# Let $\varphi : R \to R'$ be a surjective ring homomorphism. Let $I \subset R'$ be a maximal ideal. Then show that $\varphi^{-1}(I)$ is maximal ideal.

Let $$\varphi : R \to R'$$ be a surjective ring homomorphism. Let $$I \subset R'$$ be a maximal ideal. Then show that $$\varphi^{-1}(I)$$ is maximal ideal.

This question already has answer, but I couldn't follow the hint. And I want to solve the question using correspondence theorem.

So As $$\varphi$$ is a surjective ring homomorphism so by first isomorphism theorem we have $$R/ker \varphi \cong R'$$. Now from the correspondence theorem, there exist a inclusion preserving Bijection map between $$\{$$maximal ideals of $$R$$ that contains $$ker\varphi$$ $$\}$$ and $$\{$$maximal ideals of $$R'$$ $$\}$$.

Now $$I$$ is a maximal ideal of $$R'$$.now how to proceed? Thanks.

• Existence of bijections is not really enough, are you aware that you can write them down more or less explicitly? Jun 4, 2022 at 15:01
• @MatthiasKlupsch I think I am not aware. Please elaborate Jun 4, 2022 at 15:11

## 1 Answer

What @Matthias Klupsch is trying to hint at is that the correspondence you are quoting can be written down explicitly. I.e. the bijection is given as follows

$$\{\mathfrak{m}\subset R: \mathfrak{m} \text{ is max. ideal s.t.} \ker \phi \subset \mathfrak{m}\} \leftrightarrow \{ \mathfrak{n}\subset R/\ker \phi: \mathfrak{n} \text{ is max. ideal}\}$$ Let $$\pi: R\to R/\ker \phi$$ be the projection. Then the correspondence above is given by sending $$\mathfrak{m}$$ on the LHS to its image under $$\pi$$ an $$\mathfrak{n}$$ on the RHS to $$\pi^{-1}(\mathfrak{n})$$. Can you finish from here?

Response to comment: I corrected an error in the original answer. Maybe try again:)

• As $R' \cong R/ker \varphi$, so $I \subset R'$ is maximal in $R/ker \varphi$, so from the correspondence theorem you mentioned, $\varphi^{-1}(I)$ is maximal suchthat it contains $ker \varphi$. Jun 4, 2022 at 16:13
• Is this complete the proof? Jun 4, 2022 at 16:31