Looking for Abstract Proofs of Nakayama Lemma/Hamilton-Cayley Theorem

Question

I am looking for a very abstract, better if purely categorical, proof of the Nakayama lemma or the Hamilton-Cayley theorem (for modules), do you know any?

Answer 1

Here's the variation of the proof of the Cayley Hamilton theorem that pleases me the most. I came up with it after pondering the proof in Atiyah & Macdonald

Intuition For any endomorphism $\Phi$, we have a factorization of the determinant $\text{det}(\Phi)$ into the adjugate and the matrix itself: $$ \text{det}(\Phi) = \text{adj}(\Phi) \Phi$$ We want to use this to get a factorization of the characteristic polynomial $p(t)$ of $\phi$ into some polynomial analogous to the adjugate and a linear term $t - \phi $: $$p(t) = f(t)(t - \phi)$$ These two factorizations are analogous, and in fact, if we get the formality right, we can view these as corresponding factorizations in isomorphic rings.

The first question is then, "what are the two isomorphic rings I mentioned in which these are corresponding factorizations?"

Let $V$ be a finite dimensional vector space over a field $k$. One of the rings is $\text{End}_k(V)[t]$. The characteristic polynomial $p(t)$ of $\phi \in \text{End}_k (V)$ naturally lives in $\text{End}_k(V)[t]$. To see this, we view $t \text{Id}_V - \phi $ as having endomorphisms as coefficients, and then take the determinant, which is then in $\text{End}_k (V)[t]$. The other ring is $\text{End}_{k[t]}(V \otimes_k k[t])$. $\Phi := 1 \otimes t - \phi \otimes 1 $ is an element in this ring, and we have a factorization $\text{det}(\Phi) 1_{V \otimes_k k[t]} = \text{adj}(\Phi) \Phi$.

In the isomorphism $$\text{End}_k ( V \otimes_k k[t]) \cong \text{End}_k (V)[t]$$ We have corresponding elements $$\Phi \leftrightarrow t - \phi$$ and $$\text{det}(\Phi) \leftrightarrow p(t)$$ Therefore, the factorization $\text{det}(\Phi) 1_{V \otimes_k k[t]} = \text{adj}(\Phi) \Phi$ corresponds to a factorization $p(t) = f(t)(t-\phi)$ in $\text{End}_k (V) [t]$. And that's the whole idea!

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Now we have everything in place to make this formal.

Theorem: Let $V$ be a finitely generated $k$-module. If $\phi : V \rightarrow V$ is a $k$-linear map, then the evaluation homomorphism $\text{ev}_{\phi} : k[t] \rightarrow \text{End}_k (V)$ sends the characteristic polynomial $\text{char}(\phi)$ to $0$.

Proof: Let's start by constructing an isomorphism $F : \text{End}_{k} (V)[t] \rightarrow \text{End}_{k[t]} (V \otimes_k k[t])$ as follows. We have isomorphisms $$\text{End}_{k[t]} (V \otimes_k k[t]) \cong \text{Hom}_k (V, \text{Hom}_{k[t]}(k[t], V \otimes_k k[t])) \cong \text{Hom}_k(V, V \otimes_k k[t])$$

These isomorphisms can be established by creating canonical maps in both directions and showing that they are inverse to each other. Now we have a canonical map in a single direction,

$$\text{End}_k (V) \otimes_k k[t] \rightarrow \text{Hom}_k(V, V \otimes_k k[t])$$

sending $\phi \otimes t^n$ to the map sending $v$ to $\phi(v)t^n$. This is injective, and surjective since $V$ is finitely generated. Composing these isomorphisms gives an isomorphism $F : \text{End}_{k} (V)[t] \rightarrow \text{End}_{k[t]} (V \otimes_k k[t])$.

Now we argue as before. View $t - \phi$ as a $k[t]$-linear endomorphism of $V \otimes_k k[t]$. Under the isomorphism $F$, $\text{char}(\phi)$ maps to $\text{det} (t - \phi) 1_{V \otimes_k k[t]} )$ and $F ( t - \phi ) = t - \phi$. $t - \phi$ divides $\text{det}(t - \phi) 1_{V \otimes_k k[t]}$ in $\text{End}_{k[t]} (V \otimes_k k[t])$, since $\text{det} (t - \phi) 1_{V \otimes_k k[t]} = \text{adj}(t - \phi) (t - \phi)$, where $\text{adj}(t - \phi)$ is the adjugate matrix. Therefore, $t - \phi$ divides $\text{char}(\phi)$ in $\text{End}_{k}(V)[t]$. So $\text{char}(\phi)$ has $\phi$ as a root in $\text{End}_k(V)$, so that the evaluation homomorphism $\text{ev}_{\phi} : k[t] \rightarrow \text{End}_k (V)$ sends the characteristic polynomial $\text{char}(\phi)$ to $0$.