1.5 Regular elements and sequences
Let \(M\) be an \(R\)-module. An element \(x \in R\) is \(M\)-regular if it does not annihilate any element in \(M\). In colon notation, it is non a member of the ideal \(( 0 : M )\). If \(x\) is an \(R\)-regular element, we simply say it is a regular element.
Let \(M\) be an \(R\)-module. Then a weak \(M\)-regular sequence is a sequence \(x_1, \ldots , x_n\) with \(x_i \in R\) for all \(i\), such that \(x_i\) is a \(M / (x_1, \ldots , x_{i-1}) M\)-regular element for all \(i\). If \(M = R\), we say that \(x_1, \ldots , x_n\) is a weak regular sequence.
Let \(M\) be an \(R\)-module. An \(M\)-regular sequence is a weak \(M\)-regular sequence such that \(M / (x_1, \ldots , x_n) M \neq 0\). If \(M = R\), we simply call this a regular sequence.
A local ring \(R\) is regular if \(\mathfrak {m}\) is generated by a system of parameters.
The following is a nice sanity check lemma.
\(R\) is regular if and only if the minimal number of generators of its maximal ideal is equal to the (Krull) dimension of \(R\)
This follows by unwinding the definition and using the dimension theorem (the length of a system of paramaters is \(\nu (R)\) by definition).
A ring \(R\) is regular if \(R_{\mathfrak {p}}\) is regular for every \(\mathfrak {p} \in R\).
The following are equivalent Firstly, \(R\) is regular. Secondly, the zariski cotangent space is a vector space of dimension \(\dim R\).
See in adjunction_blob.txt
Every regular ring is an integral domain.
See BH 2.2.3
Let \(R\) be a regular local ring. Then \(R / I\) is regular local if and only if \(I\) is generated by a (regular) system of parameters (I.e. a generating set for \(\mathfrak {m}\)).
This proof uses the following facts:
* a Nakayama corollary
* the fact that regular rings are integral domains
* the fact that you can’t have a proper containment of integral domains with the same dimension
A local ring \(R\) is regular if and only if its maximal ideal is generated by a regular sequence.