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 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

**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

**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

**Proof.**
This can be proved directly, but it is also a consequence of Lemmas 10.14.4 and 10.12.8. Namely, we have

as desired. $\square$

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