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Let $R$ be a P.I.D. with $1$ and $M$ be an $R$-module that is annihilated by the nonzero, proper ideal $(a)$. Let $a=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ be the unique factorization of $a$. Let $M_i$ be the submodule of $M$ annihilated by $p_i^{\alpha_i}$. Prove that $M=M_1\oplus M_2\oplus \cdots \oplus M_k$.

My attempt so far:

For each $1\leq j \leq k$ define $a_j = \prod_{i\ne j} p_i^{\alpha_i}$. Let $\sum_{i=1}^{n} (a_jr_i)\cdot m_i$ be an arbitrary element of the submodule $(a_j)M$. We have $p_j^{\alpha_j}\cdot (\sum_{i=1}^{n} (a_jr_i)\cdot m_i) = (p_j^{\alpha_j}a_j(r_1 +\cdots + r_n))\cdot (m_1+\cdots +m_n) =(r_1 +\cdots +r_n)\cdot (a \cdot (m_1+\cdots +m_n)) =0$.

So $\sum_{i=1}^{n} (a_jr_i)\cdot m_i \in M_j$, so that $(a_j)M\subset M_j$. Next, let $m\in M_j$. Since $R$ is a P.I.D., we know $1= a_jx + p_j^{\alpha_j}y$ for some $x, y \in R$. So $m= 1\cdot m = (a_jx + p_j^{\alpha_j}y)\cdot m = xa_j \cdot m + yp_j^{\alpha_j} \cdot m = xa_j \cdot m +0 \in (a_j)M$. Conclude that $(a_j)M = M_j$.

Next, suppose $m\in (a_j)M\cap \sum_{t\ne j} (a_t)M$. We have $1\cdot m = xa_j \cdot m + yp_j^{\alpha_j} \cdot m= xa_j\cdot m + 0$. But note that $xa_j\cdot ((\sum_{t\ne j}a_t)\cdot m) = wa\cdot m$ for some $w\in R$, so that $xa_j\cdot ((\sum_{t\ne j}a_t)\cdot m) = 0$. It follows that $xa_j = 0$, and $m=0+0=0$. Conclude that $ (a_j)M\cap \sum_{t\ne j} (a_t)M = (0)$. Thus, $\sum_{i=1}^{k} (a_i)M$ is a direct sum.

At this point, I'm not sure how to actually show this direct sum is equal to $M$. The only thing I tried is applying the Chinese Remainder Theorem as follows, but it doesn't seem to work.

We have that $(a)M=(0)$. And since $R$ is a PID, we know that since $(p_i^{\alpha_i}, p_j^{\alpha_j})= (1) = R$ for any $i\ne j$, $(p_i^{\alpha_i})$ and $(p_j^{\alpha_j})$ are comaximal ideals.

So apply the Chinese Remainder Theorem to get $M\cong M/(p_1^{\alpha_1})M \times \cdots \times M/(p_k^{\alpha_k})M$.

I'd appreciate some help on finishing this.

1 Answers 1

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Based on your argument, I assume you assume your PID to have a $1$. From there you need only show that $(a_1,\ldots, a_k)=R$. To do this show inductively that for $k

$$(a_1,a_2,\ldots, a_j)=(p_{j+1}^{e_{j+1}}\ldots p_k^{e_k})$$

Then at the last step you'll have $(a_1,\ldots, a_{k-1},a_k)=(p_k^{e_k}, a_k)=R$ which finishes your approach.

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    Yes, it has a $1$. I appreciate the response. Could you please make explicit how you use the CRT to get that isomorphism? Also, is there any way to prove that $R = \sum_{i=1}^{k}(a_i)M$? This would be the last step in the argument I gave, and I'd like to know if my argument could work. – 2017-02-25
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    @CuriousKid7 each of the $(p_i)^{e_i}$ are coprime so you use induction on the number of factors to show $R/(a)$ is isomorphic to the sum of the indicated modules. I'm not sure what you mean when you write $R=\sum a_i M$ though, $M$ is a, $R$-torsion module and $R$ is free (as an $R$ module), so it cannot be the sum of torsion modules. – 2017-02-25
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    I'm sorry. I meant to say I WTS $M= \sum (a_i)M$. – 2017-02-25
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    @CuriousKid7 you could show that $a_i M$ annihilates $\displaystyle\oplus_{j\ne i}M_j$ and that it is an isomorphism on $M_i$ (by finding an inverse modulo $p_i^{e_i}$ using CRT or such). That amounts to the same thing in the end, just has more writing associated to the task. Generally I don't advise it, since showing the individual factors annihilate specific things is a lot cleaner. – 2017-02-25
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    OK, that does seem a lot messier. The last thing I can think of is to show that the ideal $(a_i : 1\leq i\leq k)=R$. The result would follow quickly if this is true. But is this true? Can we get this by invoking co-maximality somehow? – 2017-02-25
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    Well you can show that relatively simply: by induction show that $(a_1, a_2,\ldots ,a_j) = (p_{j+1}^{e_{j+1}}\ldots p_k^{e_k})$ for $j$a_k$ you get the unit ideal. It's still a bit more writing, but it's workable. – 2017-02-25
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    I know this may be obvious, but why exactly does adding $a_k$ produce the unit ideal? It seems to me that you get the ideal $(p_k^{\alpha_k})$. – 2017-02-25
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    @CuriousKid7 Comaximality, $(a_1,\ldots, a_{k-1}) = (p_k^{e_k})$ by the induction, and there are no $p_k$ factors in $(a_k)$ (by definition!) – 2017-02-25
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    Right! For some reason, I forgot my own definition. Thank you! – 2017-02-25