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?

  • $\begingroup$ Wikipedia thinks so: en.wikipedia.org/wiki/Regular_ring (specifically: any field is regular, and if $A$ is a regular ring, so is $A[x]$). $\endgroup$
    – Todd Trimble
    Oct 19, 2014 at 22:38
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    $\begingroup$ 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. $\endgroup$
    – user27920
    Oct 20, 2014 at 3:02
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    $\begingroup$ ...every polynomial ring in finitely many indeterminates... $\endgroup$ Nov 22, 2014 at 12:45

1 Answer 1


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 $R$).

  • 2
    $\begingroup$ There is no need to consider B, in that it is difficult to provr that B has finite global dimension without proving atmthe same time that A has finite global dimension :-). A itself has finite global dimension: this follows at once from the fact that its projective dimension as a bimodule over itself is finite, and this last can be seen by looking at a Koszul complex. $\endgroup$ Oct 20, 2014 at 3:55
  • $\begingroup$ @MarianoSuárez-Alvarez: don't you have to have a local ring for the Koszul complex of regular sequence to be free resolution of the quotient? $\endgroup$
    – Alex
    Oct 20, 2014 at 21:34
  • $\begingroup$ @Alex, the Koszul complex for a regular sequence is always exact ---whether the ring is local or not. $\endgroup$ Oct 20, 2014 at 23:15
  • $\begingroup$ @MarianoSuárez-Alvarez: Okay, now that I recall the proof, it doesn't have to be local. Thanks. $\endgroup$
    – Alex
    Oct 21, 2014 at 2:55

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