Can a non-zero non-prime ideal become prime in a smaller ring?

Question

All rings are assumed commutative and unital.

Some context (feel free to skip right to the questions below). I am trying to understand to what extent the property "being prime" for an ideal $I$ is intrinsic. For example, how robust "primeness" is under the relative point of view. To this end I consider only such ring morphisms that generically do not (need to) modify $I$. For example, if $I \lhd R$ and $R \xrightarrow{\varphi} S$ is any ring morphism, then there are two issues:

• $\varphi$ may not be injective, i.e. it may not preserve $I$ faithfully;
• Even if $\varphi$ is injective, the image $\varphi(I)$ need not be an ideal in $S$, so we need to extend scalars $S \varphi(I)$, which changes $I$ (i.e. the fact that in general the extension of a prime ideal need not be prime is a non-problem in this context).

Therefore, what I want to consider is actually the following setting. Let $R \subset S$ be an extension of rings and let $I \subset R$ such that $I \lhd S$ (hence $I \lhd R$). If $I \in \operatorname{Spec} S$, then clearly $I \in \operatorname{Spec} R$ (special case of the simple fact that preimages of prime ideals are prime), so we have robustness in one direction. I have pondered the converse of this, but I have neither been able to find a counter-example nor to prove it, although it seems too general to be true. So, I want to ask the following question:

(Q1) Suppose that $R \subset S$ is a ring inclusion and $0 \neq I \subset R$ is such that $I \lhd S$ and $I \in \operatorname{Spec} R$. Must $I \in \operatorname{Spec} S$?

(If $I = 0$, this is easily satisfied, e.g. $\mathbb{Z} \subset \mathbb{Z}[t]/(t^2)$.)

A related secondary question is:

(Q2) Is there an example of a ring inclusion $R \subset S$ such that $\dim R > \dim S$? (The idea being that such a ring extension might provide an ideal that is a counter-example to the above.)

Another potential source of a counter-example is probably a ring extension $R \subset S$ with $\dim R = \dim S = \infty$ (e.g. Nagata's example), but I haven't been able to construct a counter-example.

Answer 1

As noted (implicitly) in the comments, it is not very common that $I \subseteq R$ is also an ideal in $S$. For instance, if $R$ and $S$ are domains and $I$ is nonzero, this implies that $\operatorname{Frac} R \to \operatorname{Frac} S$ is an isomorphism, and $R \to S$ is finite if $I$ is finitely generated (e.g. when $R$ is Noetherian).

Indeed, write $K = \operatorname{Frac} R$ and $L = \operatorname{Frac} S$. If $s \in S$ is an element that is $K$-linearly independent from $1$ in $L$ and $x \in I$ is nonzero, then $sx \in L$ cannot be in $K$, contradicting the assumption that $I$ is an $S$-ideal. This proves $K = L$. For the other statement, since $I$ is an $S$-module, we get a ring homomorphism $S \to \operatorname{End}_R(I)$, which is injective since $S$ is a domain and $I$ is nonzero. If $I$ is finitely generated, then so is $\operatorname{End}_R(I)$, hence so is $S$. (I am not really sure if there is an analogue of this in the non-Noetherian case.)

But with these restrictions, you can easily produce examples. For instance, let $S = k[x]$ and let $R$ be the subring $k[x^2,x^3] \cong k[u,v]/(u^3-v^2)$. The ideal $I = (x^2,x^3) \subseteq k[x]$ is contained in $R$ (hence also an $R$-ideal). But $I$ is not prime in $k[x]$, whereas it is prime in $k[x^2,x^3]$ (as it corresponds to the origin $(u,v)$ in $k[u,v]/(u^3-v^2)$).

Answer 2

Since you didn't ask that $S$ be a domain, here is a very simple example, although it might feel as cheating and @R.vanDobbendeBruyn's example is excellent. The only value in this one is perhaps its simplicity.

Anyway, let $R$ be an arbitrary ring with a prime ideal $I\triangleleft R$ and $R'$ another arbitrary (non-zero) ring. Finally, let $S=R\oplus R'$. As $I$ is a prime ideal, $R/I\neq 0$, so $S/I\simeq R/I\oplus R'$ is not a domain, so $I$ cannot be a prime ideal in $S$.

As pointed out by @M.G., this example does not really work, because $R$ is only a subring of $S$ as non-unital rings. So, here is an improvement of (sort of) the same idea:

Let $(R,\mathfrak m,k)$ be a local ring and let $S=R[t] \big /\!\left((t^2)+t\cdot \mathfrak m\right)\simeq R[\varepsilon]$ where $\varepsilon^2=0$ and $\varepsilon\cdot\mathfrak m=0$. Then

1. $R\subseteq S$ is a subring, and
2. $\mathfrak m$ is also an ideal in $S$, but
3. $S/\mathfrak m\simeq k[t]/(t^2)\simeq k[\varepsilon]$, so $\mathfrak m$ is not prime in $S$.