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added argument to deal with injectivity

edited Nov 7, 2009 at 23:15

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Would it be particularly surprising if this were true? If I have understood what you mean correctly (so that since you want to consider integrality in full generality a version of essentialness relative to the image of R is required) I believe I have a proof:

First observe that we can reduce to R local. Indeed one can check normality locally and this preserves minimality of primes (when they survive the localization) and so we can assume R is a local reduced commutative ring with unit.

Now let P be a minimal prime ideal in R and consider the composite R \rightarrow R/P \rightarrow k(P) http://latex.mathoverflow.net/png?R%20%5Crightarrow%20R%2FP%20%5Crightarrow%20k%28P%29 and suppose that b is an integral element in k(P) over R/P and hence over R. As in Jose's comment we know that b is essential over R/P and hence over R (I don't see how to make sense of essential for non-injective morphisms otherwise, maybe I am being dense here). So by hypothesis R/P[b] is a finitely generated projective R-module and so is free since R is local. But Ann(R/P[b]) is clearly at least P so P=0 (since R is reduced) and R is in fact a domain.

I next claim that in fact R is integrally closed in its field of fractions K(R). To see this lets denote by S the integral closure of R in K(R). Then
S = \mathrm{colim} ;R[\alpha\sb 1,\ldots,\alpha\sb n] http://latex.mathoverflow.net/png?S%20%3D%20%5Cmathrm%7Bcolim%7D%20%5C%3BR%5B%5Calpha%5F1%2C%5Cldots%2C%5Calpha%5Fn%5D
where the \alpha\sb i http://latex.mathoverflow.net/png?%5Calpha%5Fi vary over all the integral elements. By hypothesis each of the modules occurring in the colimit is free so S is flat over R. In particular, it is flat, finite, and R \subseteq S http://latex.mathoverflow.net/png?R%20%5Csubseteq%20S so that it is faithfully flat over R. It now follows that S=R by the following standard argument.

Suppose a = \frac{x}{y} http://latex.mathoverflow.net/png?a%20%3D%20%5Cfrac%7Bx%7D%7By%7D is in S, where x,y are in R. Then x is in yS and yS \cap http://latex.mathoverflow.net/png?%5Ccap R = yR by faithful flatness (we prove this below) so that y divides x in R also. In particular a is in R.

Proof that yS \cap http://latex.mathoverflow.net/png?%5Ccap R = yR: since S is faithfully flat over R we get by changing base a faithfully flat map for any ideal I
R/I \rightarrow S\otimes\sb R R/I \cong S/IS http://latex.mathoverflow.net/png?R%2FI%20%5Crightarrow%20S%5Cotimes%5FR%20R%2FI%20%5Ccong%20S%2FIS
which is injective (since faitfully flat maps are always injective - this follows by using the fact that the kernel of the functor on module categories given by base changing is trivial). In particular we have that IS \cap R = I http://latex.mathoverflow.net/png?IS%20%5Ccap%20R%20%3D%20I

In fact I think this gives something stronger. We have shown that the localization at each maximal ideal is a normal domain so in particular R is normal. If R is noetherian it follows that it is a product of finitely many normal domains.

If one restricts to considering injective morphisms I don't think one has to do much more work. Suppose \bar{b} http://latex.mathoverflow.net/png?%5Cbar%7Bb%7D is integral and essential in k(P) over R/P. Let \bar{h} http://latex.mathoverflow.net/png?%5Cbar%7Bh%7D be an element of R/P[x] that \bar{b} http://latex.mathoverflow.net/png?%5Cbar%7Bb%7D satisfies. Now consider the following commutative diagram

where R/P = \bar{R} http://latex.mathoverflow.net/png?R%2FP%20%3D%20%5Cbar%7BR%7D so my diagram lines up. Since this commutes we see that b is essential over R and it is clearly integral so R[b] is free over R by hypothesis and hence R/P[\bar{b} http://latex.mathoverflow.net/png?%5Cbar%7Bb%7D] is free over R/P since one just obtains the bottom row by tensoring. One then runs the same argument as above for R/P i.e., we now know its integral closure in k(P) is flat and hence faithfully flat over R/P and so they must agree.

answered Nov 7, 2009 at 0:49

Greg Stevenson

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