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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.

Alexander
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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:)

nilsw
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  • 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 $ker \varphi$. – Alexander Jun 04 '22 at 16:13
  • Is this complete the proof? – Alexander Jun 04 '22 at 16:31