# 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 $\Gamma(V)$.

• One additional example is the ideal generated by the polynomial $x^2+1$ considered $\mathbb{R}[x]$ and $\mathbb{C}[x]$. I suggest you search for the term “geometrically irreducible”. Dec 21 '17 at 3:43
• Extensions of prime ideals pretty much never remain prime. If $A \subseteq B$ are rings, and $\mathfrak p$ is a prime of $A$, then $\mathfrak p B$ is prime in $B$ if and only if $B/\mathfrak p B \cong B \otimes_A A/\mathfrak p$ is an integral domain, which it seldom is, even if $A$ and $B$ are integral domains.
– D_S
Dec 21 '17 at 5:43
• @D_S, please, what if $p$ is a maximal ideal of $A$? In that case $B/pB = B \otimes_A k$; is there something interesting that can be said about this tensor product? When is it an integral domain? Mar 10 at 22:42
• If you ask a new question I'd be happy to answer. It is of course an integral domain if and only if $pB$ is a prime ideal of $B$.
– D_S
Mar 10 at 22:51
• @D_S, thank you! I will ask a new question. Mar 10 at 23:21

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 $(5) = (1+2\cdot i)\cdot (1-2\cdot i)$

• I had thought of the first counterexample myself but i ignored it thinking that the whole ring is still a prime idea, just not a proper one. The second counterexample though was persuasive enough! Thank you for the fast answer
– fhn
Dec 21 '17 at 1:05

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 $\Sigma$.

• Thank you very much for your answer. Is that result/proposition you mentioned (about when p remains prime in S if-f...) from algebraic number theory? I am not yet familiar with the subject (still undergrad), so could you maybe provide a link or a book for a proof on that result so that i can get a taste of what it s like? I m trying to get a feeling why it works in the case I mentioned or the last cases you mentioned, but not on the first ones with R=Z.
– fhn
Dec 21 '17 at 1:25
• Yes, it is a classic result from algebraic number theory. Unfortunately, Wikipedia doesn't give any proof, and this result is mentioned only in the French page. For me a fine reference is the small book by Pierre Samuel, Algebraic theory of numbers (Dover Books). Dec 21 '17 at 2:05