Every commutative ring is a quotient of a normal ring?

583 Views Asked by At

In the book Étale cohomology by Milne I found on p. 37 (in the context of constructing the henselization of a local ring) the following claim: "Every ring is a quotient of a normal ring". The same is used in Local rings by Nagata on p. 180.

Why is this true? Both Milne and Nagata don't explain this so it seems I'm just too blind.

Is this only true for $R$ a local ring or is it also true for any commutative ring?

Is there a global version of this statement? Is it true that for any scheme $X$ there is a locally closed embedding $X \hookrightarrow Y$ into a normal scheme $Y$? Is there an official term for such an embedding?

EDIT. Perfect answer by Pete below led me to the following additional question I should have asked above already: Is any noetherian ring a quotient of a normal noetherian ring?

1

There are 1 best solutions below

6
On BEST ANSWER

The most natural family of rings which comes to mind as having every commutative ring as a quotient is $\mathbb{Z}[\{t_i \mid i \in I\}]$, i.e., a polynomial ring in an arbitrary set of indeterminates over $\mathbb{Z}$. These rings are all UFDs -- see e.g. Corollary 15.27 of these notes -- hence integrally closed ("normal").