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Let $R$ be a local ring. By an open idempotent I mean an $R$-module $F$ equipped with a homomorphism $e : F \to R$ such that $e \otimes F = F \otimes e$ is an isomorphism $F \otimes F \cong F$ (this notion in a monoidal category is due to Drinfeld, Boyarchenko). For example, $0 \to R$ and $R \to R$ are open idempotents.

Are these the only ones? It is not hard to see this under the assumption that $F$ is finitely generated, using Nakayama. But I don't know what happens in the general case.

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up vote 5 down vote accepted

The answer is no in general. It is yes if $R$ is $\mathfrak{m}$-adically separated (e.g. noetherian), where $\mathfrak{m}$ is the maximal ideal.

For a counterexample, assume $R$ is a valuation ring (not a field) and $\mathfrak{m}=\mathfrak{m}^2$. Take $F=\mathfrak{m}$, $e=$ the inclusion in $R$. Now $e\otimes \mathfrak{m}$ is injective by flatness (over a valuation ring, flat=torsion free) and surjective because its image is $\mathfrak{m}^2$.

Now assume $R$ separated, let $k$ be the residue field, and put $\overline{F}=F\otimes k$. Then $e\otimes k$ is a linear form $\overline{F}\to k$ inducing an isomorphism $\overline{F}\otimes\overline{F}\to \overline{F}$, which immediately implies $\overline{F}=0$ or $\overline{F}=k$ (and $e\otimes k$ is an isomorphism in the second case).
If $\overline{F}=0$, then $F=\mathfrak{m}F$, hence $F=\mathfrak{m}^nF$ for all $n$, hence $e(F)\subset \mathfrak{m}^n$ for all $n$. By our assumption this implies $e=0$, and finally $F=0$.
In the other case, $e$ is surjective (its image is not contained in $\mathfrak{m}$), hence $F= R\oplus\ker(e)$. Tensoring with $F$ we get that $F\otimes \ker(e)$ must be zero, hence $\ker(e)=0$ since $R$ is a direct summand of $F$. Hence $e$ is an isomorphism. (In this case the assumption on $R$ is not used).

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Thank you for this wonderful answer. Why does torsionfree => flat hold over valuation rings? – Martin Brandenburg Mar 6 '13 at 14:27
Dear @Martin, The ring of integers of $\mathbf{C}_p$, the completion of the algebraic closure of $\mathbf{Q}_p$, has this property, and in a valuation ring, finitely generated ideals are principal, so the proof is the same as for principal ideal domains, i.e., to prove that $M$ is flat, prove that $I\otimes M\rightarrow M$ is injective for all finitely generated ideals. – Keenan Kidwell Mar 6 '13 at 14:28
Alright, thank you. – Martin Brandenburg Mar 6 '13 at 15:41

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