## 89.11 Tangent spaces of functors

Let $R$ be a ring. We write $\text{Mod}_ R$ for the category of $R$-modules and $\text{Mod}^{fg}_ R$ for the category of finitely generated $R$-modules.

Definition 89.11.1. Let $L: \text{Mod}^{fg}_ R \to \text{Mod}_ R$, resp. $L: \text{Mod}_ R \to \text{Mod}_ R$ be a functor. We say that $L$ is $R$-linear if for every pair of objects $M, N$ of $\text{Mod}^{fg}_ R$, resp. $\text{Mod}_ R$ the map

$L : \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ R(L(M), L(N))$

is a map of $R$-modules.

Remark 89.11.2. One can define the notion of an $R$-linearity for any functor between categories enriched over $\text{Mod}_ R$. We made the definition specifically for functors $L: \text{Mod}^{fg}_ R \to \text{Mod}_ R$ and $L: \text{Mod}_ R \to \text{Mod}_ R$ because these are the cases that we have needed so far.

Remark 89.11.3. If $L: \text{Mod}^{fg}_ R \to \text{Mod}_ R$ is an $R$-linear functor, then $L$ preserves finite products and sends the zero module to the zero module, see Homology, Lemma 12.3.7. On the other hand, if a functor $\text{Mod}^{fg}_ R \to \textit{Sets}$ preserves finite products and sends the zero module to a one element set, then it has a unique lift to a $R$-linear functor, see Lemma 89.11.4.

Lemma 89.11.4. Let $L: \text{Mod}^{fg}_ R \to \textit{Sets}$, resp. $L: \text{Mod}_ R \to \textit{Sets}$ be a functor. Suppose $L(0)$ is a one element set and $L$ preserves finite products. Then there exists a unique $R$-linear functor $\widetilde{L} : \text{Mod}^{fg}_ R \to \text{Mod}_ R$, resp. $\widetilde{L} : \text{Mod}^{fg}_ R \to \text{Mod}_ R$, such that

$\vcenter { \xymatrix{ & \text{Mod}_ R \ar[dr]^{\text{forget}} & \\ \text{Mod}^{fg}_ R \ar[ur]^{\widetilde{L}} \ar[rr]^{L} & & \textit{Sets} } } \quad \text{resp.}\quad \vcenter { \xymatrix{ & \text{Mod}_ R \ar[dr]^{\text{forget}} & \\ \text{Mod}_ R \ar[ur]^{\widetilde{L}} \ar[rr]^{L} & & \textit{Sets} } }$

commutes.

Proof. We only prove this in case $L: \text{Mod}^{fg}_ R \to \textit{Sets}$. Let $M$ be a finitely generated $R$-module. We define $\widetilde{L}(M)$ to be the set $L(M)$ with the following $R$-module structure.

Multiplication: If $r \in R$, multiplication by $r$ on $L(M)$ is defined to be the map $L(M) \to L(M)$ induced by the multiplication map $r \cdot : M \to M$.

Addition: The sum map $M \times M \to M: (m_1, m_2) \mapsto m_1 + m_2$ induces a map $L(M \times M) \to L(M)$. By assumption $L(M \times M)$ is canonically isomorphic to $L(M) \times L(M)$. Addition on $L(M)$ is defined by the map $L(M) \times L(M) \cong L(M \times M) \to L(M)$.

Zero: There is a unique map $0 \to M$. The zero element of $L(M)$ is the image of $L(0) \to L(M)$.

We omit the verification that this defines an $R$-module $\widetilde{L}(M)$, the unique such that is $R$-linearly functorial in $M$. $\square$

Lemma 89.11.5. Let $L_1, L_2: \text{Mod}^{fg}_ R \to \textit{Sets}$ be functors that take $0$ to a one element set and preserve finite products. Let $t : L_1 \to L_2$ be a morphism of functors. Then $t$ induces a morphism $\widetilde{t} : \widetilde{L}_1 \to \widetilde{L}_2$ between the functors guaranteed by Lemma 89.11.4, which is given simply by $\widetilde{t}_ M = t_ M: \widetilde{L}_1(M) \to \widetilde{L}_2(M)$ for each $M \in \mathop{\mathrm{Ob}}\nolimits (\text{Mod}^{fg}_ R)$. In other words, $t_ M: \widetilde{L}_1(M) \to \widetilde{L}_2(M)$ is a map of $R$-modules.

Proof. Omitted. $\square$

In the case $R = K$ is a field, a $K$-linear functor $L : \text{Mod}^{fg}_ K \to \text{Mod}_ K$ is determined by its value $L(K)$.

Lemma 89.11.6. Let $K$ be a field. Let $L: \text{Mod}^{fg}_ K \to \text{Mod}_ K$ be a $K$-linear functor. Then $L$ is isomorphic to the functor $L(K) \otimes _ K - : \text{Mod}^{fg}_ K \to \text{Mod}_ K$.

Proof. For $V \in \mathop{\mathrm{Ob}}\nolimits (\text{Mod}^{fg}_ K)$, the isomorphism $L(K) \otimes _ K V \to L(V)$ is given on pure tensors by $x \otimes v \mapsto L(f_ v)(x)$, where $f_ v: K \to V$ is the $K$-linear map sending $1 \mapsto v$. When $V = K$, this is the isomorphism $L(K) \otimes _ K K \to L(K)$ given by multiplication by $K$. For general $V$, it is an isomorphism by the case $V = K$ and the fact that $L$ commutes with finite products (Remark 89.11.3). $\square$

For a ring $R$ and an $R$-module $M$, let $R[M]$ be the $R$-algebra whose underlying $R$-module is $R \oplus M$ and whose multiplication is given by $(r, m) \cdot (r', m') = (rr', rm' + r'm)$. When $M = R$ this is the ring of dual numbers over $R$, which we denote by $R[\epsilon ]$.

Now let $S$ be a ring and assume $R$ is an $S$-algebra. Then the assignment $M \mapsto R[M]$ determines a functor $\text{Mod}_ R \to S\text{-Alg}/R$, where $S\text{-Alg}/R$ denotes the category of $S$-algebras over $R$. Note that $S\text{-Alg}/R$ admits finite products: if $A_1 \to R$ and $A_2 \to R$ are two objects, then $A_1 \times _ R A_2$ is a product.

Lemma 89.11.7. Let $R$ be an $S$-algebra. Then the functor $\text{Mod}_ R \to S\text{-Alg}/R$ described above preserves finite products.

Proof. This is merely the statement that if $M$ and $N$ are $R$-modules, then the map $R[M \times N] \to R[M] \times _ R R[N]$ is an isomorphism in $S\text{-Alg}/R$. $\square$

Lemma 89.11.8. Let $R$ be an $S$-algebra, and let $\mathcal{C}$ be a strictly full subcategory of $S\text{-Alg}/R$ containing $R[M]$ for all $M \in \mathop{\mathrm{Ob}}\nolimits (\text{Mod}^{fg}_ R)$. Let $F: \mathcal{C} \to \textit{Sets}$ be a functor. Suppose that $F(R)$ is a one element set and that for any $M, N \in \mathop{\mathrm{Ob}}\nolimits (\text{Mod}^{fg}_ R)$, the induced map

$F(R[M] \times _ R R[N]) \to F(R[M]) \times F(R[N])$

is a bijection. Then $F(R[M])$ has a natural $R$-module structure for any $M \in \mathop{\mathrm{Ob}}\nolimits (\text{Mod}^{fg}_ R)$.

Proof. Note that $R \cong R$ and $R[M] \times _ R R[N] \cong R[M \times N]$ hence $R$ and $R[M] \times _ R R[N]$ are objects of $\mathcal{C}$ by our assumptions on $\mathcal{C}$. Thus the conditions on $F$ make sense. The functor $\text{Mod}_ R \to S\text{-Alg}/R$ of Lemma 89.11.7 restricts to a functor $\text{Mod}^{fg}_ R \to \mathcal{C}$ by the assumption on $\mathcal{C}$. Let $L$ be the composition $\text{Mod}^{fg}_ R \to \mathcal{C} \to \textit{Sets}$, i.e., $L(M) = F(R[M])$. Then $L$ preserves finite products by Lemma 89.11.7 and the assumption on $F$. Hence Lemma 89.11.4 shows that $L(M) = F(R[M])$ has a natural $R$-module structure for any $M \in \mathop{\mathrm{Ob}}\nolimits (\text{Mod}^{fg}_ R)$. $\square$

Definition 89.11.9. Let $\mathcal{C}$ be a category as in Lemma 89.11.8. Let $F : \mathcal{C} \to \textit{Sets}$ be a functor such that $F(R)$ is a one element set. The tangent space $TF$ of $F$ is $F(R[\epsilon ])$.

When $F : \mathcal{C} \to \textit{Sets}$ satisfies the hypotheses of Lemma 89.11.8, the tangent space $TF$ has a natural $R$-module structure.

Example 89.11.10. Since $\mathcal{C}_\Lambda$ contains all $k[V]$ for finite dimensional vector spaces $V$ we see that Definition 89.11.9 applies with $S = \Lambda$, $R = k$, $\mathcal{C} = \mathcal{C}_\Lambda$, and $F : \mathcal{C}_\Lambda \to \textit{Sets}$ a predeformation functor. The tangent space is $TF = F(k[\epsilon ])$.

Example 89.11.11. Let us work out the tangent space of Example 89.11.10 when $F : \mathcal{C}_\Lambda \to \textit{Sets}$ is a prorepresentable functor, say $F = \underline{S}|_{\mathcal{C}_\Lambda }$ for $S \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$. Then $F$ commutes with arbitrary limits and thus satisfies the hypotheses of Lemma 89.11.8. We compute

$TF = F(k[\epsilon ]) = \mathop{\mathrm{Mor}}\nolimits _{\mathcal{C}_\Lambda }(S, k[\epsilon ]) = \text{Der}_\Lambda (S, k)$

and more generally for a finite dimensional $k$-vector space $V$ we have

$F(k[V]) = \mathop{\mathrm{Mor}}\nolimits _{\mathcal{C}_\Lambda }(S, k[V]) = \text{Der}_\Lambda (S, V).$

Explicitly, a $\Lambda$-algebra map $f : S \to k[V]$ compatible with the augmentations $q : S \to k$ and $k[V] \to k$ corresponds to the derivation $D$ defined by $s \mapsto f(s) - q(s)$. Conversely, a $\Lambda$-derivation $D : S \to V$ corresponds to $f : S \to k[V]$ in $\mathcal{C}_\Lambda$ defined by the rule $f(s) = q(s) + D(s)$. Since these identifications are functorial we see that the $k$-vector spaces structures on $TF$ and $\text{Der}_\Lambda (S, k)$ correspond (see Lemma 89.11.5). It follows that $\dim _ k TF$ is finite by Lemma 89.4.5.

Example 89.11.12. The computation of Example 89.11.11 simplifies in the classical case. Namely, in this case the tangent space of the functor $F = \underline{S}|_{\mathcal{C}_\Lambda }$ is simply the relative cotangent space of $S$ over $\Lambda$, in a formula $TF = T_{S/\Lambda }$. In fact, this works more generally when the field extension $k/k'$ is separable. See Exercises, Exercise 110.35.2.

Lemma 89.11.13. Let $F, G: \mathcal{C} \to \textit{Sets}$ be functors satisfying the hypotheses of Lemma 89.11.8. Let $t : F \to G$ be a morphism of functors. For any $M \in \mathop{\mathrm{Ob}}\nolimits (\text{Mod}^{fg}_ R)$, the map $t_{R[M]}: F(R[M]) \to G(R[M])$ is a map of $R$-modules, where $F(R[M])$ and $G(R[M])$ are given the $R$-module structure from Lemma 89.11.8. In particular, $t_{R[\epsilon ]} : TF \to TG$ is a map of $R$-modules.

Proof. Follows from Lemma 89.11.5. $\square$

Example 89.11.14. Suppose that $f : R \to S$ is a ring map in $\widehat{\mathcal{C}}_\Lambda$. Set $F = \underline{R}|_{\mathcal{C}_\Lambda }$ and $G = \underline{S}|_{\mathcal{C}_\Lambda }$. The ring map $f$ induces a transformation of functors $G \to F$. By Lemma 89.11.13 we get a $k$-linear map $TG \to TF$. This is the map

$TG = \text{Der}_\Lambda (S, k) \longrightarrow \text{Der}_\Lambda (R, k) = TF$

as follows from the canonical identifications $F(k[V]) = \text{Der}_\Lambda (R, V)$ and $G(k[V]) = \text{Der}_\Lambda (S, V)$ of Example 89.11.11 and the rule for computing the map on tangent spaces.

Lemma 89.11.15. Let $F: \mathcal{C} \to \textit{Sets}$ be a functor satisfying the hypotheses of Lemma 89.11.8. Assume $R = K$ is a field. Then $F(K[V]) \cong TF \otimes _ K V$ for any finite dimensional $K$-vector space $V$.

Proof. Follows from Lemma 89.11.6. $\square$

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