Prime ideals and epimorphism

Question

Let $\phi$:$R$$\rightarrow$$S$ be a ring epimorphism. Show that if $P\triangleleft S$ is a prime ideal (of S), then $\phi^{-1}(P)\triangleleft R$ is a prime ideal (of R).

Can someone help me with that?

Thanks in advance

Answer 1

Expanding on @Daniel Fischer's comment:

$P\subseteq S$ is prime iff $S/P$ is a domain.

Furthermore if $\phi:R\to S$ is epi then $S\cong R/\ker\phi$.

Now $S/P \cong R/(\ker\phi+\phi^{-1}(P))$ but since $0\in P$ and $\ker\phi = \phi^{-1}(0)$, $\ker\phi\subseteq\phi^{-1}(P)$ so $R/(\ker\phi+\phi^{-1}(P)) = R/\phi^{-1}(P)$ which is a domain(because it is isomorphic to a domain), and so $\phi^{-1}(P)$ is prime.

Answer 2

Clearly $\phi^{-1}(P)$ is an ideal. Suppose $ab \in \phi^{-1}(P) $, then $\phi(ab)=\phi(a)\phi(b) \in P$ ; but $P$ is prime so $\phi(a)\in P$ or $\phi(b)\in P$, i.e. $a \in \phi^{-1}(P) $ or $b \in \phi^{-1}(P) $ .

Answer 3

By definition (for me), an ideal $I$ of a commutative ring$~R$ is prime whenever $R/I$ is an integral domain. In the situation of the question the composite morphism $R\overset\phi\to S\to S/P$ has kernel $\phi^{-1}(P)$, so by the first isomorphism theorem the image of the composite morphism is isomorphic to the quotient ring $R/\phi^{-1}(P)$. But that image is a subring of the integral domain $S/P$, and therefore itself an integral domain.

Answer 4

First of all, instead of ring epimorphism, which is a quite delicate notion, probably surjective ring homomorphisms are meant (which should not be confused, as many books do, with ring homomorphisms whose underlying map of sets is an epimorphism). The answers so far didn't use surjectivity and in fact they are incomplete. A prime ideal is, by definition, proper. And in order to ensure that $\phi : R \to S$ pulls back proper ideals (or just subsets) of $S$ to proper ideals (...) of $R$, we need (or rather it suffices) that $\phi$ is surjective. Note that this is automatic if $R,S$ are unital and $\phi$ is unital, since an ideal is proper iff it does not contain $1$.

Also note that in the noncommutative case the definition of a prime ideal is a little bit more complicated, and the answers so far don't deal with this case. So here is the proof (which is, of course, trivial):

Let $\phi : R \to S$ be a surjective homomorphism of rings, $P \subseteq S$ be a prime ideal. Then $\phi^{-1}(P) \subseteq R$ is an ideal (I assume this is already known), which is proper since otherwise $1 \in \phi^{-1}(P)$, i.e. $1 = \phi(1) \in P$, a contradiction to $P \neq S$. Now let $A,B \subseteq R$ ideals of $R$ with $AB \subseteq \phi^{-1}(P)$. Since $\phi$ is surjective, we have that $\phi(A)$ and $\phi(B)$ are ideals of $S$. Clearly we have $\phi(A) \phi(B) \subseteq P$. Since $P$ is prime, it follows that $\phi(A) \subseteq P$ or $\phi(B) \subseteq P$, hence $A \subseteq \phi^{-1}(P)$ or $B \subseteq \phi^{-1}(P)$. QED

In the non-commutative case, without the assumption that $\phi$ is surjective, prime ideals don't have to pull back, see math.SE/399579.