Could you give an example of finitely generated Z[x]-module which is not a direct sum of cyclic modules?
I have no idea about the example, could you give me some ideas? Thank you.
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Consider the ideal $I=(2,X)$ in $\Bbb Z[X]$.
Concretely, it consists of polynomials with integer coefficients of the form $a_0+a_1X+\cdots$ with $a_0$ even.
It cannot be generated by a single element $P(x)$ because you can never get $2$ and $X$ both multiples of a single polynomial.
But it is also false that $I=(2)\oplus(X)$ because $(2)\cap(X)$ contains non-zero elements (i.e. $2X$). A similar argument works if you attempt in any other way to generate $I$ with more than one generator.
The situation can be replicated almost verbatim for any ring which admits non principal ideals.
Andrea Mori
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One more question, do you mean the module is Z[x} itself(Z[x] over itself)? – 6666 Nov 21 '14 at 03:03
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No, any ring as module over itself is cyclic and generated by 1. I'm considering an ideal. Recall that the $R$-submodules of a ring $R$ are precisely the ideals of $R$. – Andrea Mori Nov 21 '14 at 07:20
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But my question is about a module over Z[x], so what's the module? – 6666 Nov 21 '14 at 07:24
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The ideal $J$ is the module. – Andrea Mori Nov 21 '14 at 08:10
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Sorry, what's J? – 6666 Nov 21 '14 at 16:00
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Sorry, I meant $I$. – Andrea Mori Nov 22 '14 at 00:46
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This answer doesn't prove that $I$ is not a direct sum of cyclic modules. I find the remark "A similar argument works if you attempt in any other way to generate $I$ with more than one generator." as being rather vague. – user26857 Jul 07 '15 at 09:21
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@user26857 : you mean that I have to do all the work? – Andrea Mori Jul 07 '15 at 18:30
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I've meant exactly what I said. – user26857 Jul 07 '15 at 20:06
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@user26857, me too. And, by the way, a non-principal ideal in a domain $R$ is never free as a $R$-module because the product of any choice of subset of generators is a non-zero element in the intersection of the modules they generate singularly. This is precisely the argument given in my answer. – Andrea Mori Jul 07 '15 at 23:37
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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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You gave an interesting answer. Thank you very much. And I am interested in how you guys came up an idea of the example (2,$X$)? – 6666 Jul 07 '15 at 10:20
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@Joseph This is a well known example of non-principal ideal and as you can see to write an ideal as a direct sum of cyclic modules reduces to the ideal is principal. – user26857 Jul 07 '15 at 10:47
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Excuse me, I am a little confused that, why we can get:here 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 $\beta\in A$? – 6666 Jul 08 '15 at 03:08
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1@Ricardo if $I$ was a sum of more than one $N_\alpha$, then $N_\beta \cup N_\alpha \neq 0$ for all $\alpha \neq \beta$ would contradict $N_\beta \cup \sum_{\alpha\neq\beta} N_\alpha = 0$. So $I$ can only be a direct sum of at most one $N_\alpha$, i.e. $|A| = 1$. – Vadim May 07 '21 at 07:25
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