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

where $S$ acts on the left tensor factor: $s\cdot(s'\otimes m)=(ss')\otimes m$.

There is a canonical $R$-linear map

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

Universal property and adjunction

For every $S$-module $N$, restriction of scalars along $f$ produces an $R$-module; this is restriction of scalars . Extension of scalars is left adjoint to restriction of scalars, meaning there is a natural bijection

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

A particularly important special case is localization: if $S^{-1}R$ is the localization of R at a multiplicative set $S\subseteq R$, then extension of scalars along $R\to S^{-1}R$ recovers localization of modules :

Examples

1. Quotient base change. Let $S=R/I$ and $f:R\to R/I$ be the quotient map. Then for any $R$-module $M$,

   $$(R/I)\otimes_R M \cong M/IM.$$

   For example, with $R=\mathbb Z$, $S=\mathbb Z/n\mathbb Z$, one gets $(\mathbb Z/n)\otimes_{\mathbb Z} M \cong M/nM$.

2. Field extension. If $k\subseteq K$ is a field extension and $V$ is a $k$-vector space, then $K\otimes_k V$ is the $K$-vector space obtained by extending scalars. If $V\cong k^d$ is finite-dimensional, then $K\otimes_k V\cong K^d$.

3. Localization as extension of scalars. Let $R=k[x]$, let $S=\{1,x,x^2,\dots\}$, and set $R'=S^{-1}R\cong k[x,x^{-1}]$. For $M=R/(x)$, extension of scalars gives

   $$R'\otimes_R M \cong S^{-1}M = 0,$$

   since $x$ becomes invertible after localization but kills $M$.