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Hi, Let $M_i$ be A modules. Then we know that $Ass (\oplus M_i) = \bigcup Ass(M_i) $. We consider here isomorphisms between modules.

Now consider a stanley decomposition so $M=\oplus ^r_{i=1} u_iK[Z_i]$ where $ Z_i \subseteq \left\lbrace x_1,...,x_n \right\rbrace $, $u_i$ is a monomial in $S=K[x_1,...,x_n]$ . M is a $ K[x_1,...,x_n]$ module $Z^n $ graded and $u_iK[Z_i]$ is $K[Z_i]$ free . In this direct sum the above equality is not true because here we consider isomorphism between vector spaces. I mean by this that Ass M is not $\bigcup Ass(u_iK[Z_i])$. This happens because in the direct sum we have a vector spaces isomorphism but I don't understand the difference between the module isomorphisms and the vetor space isomorphisms.

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MathOverflow doesn't like {}'s. Try \left\lbrace and \right\rbrace. –  darij grinberg Mar 24 '11 at 11:36
Each side of $M=\oplus u_iK$ is a $K$-module. As $K$-modules we have $Ass(M)=\bigcup Ass(u_iK)=\left\brace 0 \right\brace$ because there are no non-zero primes in $K$ –  Steven Landsburg Mar 24 '11 at 12:28
we look at each side as $k[x1,...xn]$ modules. So Ass K = m because K is isomorphic with $k[x_1,...,x_n]/m$ where m is the maximal irellevant ideal. –  Andrei Mar 24 '11 at 22:28
I think That my mistake is the next one. In the stanley decompostion we have a isomorphism over K so it's a vector space isomorphism, and if we want to use the Ass formula we need to have an isomorphism over $K[X_1,...x_n]$ which will be a module isomorphism. –  Andrei Mar 31 '11 at 10:41

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