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Let $(R,\mathfrak{m},k)$ be a commutative Noetherian local ring. Is there a commutative flat $R$-algebra $S$ such that $\mathfrak{m}S=0$ or $S/\mathfrak{m}S$ has finite flat dimension over $S$? I know that if $R$ is regular of prime characteristic, then the perfect closure of $R$ satisfies that condition but in the general case I don't know anything about it.

CARLO
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1 Answers1

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If $\mathfrak{m}S=0$ then $S/\mathfrak{m}S$ has finite flat dimension over $S$, so I will understand the question as:

Let $(R,\mathfrak{m},k)$ be a commutative Noetherian local ring. Is there a commutative flat $R$-algebra $S$ such that $S/\mathfrak{m}S$ has finite flat dimension over $S$?

I don't know if you are considering the posible case $S/\mathfrak{m}S=0$ (if yes, for instance the case when $R$ is a domain, then its fraction field works). If not, the answer is affirmative if and only if $R$ is regular. If $R$ is regular, then we can take $S:=R$. Conversely, from the isomorphisms

$$Tor^R(k,k)\otimes_k(S\otimes_Rk)=Tor^R(k,S\otimes_Rk)=Tor^S(S/\mathfrak{m}S,S\otimes_Rk) $$ we deduce that the flat dimension of $k$ over $R$ is finite and so $R$ is regular.

Milou
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