If Two Ideals Are Comaximal in a Commutative Ring, then Their Powers Are Comaximal Ideals Problem 360

Let $R$ be a commutative ring and let $I_1$ and $I_2$ be comaximal ideals. That is, we have
$I_1+I_2=R.$

Then show that for any positive integers $m$ and $n$, the ideals $I_1^m$ and $I_2^n$ are comaximal. Add to solve later

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Proof.

Since $I_1+I_2=R$, there exists $a \in I_1$ and $b \in I_2$ such that
$a+b=1.$ Then we have
\begin{align*}
1&=1^{m+n-1}=(a+b)^{m+n-1}\\[6pt] &=\sum_{k=1}^{m+n-1}\begin{pmatrix}
m+n-1 \\
k
\end{pmatrix}
a^k b^{m+n-1-k}\\[6pt] &=\sum_{k=1}^{m-1}\begin{pmatrix}
m+n-1 \\
k
\end{pmatrix}
a^k b^{m+n-1-k}
+
\sum_{k=m}^{m+n-1}\begin{pmatrix}
m+n-1 \\
k
\end{pmatrix}
a^k b^{m+n-1-k}.
\end{align*}
In the third equality, we used the binomial expansion.

Note that the first sum is in $I_2^n$ since it is divisible by $b^n\in I_2^n$.
The second sum is in $I_1^n$ since it is divisible by $a^m\in I_1^n$.

Thus the sum is in $I_1^m+I_2^n$, and hence we have $1 \in I_1^m+I_2^n$, which implies that $I_1^m+I_2^n=R$. Add to solve later

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