$\mathfrak{a}_{1} + \dots + \mathfrak{a}_{n} = A \Rightarrow \mathfrak{a}_{1}^{r_{1}} + \dots + \mathfrak{a}_{n}^{r_{n}} = A$

Question

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.

Answer 1

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 $\;(i+j)^{n+m-1}\in I^n+J^m\;$

Answer 2

You have an equation $a_1+\cdots+a_n=1$, $a_i\in\mathfrak a_i$. Now raise that to a sufficiently large power.

Answer 3

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 $I^n+J^m=R$.

Answer 4

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.