# Proof of going-up theorem

Can you tell me if my proof is correct? Thanks. (I'm using propositions 5.6 and 5.10 from Atiyah-Macdonald which I proved separately.)

Theorem: Let $R$ be integral over $S$. Let $p_1 \subset p_2$ be prime in $S$. Let $q_1$ be prime in $R$ such that $q_1 \cap S = p_1$. Then there exists a prime ideal $q$ in $R$ such that $q \cap S = p_2$.

Proof:

Since $R$ is integral over $S$ and $q_1$ an ideal such that $q_1 \cap S = p_1$ we know by prop. 5.6. i) p. 61 in Atiyah-Macdonald that $R/q_1$ is integral over $S/p_1$.

Since $p_1 \subset p_2$ and $p_2$ prime in $S$, $p_2 / p_1$ is prime in $S/p_1$. Hence by prop. 5.10 p. 62 we know that there exists a prime ideal $\bar{q}$ in $R/q_1$ such that $\bar{q} \cap S/p_1 = p_2 / p_1$.

Let $\pi : R \to R/q_1$ be the projection. Then we claim that $\pi^{-1} (\bar{q})$ is prime in $R$, contains $q_1$ and $\pi^{-1} (\bar{q}) \cap S = p_2$.

(1) Since $\bar{q}$ is prime and $\pi$ is a ring homomorphism we know that $\pi^{-1}(\bar{q})$ is prime.

(2) Since $\bar{q}$ is an ideal it contains $0$. Since $\pi^{-1}(0) = q_1$ its inverse image contains $q_1$.

(3) To finish the proof we need to show $\pi^{-1} (\bar{q}) \cap S = p_2$. For this we apply $\pi^{-1}$ to $\bar{q} \cap S/p_1 = p_2 / p_1$ to get $$\pi^{-1}(\bar{q}) \cap \pi^{-1}(S/p_1) = \pi^{-1}(p_2 / p_1)$$

We remember that $S / p_1 = S / (S \cap q_1)$. We can embed $S / (S \cap q_1)$ into $R/q_1$ with $i: R/(S \cap q_1) \to R / q_1$, $r/s \mapsto r/s$. Then $i(S/(S \cap q_1)) = S/q_1$. Hence $\pi^{-1} (S/(S \cap q_1)) = \pi^{-1} (S/q_1) = S$ and similarly $\pi^{-1}(p_2 / p_1) = p_2$. Hence $\pi^{-1}(\bar{q}) \cap S = p_2$.

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It is correct, but the last part is too complicated. The ideal $\overline{q}$ can be written as $q/q_1$, where $q$ is a prime ideal of $R$ containing $q_1$, and now all follows easily. – user26857 Jul 7 '12 at 8:38