# Is every polynomial ring over any field regular?

Is every polynomial ring over any field regular?

For a field that is algebraically closed, it is true as any maximal ideal of $k[x_1,...,x_n]$ corresponds to a point $(t_1,...,t_n)$ in $\mathbb{A}^n$ and thus is generated by $n$ elements $x_i-t_i$. But when $k$ is not algebraically closed, does one still have the same conclusion?

• Wikipedia thinks so: en.wikipedia.org/wiki/Regular_ring (specifically: any field is regular, and if $A$ is a regular ring, so is $A[x]$). – Todd Trimble Oct 19 '14 at 22:38
• Theorem 5.1 in Matsumura's "Commutative Ring Theory" gives an elementary direct proof that any maximal ideal in $k[x_1,\dots,x_n]$ has $n$ generators for an arbitrary field $k$. So there is no need to first treat the algebraically closed case and to then deduce the general case from that. – user27920 Oct 20 '14 at 3:02
• ...every polynomial ring in finitely many indeterminates... – Fred Rohrer Nov 22 '14 at 12:45

Yes, this follows from the homological characterization of regularity. Let $A=k[x_1,\dots,x_n]$ and $B=K[x_1,\dots,x_n]$, where $K$ is the algebraic closure of $k$. Then for any $A$-modules $M$ and $N$ with $M$ finitely generated, $$B\otimes \operatorname{Ext}_A^*(M,N)=\operatorname{Ext}_B^*(B\otimes M,B\otimes N).$$ Since $B$ has global dimension $n$, the right hand side vanishes for $*>n$, and hence so does $\operatorname{Ext}_A^*(M,N)$. It follows that $A$ has global dimension $n$ and is thus regular. Alternatively, there are various ways to prove directly that $A$ has finite global dimension, without reducing to the algebraically closed case (for instance, you can show that the global dimension of $R[x]$ is always one more than the global dimension of $R$).