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Let $R$ be a ring (commutative noetherian with unit), and let $K(R)$ be its total ring of fractions (obtained by inverting all nonzerodivisors). Thus, $R \hookrightarrow K(R)$. Let $a \in K(R)$ be integral over $R$; in other words, $R[a]$ is a subring of $K(R)$ that is finite over $R$.

Is it necessarily true that $R[a]$ is flat over $R$? More to the point, since I'm expecting a negative answer: what is a concrete counterexample?

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4 Answers 4

up vote 15 down vote accepted

Charles, in your answer you're basically discovering the fact that the normalization is not flat (answer edited to show that it actually does provide an answer to the original question)

Let $X$ be a non-normal reduced scheme and $\sigma:\widetilde X\to X$ its normalization. Now let $\pi:\widehat X\to X$ be any finite locally projective non-isomorphism that $\sigma$ factors through, i.e., there exists a $\mu:\widetilde X\to\widehat X$ such that $\sigma=\pi\circ \mu$. Then $\pi$ is not flat. Your example is a special case of this: If $R$ and $a$ are as in your question, and we set $X=\mathrm{Spec}R$, then $\pi:\widehat X=\mathrm{Spec} R[a]\to X$ satisfies the above conditions, hence $\pi$ is not finite.

And here is a proof of the statement: Since $\sigma$ is generically an isomorphism, the same is true for $\pi$ and hence this general fiber is a reduced closed point, so its Hilbert polynomial is $1$. At the same time the fiber over a (non-normal) point where $\pi$ is not an isomorphism is either not reduced or is reducible and hence its Hilbert polynomial is different than $1$, so $\pi$ cannot be flat. $\square$


Remark
Perhaps a more interesting question is to ask for a finite non-flat morphism whose target is smooth. Obviously it will not come from an integral element of the fraction field of the target. Also, the the dimension of the players have to be at least $2$, since if the target of a morphism that is dominant on all irreducible components is a smooth curve, then the morphism is automatically flat.

My favorite example of such a morphism is to take two copies of $\mathbb A^d$, $d\geq 2$, intersecting in a single point and mapping them to $\mathbb A^d$ in the obvious way. The target is smooth, the morphism is finite, it is étale outside a single point, but it is not flat. You can try to prove this directly or to use the fact that a finite morphism whose target is smooth is flat if and only if the source of the morphism is Cohen-Macaulay. It is really easy to see that two copies of $\mathbb A^d$, $d\geq 2$, intersecting in a single point is not Cohen-Macaulay.

Addendum (to respond to Charles's questions in the comment below)
1) The iff statement above follows from the more general Theorem 18.16(b) on page 465 of Eisenbud's CA with a view toward AG.
2) To see that the example given is not even $S_2$ observe the following: take one regular function on each of the copies of $\mathbb A^d$ such that their values do not agree at the point of intersection. Together they give a regular function on the complement of the point which does not extend to the point. Since the dimension is at least $2$ this implies that the Hartog condition fails and hence this object is not normal. Since it is obviously $R_1$ (again also since its dimension is at least $2$), being normal is equivalent to being $S_2$, so it is not $S_2$ and (again since its dimension is at least $2$), it is therefore not CM. You can see that for curves this would not be a problem.

I can imagine that one may find this more complicated than computing a regular sequence, but once you work with the condition $S_2$ a little this seems an obvious observation. For the connection between $S_2$ and Hartog's condition see this and this answers to another MO question. I should also add that the part about being normal is absolutely unnecessary as the point about Hartog's theorem is the fact that normal implies $S_2$, so the argument is really just the following:

2') The failure of the Hartog condition implies that the example is not $S_2$ and hence not CM.

The reason I argued by normality was that I imagined that people might associate this condition with being normal rather than being $S_2$. I may have made it too complicated in the process. :)

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Thanks for this excellent answer to the title, if not the specific question I was asking. Two elaborations that I would appreciate: 1) Could you provide a specific reference for the "only if" part of the statement in the next-to-last sentence, and 2) what sort of techniques do you have in mind for showing "easily" that the ring in question is not Cohen-Macaulay? Obviously you can't show it's not equidimensional. I did this once for $d=2$, and as I recall, my strategy was to embed this thing in some affine space and then compute a maximal regular sequence; is there an easier way? –  Charles Staats Jan 16 '12 at 2:41
    
Charles, I would argue that I actually did provide an answer in the first paragraph, albeit after three perfect answers, although perhaps this argument deserves to be mentioned as well. (I see that the spirit of it is very similar to yours, although I did not read your proof only the statement before writing this answer). In any case, I edited the answer so it is clear that the answer belongs here and separated the remarks that are indeed inspired by the title of your question. I also added answers to your questions 1) and 2). Let me know if you have more. –  Sándor Kovács Jan 16 '12 at 4:58
    
ps: I would be interested to know whether you find this argument easier or harder than computing an actual maximal regular sequence! –  Sándor Kovács Jan 16 '12 at 4:59
    
I would say that your explanation is an order of magnitude better than computing a maximal regular sequence. Apart from being both easier and more conceptual, it also handles all $d \geq 2$ at once. My computation, so far as I recall, did not. –  Charles Staats Jan 17 '12 at 16:06
    
Thanks, that's kind. I'm glad to hear that. Cheers. –  Sándor Kovács Jan 17 '12 at 18:29

Combining insight from Mahdi's and Yusuf's answers, it looks as thought the finite map in question is virtually never flat.

Specifically, assume $R$ is an integral domain, and $K = K(R)$ is its fraction field. Let $a \in K$ be integral over $R$. Consider the finite morphism $$\operatorname{Spec} R[a] \to \operatorname{Spec} R.$$ Since $R[a]$ and $R$ have the same fraction field $K$, this morphism has degree one (is birational). If it is flat, then every fiber must have the same degree (one); in other words, this morphism is an isomorphism in every fiber. Finiteness, together with Nakayama's Lemma, then implies it is an isomorphism above every stalk. Hence, it is an isomorphism, i.e., $a \in R$.

Contrapositively, if $a \not\in R$, then the morphism is not flat.

For a specific example suggested by Yusuf's answer, consider $R = \Bbbk[x^2, x^3]$, and $a = x = x^3 / x^2 \in K(R)$.

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Geometrically: consider a twisted cubic $T \subseteq \mathbb{P}^{3}$ and its image $T' \subseteq \mathbb{P}^{2}$ under projection from a point in $\mathbb{P}^{3} - T.$ Then $T'$ is a plane cubic with either a nodal or cuspidal singularity. The projection morphism $T \rightarrow T'$ is finite, but not flat, since the fiber over the singular point has length 2 while all other fibers have length 1.

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Suppose $R$ is local and Noetherian. Then Flat $+$ Finite $=$ Free. In the case of your question, to find a counter-example it suffices to find an $a$ such that $R[a]$ is not free over $R$.

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