Let $R$ be a ring, $S\subset R$ a subring and $I\subset R$ an ideal of $R$. Prove that every ideal of $S$ need not be of the form $S\cap I$. Thoughts:
My idea to prove this is to find an explicit ring structure (in the context of the prompt) such that there exists some ideal $J$ of $S$ with $J\neq S\cap I$ for any ideal $I$ of $R$. I believe one way to do this would be to obtain some relatively small and easy to handle $R$ that has very few ideals, yet also has some proper non-trivial subring $S$ that has more ideals than $R$. That way, it would not be possible for the ideals of $S$ of the form $S\cap I$ to actually realize all ideals of $S$.
Attempt:
Consider the field $\mathbb{R}$, which has exactly the ideals $(0)$ and $(1)$. We then need to find a subring with more than two ideals. We have that $\mathbb{Z}\subset \mathbb{R}$ is a subring, and $\mathbb{Z}$ has infinitely many ideals of the form $n\mathbb{Z}=(n)$ for integers $n\geq 2$.
Does this seem reasonable?
 A: This is completely correct (as pointed out by the user Crostul already). You could have chosen $\mathbb Q$ instead of $\mathbb R$ which is a field as well but somewhat "closer" to the ring $\mathbb Z$ ($\mathbb Q$ is the quotient field of $\mathbb Z$ and so in some sense a natural extension). But this is just a matter of taste.
This argument generalizes to arbitrary fields $K$ with proper subrings $R$. Indeed, recall the standard fact that a (commutative unital) ring is a field iff its only ideals are trivial. Hence, if we have a proper subring, this ring has to have some non-trivial ideal $I$ which then is not the intersection of $R$ with the trivial ideals $(0)$ and $K$ of $K$ (as these intersections are just $(0)$ and $R$, respectively).

Claim. Let $R$ be a (commutative unital) ring. Then $R$ is a field iff its only ideals are $(0)$ and $R$.
Proof. Suppose $R$ is a field and take some ideal $I$. If $I\ne(0)$ we can find $r\ne0$ in $I$. As $R$ is a field there is some $s\in R$ such that $sr=1$. But then
$$x=x\cdot 1=x(sr)=(xs)r\in I$$
for all $x\in R$. Hence $I=R$. Conversely, suppose that $R$ is a ring with solely the ideals $(0)$ and $R$. Take $r\ne0$ in $R$ and consider the principal ideal $(r)$. As $r\ne0$ we have $(r)\ne(0)$ and hence by assumption $(r)=R$. But then $1\in(r)$ which is equivalent to $sr=1$ for some $s\in R$. This shows that every non-zero element of $R$ is invertible. $\square$
The underlying idea can be phrased as "for $r\in R$ we have $(r)=R$ iff $r$ is a unit". The proof of this proposition follows the same strategy as the one I gave above (see here for example).
So the contraposition of the claim reads as "a ring $R$ is not a field iff it has some non-trivial ideal $(0)<I<R$". Generally, one way such ideals arise is as $I=(r)$ for some non-unit $r$ (which exists due to $R$ not being a field). This should be clear from the given proof and the cited proposition about principal ideals genearted by units.
