If $I,J \subset A$ are comaximal ideals in a commutative ring $A$, i.e. $I+J=A$, then for all $n,m \in \mathbb N$ the ideals $I^n$ and $J^m$ are also comaximal.
Proof: $\emptyset= V(A)=V(I+J)=V(I)\cap V(J)=V(I^n)\cap V(J^m)=V(I^n+J^m)$ hence $I^n+J^m=A$.

Warning: I wouldn't like to be drawn into a discussion on whether this is just terminology or trivial algebraically or a big cheat or what not. All I know is that when I had to prove this result a long time ago, I came up with this proof a few months after I had started learning affine schemes and I was exhilarated at the thought that I could literally see why the result held by drawing two disjoint little doodles representing $V(I)$ and $V(J)$ inside a potato representing $Spec(A)$.

Edit (April 8th, 2016)
Here is an example of how thinking scheme-theoretically led to a proof of a purely algebraic problem.

If $I,J \subset A$ are comaximal ideals in a commutative ring $A$, i.e. $I+J=A$, then for all $n,m \in \mathbb N$ the ideals $I^n$ and $J^m$ are also comaximal.
Proof: $\emptyset= V(A)=V(I+J)=V(I)\cap V(J)=V(I^n)\cap V(J^m)=V(I^n+J^m)$ hence $I^n+J^m=A$.

Warning: I wouldn't like to be drawn into a discussion on whether this is just terminology or trivial algebraically or a big cheat or what not. All I know is that when I had to prove this result a long time ago, I came up with this proof a few months after I had started learning affine schemes and I was exhilarated at the thought that I could literally see why the result held by drawing two disjoint little doodles representing $V(I)$ and $V(J)$ inside a potato representing $Spec(A)$.

Edit (April 8th, 2016)
Here is an example of how thinking scheme-theoretically led to a proof of a purely algebraic problem.

If $I,J \subset A$ are comaximal ideals in a commutative ring $A$, i.e. $I+J=A$, then for all $n,m \in \mathbb N$ the ideals $I^n$ and $J^m$ are also comaximal.
Proof: $\emptyset= V(A)=V(I+J)=V(I)\cap V(J)=V(I^n)\cap V(J^m)=V(I^n+J^m)$ hence $I^n+J^m=A$.

Warning: I wouldn't like to be drawn into a discussion on whether this is just terminology or trivial algebraically or a big cheat or what not. All I know is that when I had to prove this result a long time ago, I came up with this proof a few months after I had started learning affine schemes and I was exhilarated at the thought that I could literally see why the result held by drawing two disjoint little doodles representing $V(I)$ and $V(J)$ inside a potato representing $Spec(A)$.

If $I,J \subset A$ are comaximal ideals in a commutative ring $A$, i.e. $I+J=A$, then for all $n,m \in \mathbb N$ the ideals $I^n$ and $J^m$ are also comaximal.
Proof: $\emptyset= V(A)=V(I+J)=V(I)\cap V(J)=V(I^n)\cap V(J^m)=V(I^n+J^m)$ hence $I^n+J^m=A$.

Warning: I wouldn't like to be drawn into a discussion on whether this is just terminology or trivial algebraically or a big cheat or what not. All I know is that when I had to prove this result a long time ago, I came up with this proof a few months after I had started learning affine schemes and I was exhilarated at the thought that I could literally see why the result held by drawing two disjoint little doodles representing $V(I)$ and $V(J)$ inside a potato representing $Spec(A)$.

Edit (April 8th, 2016)
Here is an example of how thinking scheme-theoretically led to a proof of a purely algebraic problem.