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I've found this claim:

Let $A$ be a Noetherian ring of Krull dimension $0$ . Then $A$ is a field or it has a finite number of prime ideals.

Why is this true ?

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Noetherian Krull dimension $0$ is equivalent to Artinian and Artinian rings have finitely many maximal ideals (prime ideals in a krull dimension $0$ ring are always maximal). – Seth Apr 27 '14 at 14:59
Also a field has a finite number of prime ideals, so why distinguishing the two cases? – egreg Apr 27 '14 at 15:02
up vote 2 down vote accepted

If $A$ a Noetherian ring of Krull dimension $0$, it implies that $A$ an Artinian ring and that every prime ideal is maximal.

So you only need to check that $A$ has finitely many maximal ideal:

Consider the set $\Sigma$ of all finite intersections of maximal ideals. $\Sigma$ is non-empty and thus has a minimal element $\mathfrak m_0=\bigcap_{i=0}^n\mathfrak m_i$ (because $A$ is Artinian).

Let $\mathfrak{m}$ be a maximal ideal of $A$, then $\mathfrak m \cap \mathfrak m_0\subset \mathfrak m_0$ and these elements are in $\Sigma$. By minimality of $\mathfrak{m}_0$, you get $$\mathfrak m\cap \mathfrak m_0 = \mathfrak m_0.$$ Hence, $\mathfrak m$ must contain one of the $\mathfrak m_i$ and since these ideals are maximal, it must be equal to one of the $\mathfrak m_i$.

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