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I want to prove that if $M$ is finitely generated semisimple module then it satisfies the internal cancellation property. Let $M=A\oplus B =N\oplus K$ with $A\cong N$. Since $M$ is finitely generated semisimple, then so are the summands $A,B,N$, and $K$. Consequently, each of them is the direct sum of finitely many simple submodules. Write $A=A_1\oplus \ldots \oplus A_a,\, B=B_1\oplus \ldots \oplus B_b,\, N=N_1\oplus \ldots \oplus N_n\,$ and $K=K_1\oplus \ldots \oplus K_k$, where $A_i,B_j,N_s,K_{\ell}$ are all simples and $a,b,n,k\in \mathbb{Z}^+$. By the hypothesis, \begin{align*} A_1\oplus\ldots \oplus A_a \oplus B_1\oplus \ldots \oplus B_b &= N_1 \oplus \ldots \oplus N_n \oplus K_1 \oplus \ldots \oplus K_k \\ &\cong A_1 \oplus \ldots \oplus A_a \oplus K_1 \oplus \ldots \oplus K_k. \end{align*} By Krull-Schmidt theorem, $a+b=a+k$ (hence $b=k$).

I need to show that there exists a bijection $f:\lbrace1,2,\ldots,b \rbrace \to \lbrace1,2,\ldots,b \rbrace$ such that $B_i \cong B_{f(i)}$ for each $1\leq i \leq b$.

How can I complete my proof?.

Hussein Eid
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  • FYI (since you deleted our earlier discussion. That probably wasn't a good idea) there is a version that works for general rings: if you have $\oplus A_i\cong \oplus B_i$ (only finitely many $A_i$'s and $B_i$'s) and each of the $A_i$'s is indecomposable and each of the $B_i$'s is strongly indecomposable (in the sense their endomorphism rings are local) then their lengths are equal and they pair up into isomorphic pairs. – rschwieb Sep 30 '22 at 14:12
  • A cancellation theorem usually goes "Suppose $A\oplus B\cong A\oplus C$ and (otherstuff)..... then $B\cong C$." So it's a little weird to start with $A\oplus B\cong N\oplus K$. You sure you want to do that? – rschwieb Sep 30 '22 at 14:13
  • If you reposted this from this because you wanted to shed the close vote, then I would advise you not to do that anymore. Dodging moderation is taken very seriously. Remember, closure is only mean to be a temporary thing. In your case, since you are very helpful and responsive, I expect it would not have gotten any more closure votes and they would have aged out. The question really did need lots of clarification, and if it had been closed it would have been shielded from premature answers. – rschwieb Sep 30 '22 at 14:16

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