Extension of scalars

Let $f:R\to S$ be a homomorphism of commutative rings, and let $M$ be an $R$ -module. The extension of scalars (or base change) of $M$ along $f$ is the $S$ -module

S\otimes_R M,

where $S$ acts on the left tensor factor by

s\cdot(s'\otimes m)=(ss')\otimes m.

There is a canonical $R$ -linear map

\eta_M: M \longrightarrow S\otimes_R M,\qquad m\longmapsto 1\otimes m,

where $S\otimes_R M$ is viewed as an $R$ -module via $f$ .

Universal property

For every $S$ -module $N$ , extension of scalars is left adjoint to restriction of scalars: there is a natural bijection

\mathrm{Hom}_S(S\otimes_R M,\;N)\ \cong\ \mathrm{Hom}_R(M,\;\mathrm{Res}_f N),

where $\mathrm{Res}_f N$ denotes $N$ viewed as an $R$ -module via $f$ .

If $U\subseteq R$ is a multiplicative set, extension of scalars along $R\to U^{-1}R$ recovers localization of modules:

(U^{-1}R)\otimes_R M \cong U^{-1}M.

Examples

Quotient base change. Let $S=R/I$ and let $f:R\to R/I$ be the quotient map. Then $(R/I)\otimes_R M \cong M/IM.$ In particular, $(\mathbb Z/n\mathbb Z)\otimes_{\mathbb Z} M \cong M/nM$ .

Field extension. If $k\subseteq K$ is a field extension and $V$ is a $k$ -vector space, then $K\otimes_k V$ is the corresponding $K$ -vector space. If $V\cong k^d$ , then $K\otimes_k V\cong K^d$ .

Localization. Let $R=k[x]$ , let $U=\{1,x,x^2,\dots\}$ , and set $R'=U^{-1}R\cong k[x,x^{-1}]$ . For $M=R/(x)$ , $R'\otimes_R M \cong U^{-1}M = 0,$ since $x$ becomes invertible after localization but annihilates $M$ .