# Surjective ring homomorphism

So I read an interesting theorem about surjective ring homomorphisms and I have two questions that I cannot answer myself.

Motivation: A ring hom is surjective iff it send ideals to ideas.

It turns out that a surjective ring hom also sends prime ideals to prime ideals. I was wondering would the converse be true? I cannot think of any elementary examples.

Another question I had in mind is that could a surjective ring homomorphism send a maximal ideal to a prime ideal that is not maximal? I cannot seem to think of an example either.

• The embedding of $\mathbb{Q}$ into $\mathbb{R}$ sends the unique prime ideal of $\mathbb{Q}$, namely $(0)$, to the unique prime ideal of $\mathbb{R}$ (namely, $(0)$). Commented Dec 15, 2022 at 19:55
• @ArturoMagidin That's a nice example, it's always easy to forget the basics Commented Dec 15, 2022 at 19:57
• As to your second question, no: the Lattice Isomorphism Theorem (or 4th Isomorphism Theorem) tells you that if $f\colon R\to S$ is a surjective ring homomorphism, then this induces an inclusion-preserving correspondence between ideals of $R$ that contain $\ker(f)$ and ideals of $S$. If $M$ is a maximal ideal of $R$ that contains $\ker(f)$, then it is mapped to a maximal ideal of $S$. If $M$ does not contain $\ker(f)$, then it is mapped to the same thing as $M+\ker(f)=R$, which is all of $S$ which is not prime. Commented Dec 15, 2022 at 19:57
• The image of $M$ is isomorphic to $M/(M\cap\ker(f))$. By the Second (or Third, depending on your numbering) Isomorphism Theorem, this essentially the same as $(M+\ker(f))/\ker(f)$. If $M$ does not contain $\ker(f)$, but is maximal, then since $M+\ker(f)$ is an ideal that properly contains $M$, it must be all of $R$. Commented Dec 15, 2022 at 20:07
• @ArturoMagidin Ah! I see! That's brilliant - thanks a lot! Feel free making this into an answer :) Commented Dec 15, 2022 at 20:10

An easy example of a non-surjective ring morphism that sends prime ideals to prime ideals is the embedding of a field into any extensions field: the only prime ideal of the field is $$(0)$$, which of course is mapped to the only prime ideal of the larger field. For example, $$\mathbb{Q}\hookrightarrow \mathbb{R}$$.
No. If $$f\colon R\to S$$ is a surjective ring homomorphism, and $$M$$ is a maximal ideal of $$R$$, then either $$f(M)$$ is maximal in $$S$$, or $$f(M)=S$$. To see this, note that the Lattice Isomorphism Theorem establishes an inclusion-preserving correspondence between ideals of $$S$$ and ideals of $$R$$ that contain $$\ker(f)$$. If $$M$$ contains $$\ker(f)$$, then because it is maximal in $$R$$ it follows by this correspondence that its image is maximal in $$S$$. If $$M$$ does not contain $$\ker(f)$$, then since $$M$$ is maximal and $$M\lt M+\ker(f)\leq R$$ with $$M+\ker(f)$$ an ideal, it follows that $$M+\ker(f)=R$$. Thus, $$f(M) = f(M+\ker(f)) = f(R) = S$$. So the image of $$M$$ is either maximal or the whole ring; it cannot be a non-maximal prime, if $$f$$ is surjective.