A nonregular local ring

Question

Consider the ring of the formal power series $k[[T_1,\ldots,T_n]]$ ($k$ algebraically closed) where $\mathfrak m$ is the maximal ideal. If $f\in\mathfrak m^2$, why $$\frac{k[[T_1,\ldots,T_n]]}{(f)}$$ is a nonregular local ring?

Answer 1

Recall that, for $S$ a commutative ring with $1$, an element in $S[[x]]$ is a unit if and only if its constant term is a unit (see the wikipedia page for instance). Since $$R:=k[[T_1,\dots,T_n]]=k[[T_1,\dots,T_{n-1}]][[T_n]],$$ by induction an element in $R$ is a unit if and only if it has nonzero constant term, and $R$ is a local ring with maximal ideal $m:=(T_1,\dots,T_n)$. Hence $R/(f)$ is also a local ring with maximal ideal $m/(f)$.

Let's compute dimensions. It's proved in Atiyah (Cor. 11.18 on page 122) that for $A$ a Noetherian local ring and $x \in A$ a non-zero divisor, $\dim A/(x)=(\dim A) -1$. Since $R$ is a Noetherian domain, this implies $$\dim R/(f)=(\dim R)-1.$$

Now suppose that $y_1,\dots, y_{k}$ generate $m/(f)$. Since $f \in m^2$, we can consider $$m/(f)/(m^2/(f)) \simeq m/m^2$$ as a $k$-vector space. This vector space is spanned by the images of $y_1,\dots,y_k$. By Nakayama's Lemma in local ring form, $m \subset R$ has a generating set of size $k$. This implies $k \geq \dim R$ (taking for granted that $R$ is regular). Hence, a minimal generating set for $m/(f)$ has size greater than $\dim R/(f)$ and $R/(f)$ is not regular.