Prime ideals in subring are prime ideals in the bigger ring? I thought of this question through an excercise in algebraic geometry where the rings were $\Gamma(V)$ and $\mathcal{O}_P(V)$, although my question is more general. If $R\subset S$ two (commutative with 1) rings (R subring of S) and I prime ideal in R, then consider I'=(I) the ideal in S, the ideal generated by I. Is that ideal prime? I would think that the answer is positive even in this general case but i can't seem to find any obvious and fast proof. Any help would be appreciated. Just for reference the question came up from excercise 2.18 in fulton algebraic curves in which long story short i had to prove that there is a 1-1 correspondence between prime ideals in $\mathcal{O}_P(V)$ and prime ideals in $\Gamma(V)$.
 A: As I don't have enough reputation for writing comments, I'll post it here.
It is not true in the general case. Take for example $\mathbb{Z}$ and the ideal generated by $(2)$ then this ideal in $\mathbb{Q}$ is just $\mathbb{Q}$.
You can find a plenty of examples in algebraic number theory. 
For instance  $\mathbb{Z} \subset \mathbb{Z}[i]$ but $(5)$ is not a prime ideal in $\mathbb{Z}[i]$ as it can be factored $(5) = (1+2\cdot i)\cdot (1-2\cdot i)$
A: As mentioned in @Asquire's answer, you can have counter-examples taking for $R=\mathbf Z$ and for $S$ the ring of integers of some quadratic extension. It is know that an odd prime number $p$ remains prime in $S$ (one says it is inert in this case)if and only if the discriminant of the extension is not a square modulo $p$. In the other cases, it is either the product of two prime ideals or the squre of a prime ideal.
Nevertheless the property is true in the case that $S$ is a polynomial ring $R[X_1,\dots,X_n]$.
It is also true in the case that $s$ is a ring of fractions   $\Sigma^{-1} R$, for prime ideals which do not intersect the multiplicative system $\Sigma$.
