Let $R$ be a commutative ring and $A$ and $B$ be subrings of $R$. Suppose also that an ideal $I$ of $R$ is contained in both $A$ and $B$ (so $I$ is an ideal of both $A$ and $B$).
I have two questions:
If $A$ and $B$ are isomorphic (as rings), does this mean $A/I$ and $B/I$ are isomorphic?
If $A/I$ and $B/I$ are isomorphic, does this mean $A$ and $B$ are isomorphic?
I have thought about this for a few days and I think both should be false, but I haven't been able to construct any counterexamples to these. This isn't part of the course I'm doing now, but I thought these questions are rather interesting.
Note: I consider all rings (and subrings) to be unital. Also, an isomorphism of rings should send $1$ to $1$.
Both answers are no.
For (1) consider the ring $R = k[x, y] \oplus k[x, y]$ where $k$ is any field you like. Take the subrings $A = k[x^2, x^3, y] \oplus k[x, y]$ (so in the left factor monomials can't have $x$-degree $1$) and $B = k[x, y] \oplus k[x^2, x^3, y]$. These are clearly isomorphic. Let $I = (x^2, x^3) \oplus (0)$. Then $A/I \simeq k[y] \oplus k[x, y]$ and $B/I \simeq k[x, y]/x^2 \oplus k[x^2, x^3, y]$. You might just believe by inspection that these are non-isomorphic. One way to see it rigorously is to note that there are no nonzero elements in $A/I$ that square to zero.
For (2) consider the ring $R = k[x] \oplus k[x, y]/(x^2, xy)$. Let $A = k[x] \oplus k[y]$ and $B = k[x^2, x^3] \oplus k[x,y]/(x^2, xy)$. Looking at square zero elements one sees these are nonisomorphic. Both contain the ideal $I = (x^2, x^3) \oplus (y)$ and their quotients by this ideal are $A/I \simeq k[x]/x^2 \oplus k$ and $B/I \simeq k \oplus k[x]/x^2$ and these are clearly isomorphic.