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⟨2,X⟩≇⨁cyclic modules

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hbghlyj Posted 2023-2-21 08:48 |Read mode
Show that $⟨2,X⟩$ is not $ℤ[X]$-linearly isomorphic to a direct sum of cyclic modules.

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Czhang271828 Posted 2023-2-21 13:37
Hint: If $M\subset A$ an additive subgroup is an (left/right) $A$-module, then $M$ is an (left/right) ideal of $A$ by definition.

(Here we need to prove $(X,2)$ is not isomorphic to the direct sum of some principal ideals....)

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 Author| hbghlyj Posted 2023-2-27 00:46
MSE
The ideal $I=(2,X)$ of $\mathbb Z[X]$ is not a direct sum of (non-zero) cyclic $\mathbb Z[X]$-modules.
Let's suppose that $I=(2,X)$ is a direct sum of (non-zero) cyclic $\mathbb Z[X]$-modules. Then there exists a family $(N_{\alpha})_{\alpha\in A}$ of (non-zero) cyclic submodules of $I$ such that $I=\sum_{\alpha\in A}N_{\alpha}$, and $N_{\beta}\cap\sum_{\alpha\ne\beta}N_{\alpha}=0$ for all $\beta\in A$. In particular, $N_{\alpha}$ are principal ideals in $\mathbb Z[X]$. But $N_{\alpha}\cap N_{\beta}\ne 0$ for $\alpha\ne\beta$ (if $x_{\alpha}\in N_{\alpha}$ and $x_{\beta}\in N_{\beta}$, then $x_{\alpha}x_{\beta}\in N_{\alpha}\cap N_{\beta}$). This leads us to the conclusion that $|A|=1$, that is, $I$ is principal, a contradiction

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 Author| hbghlyj Posted 2023-2-27 01:02
I think the main point of 3# is "Two principal ideals of $\Bbb Z[X]$ are not disjoint so not isomorphic to direct sum"

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