# 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. $\mathrm{\square}$

Title | global dimension of a subring |
---|---|

Canonical name | GlobalDimensionOfASubring |

Date of creation | 2013-03-22 19:05:09 |

Last modified on | 2013-03-22 19:05:09 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 13D05 |

Classification | msc 16E10 |

Classification | msc 18G20 |