Let $R$ be a commutative ring with identity $1.$ Let $A$ and $B$ be two maximal ideals of $R.$ Prove that $AB=A\cap B.$
My solution is as follows:
Let $x\in AB$ then, it means, $\exists a\in A,b\in B$ such that $x=ab.$ But $ab\in A,B$ as $A$ and $B$ are two ideals of $R$. Thus, we have $x=ab\in A\cap B.$ Hence, $AB\subseteq A\cap B.$
Let $r\in A\cap B$. We proceed to show $A\cap B\subseteq AB.$
$A,B$ is a maximal ideal and $R$ is a commutative ring with identity $1,$ so, $R/A,R/B$ is a field.
This means $r+A\in R/A$ and $r+B\in R/B.$ Since $R/A,R/B$ is a field $\exists x+A\in R/A$ and $y+B\in R/B$ such that $$(x+A)(r+A)=1+A=(r+A)(x+A)$$ and $$(y+B)(r+B)=1+B=(r+B)(y+B)$$ holds.
We have, $xr+A=1+A$ and $yr+B=1+B$.
This means, $xr-1\in A$ and $yr-1\in B$. But $xr\in A$ and $yr\in B$ as, $A,B$ are ideals. This means, $1\in A,B.$
Now, $r\in A,1\in B\implies r\in AB.$ So, $A\cap B\subseteq AB.$ Hence, $AB=A\cap B.$
Does the way I presented the proof above looks good? Is it readable enough? Or is there anything that I can do to make the proof look better, or perhaps improve the proof? I want to improve the readability of my arguments if it seems necessary. Any suggestions regarding this will be greatly appreciated. Lastly, if there's anything wrong with my solution, please feel free to point them out.
If your proof involves showing that a maximal ideal contains $1$, you're doing something wrong (unless it's a proof by contradiction).
Instead, try to show that $1 \in A+B$ (or equivalently $A+B=R$) and work with that. To show this, note that $A+B$ is an ideal that contains $A$ and $A$ is a maximal ideal.
Edit: Here's the concrete mistake: if $r \in A\cap B$, then it is zero mod $A$ and mod $B$ and hence doesn't have an inverse in the field $R/A$ or $R/B$.