# 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 $\mbox{gl dim}(S)$).

Assume that $\mbox{gl dim}(S)=n<\infty$. If $S$ is free as a $R$-module, then $\mbox{gl dim}(R)\leq n+1$.

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

 $0\rightarrow K\rightarrow P_{n}\rightarrow\cdots\rightarrow P_{0}\rightarrow M% \rightarrow 0,$

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 $\mbox{gl dim}(R)\leq n+1$).

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

 $0\rightarrow K\otimes_{R}S\rightarrow P_{n}\otimes_{R}S\rightarrow\cdots% \rightarrow P_{0}\otimes_{R}S\rightarrow M\otimes_{R}S\rightarrow 0,$

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 $\mbox{gl dim}(S)=n$). 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

 $S\simeq\bigoplus_{i\in I}R,$

for some index set $I$. Finally we have

 $K\otimes_{R}S\simeq K\otimes_{R}\big{(}\bigoplus_{i\in I}R\big{)}\simeq% \bigoplus_{i\in I}\big{(}K\otimes_{R}R\big{)}\simeq\bigoplus_{i\in I}K.$

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. $\square$

Title global dimension of a subring GlobalDimensionOfASubring 2013-03-22 19:05:09 2013-03-22 19:05:09 joking (16130) joking (16130) 4 joking (16130) Theorem msc 13D05 msc 16E10 msc 18G20