi$$, and $$A$$ is invertible over $$L$$, it is also invertible over $$M$$, implying that $$\varphi'$$ is an isomorphism.

We therefore have that $$V \otimes_K M$$ and $$W \otimes_K M$$ are isomorphic as $$A \otimes_K M$$-modules and that $$M$$ is finitely generated over $$K$$. Therefore, it suffices to establish the desired result for finitely generated field extensions.

(The reduction from the finitely generated case to the finite case is currently missing. Feel free to email me to contribute a solution.)

Suppose now that $$L/K$$ is finite with degree $$n$$. $$V \otimes_K L$$ has a natural $$A$$-module structure inherited from $$V$$, namely, such that $$a(v \otimes_k \ell) = (av) \otimes_k \ell$$ and so on, which is compatible with the $$A \otimes_K L$$-module structure when $$A$$ is identified as a subalgebra of $$A \otimes_K L$$ in the obvious way. Let $$\ell_1, \ldots, \ell_n$$ be a basis for $$L$$ over $$K$$. Define $$\gamma : V \otimes_K L \to V^n$$ by $\gamma(v \otimes_k \ell_i) = (0, 0, \ldots, v, \ldots, 0, 0)$ where $$v$$ is the $$i$$th entry of the RHS. This is an isomorphism of the $$K$$-vector spaces $$V \otimes_K L$$ and $$V^n$$, and $\gamma(a(v \otimes_k \ell_i)) = \gamma((av) \otimes_k \ell_i) = (0, 0, \ldots, av, \ldots, 0, 0)$ showing that $$\gamma$$ is an isomorphism of $$A$$-modules. Since we are given that $$V \otimes_K L$$ and $$W \otimes_K L$$ are isomorphic as $$A \otimes_K L$$-modules, they are also isomorphic as $$A$$-modules, and since $$V \otimes_K L$$ and $$W \otimes_K L$$ are respectively isomorphic to $$V^n$$ and $$W^n$$ as $$A$$-modules, it follows that $$V^n$$ and $$W^n$$ are isomorphic as $$A$$-modules.

Write $$V = V_1 \oplus \ldots \oplus V_p$$ and $$W = W_1 \oplus \ldots \oplus W_q$$ where the summands are indecomposable. Then $$V^n = V_1^n \oplus \ldots \oplus V_p^n$$ and $$W^n = W_1^n \oplus \ldots \oplus W_q^n$$. By the Krull–Schmidt theorem, these decompositions are unique. Since $$V^n \cong W^n$$, this implies that $$V_1$$ is isomorphic to some $$W_i$$, so the $$n$$ copies of $$V_1$$ in $$V$$ can be identified with the $$n$$ copies of $$W_i$$ in $$W$$. Then the remaining summands in $$V$$ are isomorphic to the remaining summands in $$W$$ in some order, so $$V_2 \cong W_j$$ for some $$j \ne i$$, and so on. This establishes that $$p = q$$ and that there exists a permutation $$\sigma$$ such that $$V_i \cong W_{\sigma(i)}$$ for each $$i$$, so that $$V \cong W$$ as desired.

2. Here $$W \otimes_K L \cong V \otimes_K L \oplus Y$$, but we can do the reduction from arbitrary $$L/K$$ to finitely generated $$L/K$$ in a similar way. If $$\dim_K W = m$$ and $$\dim_K V = n$$, then $$\dim_L W \otimes_K L = m$$ and $$\dim_L V \otimes_K L = n$$, which implies $$\dim_L Y = m - n$$. Thus,

• If $$\{w_1, \ldots, w_m\}$$ is a $$K$$-basis for $$W$$, then an $$L$$-basis for $$W \otimes_K L$$ is given by $$\{w_1 \otimes_K 1, \ldots, w_m \otimes_K 1\}$$ as in part (i)
• If $$\{v_1, \ldots, v_n\}$$ is a $$K$$-basis for $$V$$ and $$\{y_1, \ldots, y_{m-n}\}$$ is an $$L$$-basis for $$Y$$, then an $$L$$-basis for $$V \otimes_K L \oplus Y$$ is given by $$\{(v_1 \otimes_K 1, 0), \ldots, (v_n \otimes_K 1, 0), (0, y_1), \ldots, (0, y_{m-n})\}$$

Using these bases we can write the isomorphism $$\varphi : W \otimes_K L \to V \otimes_K L \oplus Y$$ as an $$m$$ by $$m$$ matrix $$X$$ and we can again form the finitely generated field extension $$M/K$$ generated by the entries of this matrix. The matrix $$X$$ has an inverse $$X^{-1}$$ whose entries also all lie in $$M$$. Letting $$\varphi'$$ be the restriction of $$\varphi$$ to the domain $$W \otimes_K M$$, we find as in part (i) that the image is contained within $$V \otimes_K M \oplus Y$$.

To study the form that $$\im \varphi'$$ takes, let $$v \in V$$ be arbitrary. Write $$v = k_1 v_1 + \ldots + k_n v_n$$ where $$k_1, \ldots, k_n \in K$$. Let $$\mu \in M$$. Then, with respect to the basis of $$V \otimes_K L \oplus Y$$ given above, $$(v \otimes_K \mu, 0)$$ can be written in coordinates as $$(\mu k_1, \ldots, \mu k_n, 0, \ldots, 0)$$. As all entries of this coordinate vector lie in $$M$$, this implies that $$\varphi^{-1}((v \otimes_K \mu, 0)) = X^{-1}(\mu k_1, \ldots, \mu k_n, 0, \ldots, 0)$$ lies in $$W \otimes_K M$$. As this holds for all pure tensors $$v \otimes_K \mu$$, by linearity, the image of $$\varphi'$$ contains $$(t, 0)$$ for all $$t \in V \otimes_K M$$. This implies that $$\im \varphi'$$, in fact, takes the form $$V \otimes_K M \oplus Y'$$ where $$Y'$$ is a subspace of $$Y$$. As $$A \otimes_K M$$ is a subalgebra of $$A \otimes_K L$$, the map $$\varphi'$$ is compatible with the $$A \otimes_K M$$-module structures of $$W \otimes_K M$$ and its image $$V \otimes_K M \oplus Y'$$, and $$Y'$$ has the $$A \otimes_K M$$-module structure inherited from that of $$V \otimes_K M \oplus Y'$$ by projection. So the hypotheses of the theorem hold for $$M$$, a finitely generated extension of $$K$$.

(Again the reduction from the finitely generated case to the finite case is missing.)

As in part (i), assume $$L/K$$ is a finite extension with degree $$n$$. The $$A$$-module isomorphism between $$V \otimes_K L$$ and $$V^n$$, and between $$W \otimes_K L$$ and $$W^n$$, imply that $$V^n \oplus Y \cong W^n$$ as $$A$$-modules. Write $$V = V_1 \oplus \ldots \oplus V_p, W = W_1 \oplus \ldots \oplus W_q, Y = Y_1 \oplus \ldots \oplus Y_r$$ where the summands are indecomposable (here $$Y$$ is regarded as an $$A$$-module), so that by the Krull–Schmidt theorem, the summands on the right-hand sides of the following two equations are the same up to isomorphism and reordering: \begin{align*} V \otimes_K L \oplus Y &\cong V_1^n \oplus \ldots \oplus V_p^n \oplus Y_1 \oplus \ldots \oplus Y_r \\ W \otimes_K L &\cong W_1^n \oplus \ldots \oplus W_q^n \end{align*} Using similar arguments as in part (i), we find that $$r = n(q - p)$$, that the sum $$Y_1 \oplus \ldots \oplus Y_r$$ can be grouped as $$Y_1^n \oplus \ldots Y_{q-p}^n$$, and that there is a permutation of $$\{W_1, \ldots, W_q\}$$ such that each element is isomorphic to a corresponding element in $$\{V_1, \ldots, V_p, Y_1, \ldots, Y_{q-p}\}$$. Therefore, $$W_1 \oplus \ldots \oplus W_q \cong V_1 \oplus \ldots \oplus V_p \oplus Y_1 \oplus \ldots \oplus Y_{q-p}$$, or $$W \cong V \oplus Y'$$ as desired (where $$Y' \cong Y_1 \oplus \ldots \oplus Y_{q-p}$$).

Problem 3.8.5

1. Suppose $$A$$ is the internal direct sum of submodules $$A_1, A_2$$. Let $$a_1 \in A_1, a_2 \in A_2$$. Then $$a_1 a_2 \in A_2$$ and $$a_2 a_1 \in A_1$$, but $$a_1 a_2 = a_2 a_1$$ and $$A_1 \cap A_2 = \{0\}$$, so $$a_1 a_2 = 0$$. The constant function 1 belongs to $$A$$, so write $$1 = f + g$$ where $$f \in A_1, g \in A_2$$. Since $$fg = 0$$, it follows that for all $$x \in \mathbb{R}$$, we have either $$f(x) = 0$$ and $$g(x) = 1$$, or $$f(x) = 1$$ and $$g(x) = 0$$. Since $$f, g$$ are continuous, it follows that one of $$f$$ is the constant function 1 while the other is identically 0. Without loss of generality, suppose $$f = 1$$. Since $$1 \in A_1$$ and $$A_1$$ is a submodule, this easily implies $$A_1 = A$$, hence $$A_2 = \{0\}$$. Therefore $$A$$ is indecomposable.

A proof that $$M$$ is indecomposable can be found here.

2. $$A$$ contains a cyclic vector (for example the constant function 1) but the intermediate value theorem implies that every element of $$M$$ vanishes at some point, so no element of $$M$$ can be cyclic. Therefore $$A \not\cong M$$.

An explicit isomorphism between $$A \oplus A$$ and $$M \oplus M$$ is given here.