local Artin algebras - MathOverflow

local Artin algebras

2010-10-27T22:10:27Z

Given a commutative Artin algebra $A$ over an algebraically closed field $k$ one has a decomposition $A=A_1\oplus\ldots\oplus A_n$ into local Artin subalgebras, see for example Atiyah-McDonald, Introduction To Commutative Algebra, Theorem 8.7. The subalgebras $A_i$ are uniquely determined up to the isomorphism.

The question is as follows. Are the inclusions $A_i\subset A$ uniquely determined as well? They should be, but I cannot find an accurate proof.

UPD: So, the inclusions are not necessarily unique. But may there exist an infinite number of inclusions? Or the number of inclusions is necessarily finite?

Motivation: If there is a finite number of ways for the embedding $A_i\to A$ then the connected group of unity $(Aut A)^{\circ}$ of the automorphism group of algebra $A$ stabilizes the subalgebra $A_i$.

Answer by Bugs Bunny for local Artin algebras

2010-10-27T23:28:36Z

Yes, the decomposition is unique. The uniqueness of inclusions is a moot point because rings may have nontrivial endomorphisms.

The proof goes like this: consider decompositions of $1$ into the sums of orthogonal idempotents $1=\sum_i p_i$. Orthogonality means that $p_ip_j=0$ whenever $i\neq j$. From general nonsense (commutativity will be needed) you can find unique maximal decomposition and then $A_i = p_iA$.

Answer by Sándor Kovács for local Artin algebras

2010-10-27T23:35:43Z

As soon as there exist $i\neq j$ such that $A_i\simeq A_j$, then the embedding is not unique, because $A_i\oplus A_j$ will have many different ways to be written as $A_i\oplus A_j$. On the other hand if for any $i$ there does not exist a non-trivial $A$-algebra homomorphism $A_i\to \oplus_{j\neq i}A_j$ then the embedding $A_i\hookrightarrow A$ is unique, because then any copy of $A_i$ in $A$ would be contained in the kernel of the projection $A\to \oplus_{j\neq i}A_j$ which is that copy of $A_i$. I realize that these don't cover all possibilities, but I will leave it for you to work out the intermediate cases in case you are interested. Then again, why would you want the embeddings to be unique? Those are not natural, the natural maps here are the projections $A\to A_i$.

Answer by unknown for local Artin algebras

2010-10-28T00:24:47Z

Bugs Bunny's answer made me realize that what you are after is true. You have to formulate it more clearly though, I think. Let me then add few things to his answer.

Consider the canonical decomposition $A\simeq A_1\times\ldots\times A_n$ of the Artin ring $A$ into the product of local Artin rings $A_i$. The identity $1_A=(1,\ldots, 1)$ can be written as the sum $(1,0,\ldots,0)+\ldots+(0,\ldots,0,1)$, where we call the $i$-th summand $p_i$, following Bags Bunny. We have $p_i^2=p_i$, $p_ip_j=0$ for $i\neq j$, and $1=p_1+\ldots+p_n$. Using these projectors $p_i$ now you can decompose the abelian group $A_0$ underlying the ring $A$ into the sum of the abelian groups $p_iA$: $A_0\simeq p_1A\oplus\ldots\oplus p_nA$. This decomposition has the property that the distinguished element $1_A$ decomposes as $(p_1\cdot 1_A,\ldots,p_n\cdot 1_A)$, moreover $p_iA$ is identified, as $A$-module, to $A_i$ and the distinguished elements $p_i 1_A$ and $1_{A_i}$ correspond to each other under this identification.

Now, in what sense this decomposition of $A_0$ is unique?

Assume that you are given a decomposition of $A_0\simeq M_1\oplus\ldots\oplus M_n$ into the sum of abelian groups $M_i$ together with the data of isomorphisms of $A$-modules $\varphi_i:A_i\rightarrow M_i$, for all $i$, such that the distinguished element $1_A$ of $A_0$ decomposes as $\varphi_1(1_{A_1})+\ldots+\varphi_n(1_{A_n})$, then the decomoposition $A_0\simeq M_1\oplus\ldots\oplus M_n$ coincide with that described in the first paragraph.