local Artin algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T06:38:49Z http://mathoverflow.net/feeds/question/43888 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43888/local-artin-algebras local Artin algebras Alexander 2010-10-27T22:10:27Z 2010-10-28T00:24:47Z <p>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 <em>Atiyah-McDonald, Introduction To Commutative Algebra, Theorem 8.7</em>. The subalgebras $A_i$ are uniquely determined up to the isomorphism.</p> <p>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.</p> <p>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?</p> <p>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 $A_i$.</p> http://mathoverflow.net/questions/43888/local-artin-algebras/43902#43902 Answer by Bugs Bunny for local Artin algebras Bugs Bunny 2010-10-27T23:28:36Z 2010-10-27T23:28:36Z <p>Yes, the decomposition is unique. The uniqueness of inclusions is a moot point because rings may have nontrivial endomorphisms.</p> <p>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 $A_i = p_iA$.</p> http://mathoverflow.net/questions/43888/local-artin-algebras/43904#43904 Answer by Sándor Kovács for local Artin algebras Sándor Kovács 2010-10-27T23:35:43Z 2010-10-27T23:35:43Z <p>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 <em>that</em> 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 $A\to A_i$. </p> http://mathoverflow.net/questions/43888/local-artin-algebras/43917#43917 Answer by unknown for local Artin algebras unknown 2010-10-28T00:24:47Z 2010-10-28T00:24:47Z <p>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.</p> <p>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.</p> <p>Now, in what sense this decomposition of $A_0$ is unique?</p> <p>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. </p>