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A morphism of schemes is formally smooth and locally of finite presentation iff it is smooth.

What happens if we drop the finitely presented hypothesis? Of course, locally of finite presentation is part of smoothness, so implicilty I am asking for the flatness to fail.

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2 Answers 2

up vote 19 down vote accepted

Here's an elementary example. For any field $k$, consider the ring $k[t^q|q\in\mathbb Q_{>0}]$, which I'll abbreviate $k[t^q]$. I claim that the natural quotient $k[t^q]\to k$ given by sending $t^q$ to $0$ is formally smooth but not flat, and therefore not smooth.

First let's show it's formally smooth. Let $A$ be a ring with square-zero ideal $I\subseteq A$, and suppose we have maps $f:k[t^q]\to A$ and $g:k\to A/I$ making the following square commute (I drew it backwards because you're probably thinking of Spec of everything)

$$ \begin{array}{ccc} A/I & \xleftarrow g & k \\ \uparrow & & \uparrow\\ A & \xleftarrow f & k[t^q] \end{array} $$

We'd like to show that there's a map $k\to A$ filling the diagram in. For any $q\in \mathbb Q_{>0}$, note that $f(t^q)\in I$ by commutativity of the square, so $f(t^{2q})\in I^2=0$. But every $q$ is of the form $2q'$ for some $q'$, so we've shown that $f(t^q)=0$ for all $q\in \mathbb Q_{>0}$. So $f$ factors through $k$, as desired.

Now let's show that $k$ is not flat over $k[t^q]$. Consider the exact sequence $$0\to (t)\to k[t^q]\to k[t^q]/(t)\to 0.$$ When you tensor with $k$, you get $$0\to k\to k\to k\to 0,$$ which is obviously not exact. So $k$ is not flat over $k[t^q]$.

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Great, crisp answer, Anton! And as a bonus it shows, as Mabli has just remarked, that even formally étale doesn't imply flat (since your example is formally étale). –  Georges Elencwajg Jan 17 '10 at 22:39
I'm just going to point out a small extension of this, and simultaneously advertise a little. In general, if $R$ is any ring and $I \subset R$ an ideal with $I^2 =I$, then $R \to R/I$ is formally \'etale (but probably not flat, i.e. unless $I$ is a direct factor). This follows by the same argument, and is Prop. 2.20 in Note that if $R$ is noetherian, then in this situation $R /I$ is always flat over $R$ (as a formally smooth algebra over a noeth. ring is flat) because $I$ is automatically a direct factor. –  Akhil Mathew Sep 5 '11 at 23:13

This paper by Fishel-Grojnowski-Teleman shows that the "loop Grassmannian" G((z))/G[[z]] is formally smooth, but does not satisfy Hodge decomposition, hence is not smooth:

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