global dimension of a subring

Let $S$ be a ring with identity and $R\subset S$ a subring, such that $R$ is contained in the center of $S$. In this case $S$ is a (left) $R$-module via multiplication. Throughout by modules we will understand left modules and by global dimension we will understand left global dimension (we will denote it by $\text{gl dim}(S)$).

Proposition. Assume that . If $S$ is free as a $R$-module, then $\text{gl dim}(R)\le n+1$.

Proof. Let $M$ be a $R$-module. Then, there exists exact sequence

$$0\to K\to P_n\to \mathrm{\cdots }\to P_0\to M\to 0,$$

of $R$-modules, where each $P_i$ is projective (module $K$ is just a kernel of a map $P_n\to P_{n-1}$). We will show, that $K$ is also projective (and since $M$ is arbitrary, it will show that $\text{gl dim}(R)\le n+1$).

Since $S$ is free as a $R$-module, then the extension of scalars $(-{\otimes }_{R}S)$ is an exact functor from the cateogry of $R$-modules to the category of $S$-modules. Furthermore for any projective $R$-module $M$, the $S$-module $M{\otimes }_{R}S$ is projective (in the category of $S$-modules). Thus we have following exact sequence of $S$-modules

$$0\to K{\otimes }_{R}S\to P_n{\otimes }_{R}S\to \mathrm{\cdots }\to P_0{\otimes }_{R}S\to M{\otimes }_{R}S\to 0,$$

where each $P_i{\otimes }_{R}S$ is a projective $S$-module. But projective dimension of $M{\otimes }_{R}S$ is at most $n$ (since $\text{gl dim}(S)=n$). Thus $K{\otimes }_{R}S$ is a projective $S$-module (please, see this entry (http://planetmath.org/ExactSequencesForModulesWithFiniteProjectiveDimension) for more details).

Note that the restriction of scalars functor also maps projective $S$-modules into projective $R$-modules. Thus $K{\otimes }_{R}S$ is a projective $R$-module. But $S$ is free $R$-module, so

$$S\simeq \underset{i\in I}{\oplus }R,$$

for some index set $I$. Finally we have

$$K{\otimes }_{R}S\simeq K{\otimes }_R\left(\underset{i\in I}{\oplus }R\right)\simeq \underset{i\in I}{\oplus }\left(K{\otimes }_{R}R\right)\simeq \underset{i\in I}{\oplus }K.$$

This shows, that $K$ is a direct summand of a projective $R$-module $K{\otimes }_{R}R$ and therefore $K$ is projective, which completes the proof. $\mathrm{\square }$