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93.4 Finite projective modules

This section is just a warmup. Of course finite projective modules should not have any “moduli”.

Example 93.4.1 (Finite projective modules). Let $\mathcal{F}$ be the category defined as follows

  1. an object is a pair $(A, M)$ consisting of an object $A$ of $\mathcal{C}_\Lambda $ and a finite projective $A$-module $M$, and

  2. a morphism $(f, g) : (B, N) \to (A, M)$ consists of a morphism $f : B \to A$ in $\mathcal{C}_\Lambda $ together with a map $g : N \to M$ which is $f$-linear and induces an isomorpism $N \otimes _{B, f} A \cong M$.

The functor $p : \mathcal{F} \to \mathcal{C}_\Lambda $ sends $(A, M)$ to $A$ and $(f, g)$ to $f$. It is clear that $p$ is cofibred in groupoids. Given a finite dimensional $k$-vector space $V$, let $x_0 = (k, V)$ be the corresponding object of $\mathcal{F}(k)$. We set

\[ \mathcal{D}\! \mathit{ef}_ V = \mathcal{F}_{x_0} \]

Since every finite projective module over a local ring is finite free (Algebra, Lemma 10.78.2) we see that

\[ \begin{matrix} \text{isomorphism classes} \\ \text{of objects of }\mathcal{F}(A) \end{matrix} = \coprod \nolimits _{n \geq 0} \{ *\} \]

Although this means that the deformation theory of $\mathcal{F}$ is essentially trivial, we still work through the steps outlined in Section 93.3 to provide an easy example.

Lemma 93.4.2. Example 93.4.1 satisfies the Rim-Schlessinger condition (RS). In particular, $\mathcal{D}\! \mathit{ef}_ V$ is a deformation category for any finite dimensional vector space $V$ over $k$.

Proof. Let $A_1 \to A$ and $A_2 \to A$ be morphisms of $\mathcal{C}_\Lambda $. Assume $A_2 \to A$ is surjective. According to Formal Deformation Theory, Lemma 90.16.4 it suffices to show that the functor $\mathcal{F}(A_1 \times _ A A_2) \to \mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)$ is an equivalence of categories.

Thus we have to show that the category of finite projective modules over $A_1 \times _ A A_2$ is equivalent to the fibre product of the categories of finite projective modules over $A_1$ and $A_2$ over the category of finite projective modules over $A$. This is a special case of More on Algebra, Lemma 15.6.9. We recall that the inverse functor sends the triple $(M_1, M_2, \varphi )$ where $M_1$ is a finite projective $A_1$-module, $M_2$ is a finite projective $A_2$-module, and $\varphi : M_1 \otimes _{A_1} A \to M_2 \otimes _{A_2} A$ is an isomorphism of $A$-module, to the finite projective $A_1 \times _ A A_2$-module $M_1 \times _\varphi M_2$. $\square$

Lemma 93.4.3. In Example 93.4.1 let $V$ be a finite dimensional $k$-vector space. Then

\[ T\mathcal{D}\! \mathit{ef}_ V = (0) \quad \text{and}\quad \text{Inf}(\mathcal{D}\! \mathit{ef}_ V) = \text{End}_ k(V) \]

are finite dimensional.

Proof. With $\mathcal{F}$ as in Example 93.4.1 set $x_0 = (k, V) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$. Recall that $T\mathcal{D}\! \mathit{ef}_ V = T_{x_0}\mathcal{F}$ is the set of isomorphism classes of pairs $(x, \alpha )$ consisting of an object $x$ of $\mathcal{F} $ over the dual numbers $k[\epsilon ]$ and a morphism $\alpha : x \to x_0$ of $\mathcal{F}$ lying over $k[\epsilon ] \to k$.

Up to isomorphism, there is a unique pair $(M, \alpha )$ consisting of a finite projective module $M$ over $k[\epsilon ]$ and $k[\epsilon ]$-linear map $\alpha : M \to V$ which induces an isomorphism $M \otimes _{k[\epsilon ]} k \to V$. For example, if $V = k^{\oplus n}$, then we take $M = k[\epsilon ]^{\oplus n}$ with the obvious map $\alpha $.

Similarly, $\text{Inf}(\mathcal{D}\! \mathit{ef}_ V) = \text{Inf}_{x_0}(\mathcal{F})$ is the set of automorphisms of the trivial deformation $x'_0$ of $x_0$ over $k[\epsilon ]$. See Formal Deformation Theory, Definition 90.19.2 for details.

Given $(M, \alpha )$ as in the second paragraph, we see that an element of $\text{Inf}_{x_0}(\mathcal{F})$ is an automorphism $\gamma : M \to M$ with $\gamma \bmod \epsilon = \text{id}$. Then we can write $\gamma = \text{id}_ M + \epsilon \psi $ where $\psi : M/\epsilon M \to M/\epsilon M$ is $k$-linear. Using $\alpha $ we can think of $\psi $ as an element of $\text{End}_ k(V)$ and this finishes the proof. $\square$

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