On the structure of subrings of $R/I$ I was thinking about the following problem: Let $I$ be an ideal of $R$ (commutative and contains $1$). Now consider a subring $S < R/I$. Can we say that the subring $S$ is of the form $S'/I$ where $S'$ is a subring of $R$? Can somebody add other conditions to make this true?
Intuitively I think that the answer to the first question is negative because it would make the theorem about the correspondence of the ideals of $R$ and $R/I$ redundant. So in order to find a counterexample we need to find $S$ which is not an ideal in $R/I$. But I cannot think of any right now. The second question seems more tricky but I don't know how to think about that.  
 A: Let's look at this a little more abstractly. Given a pair of monoids $M, N$ and a homomorphism $f:M\to N$, we note that for any submonoid $N'\subseteq N$, $M'=f^{-1}(N')$ is a submonoid of $M$. For, if $a,b\in M'$, then $f(ab)=f(a)f(b)\in N'$ implies $ab\in M'$, and $f(1_M)=1_N\in N'$ means $1_M\in M'$.
There are three things to observe. Firstly, a ring is a monoid with respect to each of its operations. Secondly, ring homomorphisms are monoid homomorphisms when considering either of the operations. Thirdly, a subset of a ring $R$ that is a submonoid of both $(R,+)$ and $(R,\cdot)$ is a subring of $R$ (and conversely). 
Therefore, given a homomorphism $f:R\to S$ of rings and a subring $S'\subseteq S$, the set $R'=f^{-1}(S')$ is a submonoid of $(R,+)$ and a submonoid of $(R,\cdot)$. We conclude that $R'$ is a subring of $R$.
Of course, this isn't a complete answer to your question, since you still need to verify that $f(R')=S'$ when $f$ is surjective--it simply does not hold when $f$ is not surjective.
A: $R/I$ is the image of the canonical surjection $R\to R/I$, so basically your question translates as

Having a surjective ring homomorphism $f:A\to B$ and a subring $S<B$, can we always find an $S'< A$ such that $f(S')=S$?

$\,$

 And the answer is yes: $S':=f^{-1}(S)$. 

