Exercise concerning quotient rings and ideals

Question

I need help proving the following:

Let $A$ and $B$ be rings with $\beta\in B\subseteq A$ and suppose the ideal $\beta A$ is a prime ideal. If the "/" denotes the quotient set and the equality $A/\beta A\ =\ B/(B\cap\beta A)$ holds, then show that $A=B+\beta A$.

It is obvious that $B+\beta A\subseteq A$ and only the reverse equality is needed. I'd would appreciate any help on that.

I stumbled on this problem while studying a proof. Hopefully, I have provided all the necessary information needed to answer the question.

Answer 1

Notice that we have $B/(B \cap \beta A) = (B+\beta A)/\beta A$. Hence your given equality gives you $(B+\beta A)/\beta A = A/\beta A$ and thus $A/(B+\beta A) \cong (A/\beta A)/((B+\beta A)/\beta A)=0$.

This is precisely what you need.

Answer 2

Let me denote $\beta A$ by $I$. Then you have given that $A/I=B/(B\cap I)$. But what does this mean?

Since $B\subseteq A$ you can define a ring homomorphism $B/(B\cap I)\to A/I$ by $b\bmod (B\cap I)\mapsto b\bmod I$ which is in fact injective. Your hypothesis shows that this homomorphism is also surjective. Now let $a\in A$. Then $a\bmod I=b\bmod I$ for some $b\in B$. This shows that $a-b\in I$ hence $a\in B+I$. Thus you showed $A\subseteq B+I$ hence equality.