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> Is every polynomial ring over any field regular?
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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?



Is every polynomial ring over any field regular?
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 $R$).


