# local Artin algebras

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 $A_i$.

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You have to be careful when considering the sum A_1 + ... + A_n. Probably if you are interested in commutative rings then you should call it a product. This would also make clear that the you do not have, in general, maps from A_i to A, but rather from A to A_i (A_i in the way you are looking at it is NOT subring) Considering all of this the answer to your question I think should still be no: if you take a field k and consider k \times k then you would have at least three ways to embed k in k \times k. In more geometric terms the factors A_i, when not further decomposable correspond to –  unknown Oct 27 '10 at 22:36
... the connected components of Spec(A) –  unknown Oct 27 '10 at 22:36
Yes, the canonical way to consider the decomposition is to take the maximal ideals $m_1,...,m_n$ of $A$ and then $A$ is isomorphic to the product of the localizations $A_{m_1}\times ... \times A_{m_n}$. –  Qing Liu Oct 27 '10 at 22:47
Let $k$ be an infinite field of characteristic $p>0$. Then there are infinitely many field homomorphisms $\varphi_\alpha:k\rightarrow k$ where $\alpha$ belongs to some infinite indexing set I. Then $k$ admits infinitely many "embedding" in $k\times k$ that split the projection into the first factor: take the maps $x\rightarrow (x, \varphi_\alpha(x))$ for $\alpha\inI$. –  unknown Oct 27 '10 at 23:20
Here the $k$-algebra is considered. I.e. it is a vector space over $k$ embedded with multiplication which respects the one by elements of $k$. –  Alexander Oct 27 '10 at 23:40

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 $A_i = p_iA$.

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Where may I find something about this decomposition? And why are these $A_i$ local? Thanks for the reply. –  Alexander Oct 27 '10 at 23:36
It is called Pierce decompositi. It is probably discussed in any reasonable Ring Theory book (try Herstein). Locality can be argued. I am too drunk now to think of the simplest argument but one way is to say that the radical $J(A_i)$ is nilpotent, hence $A_i/J(A_i)$ is simple as idempotent can be lifted modulo nilpotent ideal, hence $A_i/J(A_i)$ is field (simple commutative ring). Now you can find an inverse for each $x\in A_i\setminus J(A_i)$ by a direct computation... –  Bugs Bunny Oct 28 '10 at 0:05
+1 for implanting the image of Bugs Bunny drunkenly answering a math question. –  Todd Trimble Oct 28 '10 at 18:14

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 $A\to A_i$.

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If you define $A_i$ as $p_iA$ then the embedding is canonical. You are right that it is still unnatural as it is not a map of rings. –  Bugs Bunny Oct 27 '10 at 23:54
Right. Actually my first reaction was: What do you really mean by "unique"? –  Sándor Kovács Oct 28 '10 at 0:25
Exactly, the question is not clear as it is formulated. I propose below an interpretation of the correct question to ask. –  unknown Oct 28 '10 at 0:39

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.

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