Let $A$ be a ring, $B$ a subring of $A$ and $p$ a prime ideal of $A$. Let $f: B \rightarrow A$ the natural homomorphism. Then, $B/f^{-1}(p)$ is a subring of $A/p$.
I'm trying to prove the above, and I don't really think I understand the question. Does $B/f^{-1}(p)$ is a subring of $A/p$ mean that $B/f^{-1}(p)$ is literally a subring of $A/p$? Or is it $B/f^{-1}(p)$ is isomorphic to one of the subrings of $A/p$?
On
The below is a bit formal, but it encodes the idea that the function $f$ which makes $B$ a subring of $A$ can be used to "naturally" make $B/f^{-1}(p)$ a subring of $A/p$.
It means the following: First, that $B/f^{-1}(p)$ is a ring.
Then, define a map $\bar{f}:B/f^{-1}(p)\to A/p$ as follows. For $\bar{b}\in B/f^{-1}(p)$, let $b\in B$ be any element that maps to $\bar{b}$ under the projection $\pi _B:B\to B/f^{-1}(p)$. let $\bar{f}(\bar{b})=\pi _A(f(b))$ where $\pi _A:A\to A/p$ is the projection. Show that $\bar{f}$ is an injective ring homorphism ( ie. that $\bar{f}$ is an isomorphism of $B/f^{-1}(p)$ onto a subring of $A/p$ ).
It is not literally a subring because it is not even a subset of $A/P$.
But you can show it is isomorphic to $\pi(Im(f))/P$ where $\pi$ is the projection of $A$ to$A/P$ via the first isomorphism theorem.