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Let $R$ be a (right noetherian) ring. Is there always a right $R$-module which is both flat and injective? If $R$ is an integral domain, then the answer is indeed yes, as the quotient field is such.

Sorry if this is too silly, but so far I'm stuck.

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    $\begingroup$ The zero module is both flat and injective. $\endgroup$ – Jason Starr Jun 18 '14 at 18:21
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Here's a counterexample. Let $k$ be a field and $R=k[x,y]/(x^2,xy,y^2)$; this is a local ring with maximal ideal $m=(x,y)$. Suppose $F$ is a flat $R$-module. Then there is a short exact sequence $$0\to m\otimes F\to F\to F/m\to 0$$

But $m\cong R/m\oplus R/m$ (with $x$ and $y$ generating the summands), so this sequence can be rewritten as $$0\to F/m\oplus F/m\stackrel{(x,y)}\to F\to F/m\to 0$$

Choose a basis for $F/m$ over $R/m=k$ and lift it to $F$. By exactness of the sequence above, this lift will be a basis for $F$ over $R$.

Thus any flat $R$-module is free. Since $R$ is not self-injective, no flat $R$-module can be injective. More generally, a similar argument shows that flat modules over any Artinian local ring are free (replace this short exact sequence by the filtration by powers of the maximal ideal), so any such ring that is not self-injective is a counterexample.

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