How to prove this ring is regular?

208 Views Asked by At

Let $R$ be a DVR, $\pi$ be the uniformizer, and $n\geq 2$ an integer. I need to prove the ring $R[T_1,\dots,T_n]/(T_1\cdots T_n-\pi)$ is regular.

I know a polynomial ring of a regular ring is regular, hence $R[T_1,\dots,T_n]$ is regular. But how to prove this quotient ring is also regular? Could you give a solution? (Any partial solutions are also welcom) And any reference would help a lot!

1

There are 1 best solutions below

0
On BEST ANSWER

Let $M$ be a maximal ideal in $R[T_1,\dots,T_n]$ such that $T_1\cdots T_n-\pi\in M$. There are two cases:

  1. $M\cap R=(0)$.

Let $S=R\setminus\{0\}$. Then $M'=S^{-1}M$ is a maximal ideal of $S^{-1}R[T_1,\dots,T_n]=K[T_1,\dots,T_n]$, where $K$ is the field of fractions of $R$. We have $$R[T_1,\dots,T_n]_M/(T_1\cdots T_n-\pi)_M\simeq K[T_1,\dots,T_n]_{M'}/(T_1\cdots T_n-\pi)_{M'}.$$

But $$K[T_1,\dots,T_n]/(T_1\cdots T_n-\pi)\simeq K[T_1,\dots,T_{n-1}]\left[\frac{1}{T_1\cdots T_{n-1}}\right]$$ which is obviously a regular ring.

  1. $M\cap R=(\pi)$.

Then there is $i$ such that $T_i\in M$. Suppose $i=n$. Then $M'=M/(\pi,T_n)$ is a maximal ideal of $R/(\pi)[T_1,\dots,T_{n-1}]$ which is a polynomial ring over a field. It follows that $M'=(f_1,\dots,f_{n-1})$, where $f_1,\dots,f_{n-1}$ is a regular sequence on $R/(\pi)[T_1,\dots,T_{n-1}]$. It follows that $M=(\pi,T_n,f_1,\dots,f_{n-1})$. But $\pi,T_n,f_1,\dots,f_{n-1}$ is also a regular sequence on $R[T_1,\dots,T_n]$. In particular, they form a basis of $M/M^2$ over $R[T_1,\dots,T_n]/M$. If $\pi-(T_1\cdots T_{n-1})T_n\in M^2$ it follows that $\pi$ and $T_n$ are linearly dependent, a contradiction.