Definition 10.14.1. Let \varphi : R \to S be a ring map. Let M be an S-module. Let R \to R' be any ring map. The base change of \varphi by R \to R' is the ring map R' \to S \otimes _ R R'. In this situation we often write S' = S \otimes _ R R'. The base change of the S-module M is the S'-module M \otimes _ R R'.
10.14 Base change
We formally introduce base change in algebra as follows.
If S = R[x_ i]/(f_ j) for some collection of variables x_ i, i \in I and some collection of polynomials f_ j \in R[x_ i], j \in J, then S \otimes _ R R' = R'[x_ i]/(f'_ j), where f'_ j \in R'[x_ i] is the image of f_ j under the map R[x_ i] \to R'[x_ i] induced by R \to R'. This simple remark is the key to understanding base change.
Lemma 10.14.2. Let R \to S be a ring map. Let M be an S-module. Let R \to R' be a ring map and let S' = S \otimes _ R R' and M' = M \otimes _ R R' be the base changes.
If M is a finite S-module, then the base change M' is a finite S'-module.
If M is an S-module of finite presentation, then the base change M' is an S'-module of finite presentation.
If R \to S is of finite type, then the base change R' \to S' is of finite type.
If R \to S is of finite presentation, then the base change R' \to S' is of finite presentation.
Proof. Proof of (1). Take a surjective, S-linear map S^{\oplus n} \to M \to 0. By Lemma 10.12.3 and 10.12.10 the result after tensoring with R^\prime is a surjection {S^\prime }^{\oplus n} \to M^\prime \rightarrow 0, so M^\prime is a finitely generated S^\prime -module. Proof of (2). Take a presentation S^{\oplus m} \to S^{\oplus n} \to M \to 0. By Lemma 10.12.3 and 10.12.10 the result after tensoring with R^\prime gives a finite presentation {S^\prime }^{\oplus m} \to {S^\prime }^{\oplus n} \to M^\prime \to 0, of the S^\prime -module M^\prime . Proof of (3). This follows by the remark preceding the lemma as we can take I to be finite by assumption. Proof of (4). This follows by the remark preceding the lemma as we can take I and J to be finite by assumption. \square
Let \varphi : R \to S be a ring map. Given an S-module N we obtain an R-module N_ R by the rule r \cdot n = \varphi (r)n. This is sometimes called the restriction of N to R.
Lemma 10.14.3. Let R \to S be a ring map. The functors \text{Mod}_ S \to \text{Mod}_ R, N \mapsto N_ R (restriction) and \text{Mod}_ R \to \text{Mod}_ S, M \mapsto M \otimes _ R S (base change) are adjoint functors. In a formula
\mathop{\mathrm{Hom}}\nolimits _ R(M, N_ R) = \mathop{\mathrm{Hom}}\nolimits _ S(M \otimes _ R S, N)
Proof. If \alpha : M \to N_ R is an R-module map, then we define \alpha ' : M \otimes _ R S \to N by the rule \alpha '(m \otimes s) = s\alpha (m). If \beta : M \otimes _ R S \to N is an S-module map, we define \beta ' : M \to N_ R by the rule \beta '(m) = \beta (m \otimes 1). We omit the verification that these constructions are mutually inverse. \square
The lemma above tells us that restriction has a left adjoint, namely base change. It also has a right adjoint.
Lemma 10.14.4. Let R \to S be a ring map. The functors \text{Mod}_ S \to \text{Mod}_ R, N \mapsto N_ R (restriction) and \text{Mod}_ R \to \text{Mod}_ S, M \mapsto \mathop{\mathrm{Hom}}\nolimits _ R(S, M) are adjoint functors. In a formula
\mathop{\mathrm{Hom}}\nolimits _ R(N_ R, M) = \mathop{\mathrm{Hom}}\nolimits _ S(N, \mathop{\mathrm{Hom}}\nolimits _ R(S, M))
Proof. If \alpha : N_ R \to M is an R-module map, then we define \alpha ' : N \to \mathop{\mathrm{Hom}}\nolimits _ R(S, M) by the rule \alpha '(n) = (s \mapsto \alpha (sn)). If \beta : N \to \mathop{\mathrm{Hom}}\nolimits _ R(S, M) is an S-module map, we define \beta ' : N_ R \to M by the rule \beta '(n) = \beta (n)(1). We omit the verification that these constructions are mutually inverse. \square
Lemma 10.14.5. Let R \to S be a ring map. Given S-modules M, N and an R-module P we have
\mathop{\mathrm{Hom}}\nolimits _ R(M \otimes _ S N, P) = \mathop{\mathrm{Hom}}\nolimits _ S(M, \mathop{\mathrm{Hom}}\nolimits _ R(N, P))
Proof. This can be proved directly, but it is also a consequence of Lemmas 10.14.4 and 10.12.8. Namely, we have
\begin{align*} \mathop{\mathrm{Hom}}\nolimits _ R(M \otimes _ S N, P) & = \mathop{\mathrm{Hom}}\nolimits _ S(M \otimes _ S N, \mathop{\mathrm{Hom}}\nolimits _ R(S, P)) \\ & = \mathop{\mathrm{Hom}}\nolimits _ S(M, \mathop{\mathrm{Hom}}\nolimits _ S(N, \mathop{\mathrm{Hom}}\nolimits _ R(S, P))) \\ & = \mathop{\mathrm{Hom}}\nolimits _ S(M, \mathop{\mathrm{Hom}}\nolimits _ R(N, P)) \end{align*}
as desired. \square
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