# Cohen–Macaulayness of $R=k[x_1, \dots,x_n]/\mathfrak p$

I'm looking for some help for the following question:

Let $k$ be a field and $R=k[x_1, \dots ,x_n]$. Show that $R/\mathfrak p$ is Cohen–Macaulay if $\mathfrak p$ is a prime ideal with $\operatorname{height} \mathfrak p \in\lbrace 0,1,n-1,n \rbrace$.

My proof: If $\operatorname{height} p=0$ then for all $q\in \operatorname{Max}(R)$ $\operatorname{height} pR_q=0$ therefore $pR_q\in \operatorname{Min}(R_q)=\lbrace 0\rbrace$, so $pR_p=(0)$ and it's done but If $\operatorname{height} pR_q=1$ then $\operatorname{grade} pR_p=\operatorname{height} pR_p=1$.Thus there exist regular sequence $z\in pR_q$ and we have $\dim R_q/zR_q=\dim R_q-\operatorname{height}zR_q=n-1$.since that $R_q/zR_q$ is Cohen-Macaulay then $\operatorname{depth}R_q/zR_q=\dim R_q/zR_q=n-1$.here,we know that $‎\exists y_1,...,y_{n-1}\in qR_q$ s.t $y_1,...,y_{n-1}$ is $R_q/zR_q$-sequence.Now, if we can make a $R_q/pR_q$-sequence has a length $n-1$ It's done because $\dim R_q/pR_q=\dim R_q-\operatorname{height}pR_q=n-1$. A similar argument works for n,n-1.

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Thanks.It's done. –  Angel Dec 26 '12 at 22:18
Hint. $R$ integral domain, $\operatorname{ht}(p)=0\Rightarrow p=(0)$. If $\operatorname{ht}(p)=1$, then $p$ is principal. If $\operatorname{ht}(p)=n−1,n$, then $\dim R/p=1,0$.