# Example of finitely generated Z[x]-module which is not a direct sum of cyclic modules

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.

## 2 Answers

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.

• 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 7 '15 at 10:20
• @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 7 '15 at 10:47
• 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 8 '15 at 3:08
• @Joseph Isn't this the definition of (internal) direct sum? – user26857 Jul 8 '15 at 5:35
• Thank you, and I think I need to think more about it. – 6666 Jul 8 '15 at 12:11

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.

• One more question, do you mean the module is Z[x} itself(Z[x] over itself)? – 6666 Nov 21 '14 at 3:03
• 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 7:20
• But my question is about a module over Z[x], so what's the module? – 6666 Nov 21 '14 at 7:24
• The ideal $J$ is the module. – Andrea Mori Nov 21 '14 at 8:10
• Sorry, what's J? – 6666 Nov 21 '14 at 16:00