# closed embedding is open iff ideal sheaf idempotent

Let $$f: \mathrm{Spec}(R) \to \mathrm{Spec}(A)$$ be a closed embedding of affine Noetherian schemes, given by the ideal $$I = \mathrm{ker}(A \twoheadrightarrow R)$$. If $$I = I^2$$, then it's not too hard to show that $$I = (e)$$ for some idempotent element $$e \in A$$. This gives a partition of $$\mathrm{Spec}(A)$$ into disjoint closed subsets $$V(e)$$ and $$V(1-e)$$, so $$V(I) = V(e)$$ is clopen. Conversely, if $$im(f) =V(I)$$ is open, then we can find an idempotent element $$e \in A$$ such that $$V(e) = V(I)$$. To show that $$I=I^2$$, I want to conclude that $$I = (e)$$, but a priori we only know their radicals are equal. Is this even true? More generally, is it true that $$I = I^2$$?

What I'm actually trying to prove is that a finite flat group scheme is etale iff the unit section is open.

Suppose $$\text{Spec}(A/I)\rightarrow \text{Spec}(A)$$ is an open immersion. Then in particular it is flat, hence $$A\rightarrow A/I$$ is a flat morphism.

Hence we have an exact sequence $$0\rightarrow I \otimes_A A/I\rightarrow A\otimes_A A/I\rightarrow A/I\otimes_A A/I \rightarrow 0$$ That reduces to $$0\rightarrow I/I^2\rightarrow A/I\overset{\text{id}}{\rightarrow} A/I \rightarrow 0$$ So $$I=I^2$$ and as $$I$$ is finitely generated (because $$A$$ is noetherian) we get $$I=(e)$$ with $$e^2=e$$ (as discussed in here).

For the converse it is not enough to show that $$\text{Im}(f)$$ is open to conclude that $$f$$ is an open immersion. There are some immersions with open image that are not open immersions (for example $$X_\text{red} \rightarrow X$$). But the result is still true because you can prove that $$A/(e)\cong A_{(1-e)}$$ when $$e$$ is idempotent and a localization correspond to an open immersion.

• Thank you! I should have thought of flatness. You're right about the converse. I was implicitly using that the image was the basic open $D(1-e)$. For future reference the proof that $I = I^2$ implies $I = (e)$ with $e$ idempotent also follows easily from Nakayama's lemma as mentioned here. – ggg Feb 13 at 9:58