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 $I_1^m+I_2^n=R$.

Commuting Matrices $AB=BA$ such that $A-B$ is Nilpotent Have the Same Eigenvalues
Let $A$ and $B$ be square matrices such that they commute each other: $AB=BA$.
Assume that $A-B$ is a nilpotent matrix.
Then prove that the eigenvalues of $A$ and $B$ are the same.
Proof.
Let $N:=A-B$. By assumption, the matrix $N$ is nilpotent.
This […]

Nilpotent Element a in a Ring and Unit Element $1-ab$
Let $R$ be a commutative ring with $1 \neq 0$.
An element $a\in R$ is called nilpotent if $a^n=0$ for some positive integer $n$.
Then prove that if $a$ is a nilpotent element of $R$, then $1-ab$ is a unit for all $b \in R$.
We give two proofs.
Proof 1.
Since $a$ […]

Ring Homomorphisms and Radical Ideals
Let $R$ and $R'$ be commutative rings and let $f:R\to R'$ be a ring homomorphism.
Let $I$ and $I'$ be ideals of $R$ and $R'$, respectively.
(a) Prove that $f(\sqrt{I}\,) \subset \sqrt{f(I)}$.
(b) Prove that $\sqrt{f^{-1}(I')}=f^{-1}(\sqrt{I'})$
(c) Suppose that $f$ is […]

If a Prime Ideal Contains No Nonzero Zero Divisors, then the Ring is an Integral Domain
Let $R$ be a commutative ring. Suppose that $P$ is a prime ideal of $R$ containing no nonzero zero divisor. Then show that the ring $R$ is an integral domain.
Definitions: zero divisor, integral domain
An element $a$ of a commutative ring $R$ is called a zero divisor […]

Ideal Quotient (Colon Ideal) is an Ideal
Let $R$ be a commutative ring. Let $S$ be a subset of $R$ and let $I$ be an ideal of $I$.
We define the subset
\[(I:S):=\{ a \in R \mid aS\subset I\}.\]
Prove that $(I:S)$ is an ideal of $R$. This ideal is called the ideal quotient, or colon ideal.
Proof.
Let $a, […]

If Every Proper Ideal of a Commutative Ring is a Prime Ideal, then It is a Field.
Let $R$ be a commutative ring with $1$.
Prove that if every proper ideal of $R$ is a prime ideal, then $R$ is a field.
Proof.
As the zero ideal $(0)$ of $R$ is a proper ideal, it is a prime ideal by assumption.
Hence $R=R/\{0\}$ is an integral […]

Every Prime Ideal is Maximal if $a^n=a$ for any Element $a$ in the Commutative Ring
Let $R$ be a commutative ring with identity $1\neq 0$. Suppose that for each element $a\in R$, there exists an integer $n > 1$ depending on $a$.
Then prove that every prime ideal is a maximal ideal.
Hint.
Let $R$ be a commutative ring with $1$ and $I$ be an ideal […]