Lemma 10.50.17. Let $A$ be a valuation ring. Ideals in $A$ correspond $1 - 1$ with ideals of $\Gamma$. This bijection is inclusion preserving, and maps prime ideals to prime ideals.

Proof. Omitted. $\square$

Comment #8760 by Zhenhua Wu on

the zero prime ideal of $A$ doesn't correspond to any ideal of $\Gamma$ because by the definition here we don't allow zero ideal in $\Gamma$, unless you allow $\emptyset$ to be an ideal of $\Gamma$.

Comment #8874 by Zhenhua Wu on

Sorry for the last comment. Actually from the definition of ideals of $\Gamma$ we can see that $\emptyset,\Gamma_{> 0}$ and $\Gamma_{\geq0}$ can all be ideals. They correspond to $\{0\},\mathfrak{m}$ and $A$ respectively. But $A$ is not a prime ideal, so we shouldn't allow $\Gamma_{\geq0}$ to be a prime ideal of $\Gamma$.

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