# If $\operatorname{Idem}B\rightarrow \operatorname{Idem}(B/mB)$ is surjective then $B$ is a product of local rings

Let $$A$$ be a local ring with maximal ideal $$m$$ and $$B$$ a finite $$A$$-algebra (by finite I mean that $$B$$ is a finitely generated $$A$$-module). If we denote by $$\operatorname{idem}B$$ (respectively $$\operatorname{idem}(B/mB)$$) the set of idempotent elements of $$B$$ (respectively of $$B/mB$$) then the map $$\operatorname{idem}B\rightarrow \operatorname{idem}(B/mB),\ x\mapsto \overline{x}$$ is injective.
My question is: how do I prove that if this map is surjective then $$B$$ is isomorphic to a product of local rings ?
If $$m_1,...,m_r$$ are the maximal ideals of $$B$$ then the $$\overline{m_i}=m_i/mB$$ are the maximal ideals of $$B/mB$$ and I know that $$B/mB\rightarrow \prod_{i=1}^r(B/mB)_{\overline{m_i}}$$ is an isomorphism.
So I tried proving that the canonical morphism of rings $$B\rightarrow \prod_{i=1}^rB_{m_i}$$ is an isomorphism but I couldn't do it.
Any help would be appreciated!

• $B/mB$ is a f.d. algebra over a field $A/m$. Use Pierce decomposition. Nov 22, 2021 at 12:50

## 1 Answer

I see you have also noticed that Raynaud did not explain this detail in Ch. 1 in his book on henselian local rings.

Indeed, there is more to check. Let $$e_1, \ldots, e_r$$ be the orthogonal idempotents of $$B/\mathfrak{m}B$$ that provide the decomposition as a direct product of local rings (so under this decomposition, $$e_i$$ is just $$1$$ in the $$i$$-th coordinate and $$0$$ everywhere else). Let $$\tilde{e}_i$$ be lifts of those idempotents to idempotents of $$B$$. The KEY POINT is that we need these idempotents to define a decomposition of $$B$$, which is not true if you are just given some random collection of idempotents.

Claim 1: The $$\tilde{e}_i$$ are orthogonal. Proof: For $$i \neq j$$, $$\tilde{e}_i\tilde{e}_j$$ is a lift to $$B$$ of the idempotent $$e_ie_j = 0$$. A product of idempotents is idempotent, so by injectivity (proved in Raynaud's book), $$\tilde{e}_i\tilde{e}_j = 0$$ ($$0$$ is an idempotent reducing to $$0$$ modulo $$\mathfrak{m}$$).

Claim 2: $$\sum_{i=0}^r \tilde{e}_i \in B^\times$$. Proof: this sum is congruent to $$1$$ modulo $$\mathfrak{m}B$$, therefore it is $$1$$ modulo the Jacobson radical of $$B$$. This implies that it is a unit (it is not in any maximal ideal of $$B$$; NB we have used going-up as usual).

Therefore, we can write $$B = \prod_i B\tilde{e}_i$$. You can check that this is indeed a decomposition as a product of local rings.

For more practice with these ideas (about going from just having the lift of idempotents thanks to being over a henselian local ring to knowing that the lift has properties that give you a good structure theory of the thing you started with), you can look at Bourbaki, algebre commutative, Ch. III, section 4, exercise 5. This exercise was exploited by Rouquier (resp. Bellaiche--Chenevier) to deal with residually absolutely irreducible (resp. residually multiplicity-free) pseudorepresentations over henselian local ring.