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I have to prove the following :

Let $A$ be a commutative ring with unity and let $\mathfrak{a}_{i}$ be ideals in $A$. Assume that $\mathfrak{a}_{1} + \dots + \mathfrak{a}_{n} = A$. Let $r_{i}$ be positive integers. Prove that $\mathfrak{a}_{1}^{r_{1}} + \dots + \mathfrak{a}_{n}^{r_{n}} = A$.

I must confess I have absolutely no idea how to do that. Usually I try to come with what I have already done but here... nothing.

user26857
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Nerra
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4 Answers4

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If $I+J=R$, then $I^2+J \supseteq I^2+IJ+JI+J^2=(I+J)^2=R^2=R$ and hence $I^2+J=R$. By induction we get $I^{2^n}+J=R$ and hence $I^n+J=R$. Exchanging $I$ and $J$ implies $I^n+J^m=R$.

Martin Brandenburg
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Prove for two and then use induction:

$$I+J=A\implies \forall\,n,m\in\Bbb N\;,\;\;I^n+J^m=A$$

Because

$$1=i+j\;,\;\;i\in I\;,\;\;j\in J\implies 1=(i+j)^{n+m-1}=\sum_{k=0}^{n+m-1}\binom{n+m-1}k i^kj^{n+m-1-k}$$

Observe that in the rightmost expression, we have that

$$k<n\implies n+m-1-k>n+m-1-n=m-1$$

so either $\;i^k\in I^n\;$ , or $\;j^{n+m-1-k}\in J^m\;$, and thus $\;(i+j)^{n+m-1}\in I^n+J^m\;$

Timbuc
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Here is yet another proof. Let $\mathfrak{p}$ be any prime ideal of $A$. Then $\mathfrak{a}_1^{r_1} + \dots + \mathfrak{a}_n^{r_n} \subseteq \mathfrak{p} \iff \forall i : \mathfrak{a}_i^{r_i} \subseteq \mathfrak{p} \iff \forall i : \mathfrak{a}_i \subseteq \mathfrak{p} \iff \mathfrak{a}_1 + \dots + \mathfrak{a}_n \subseteq \mathfrak{p}$. The claim follows since any proper ideal of $A$ is contained in a prime ideal.

Dune
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  • Yeah, it is always a good idea to use the axiom of choice for something which is just a two-line calculation! ;-) See also http://mathoverflow.net/questions/59071/what-elementary-problems-can-you-solve-with-schemes/59087#59087 and the comments there. – Martin Brandenburg Oct 20 '14 at 08:10
  • @Martin: Of course one can always look for the most elementary proof but such prime ideal arguments are very convenient in my opinion. One should at least be aware of them. – Dune Oct 20 '14 at 08:45
  • How do you prove this then for ideal sheaves on a ringed space (for example scheme)? A huge advantage of constructive proofs is that they are "platform independent". – Martin Brandenburg Oct 20 '14 at 09:09
  • @Martin: I did never disagree with you in that point. Nevertheless it is no harm to be aware of such arguments. – Dune Oct 20 '14 at 09:19
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You have an equation $a_1+\cdots+a_n=1$, $a_i\in\mathfrak a_i$. Now raise that to a sufficiently large power.

Pedro
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  • Thank you for your answer. I thought about doing that but had no idea it could work out. – Nerra Oct 19 '14 at 17:21