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let A be a ring , I shall prove that :

($ Nil(A)=0 $ and Every prime ideal of $A$ is maximal) $\Rightarrow$ ($A_m$ is a field for each maximal ideal $m$ )

where $ Nil(A) $ is the nilradical of A

now reading a proof somewhere , it said :

All primes of $A$ are maximal $\Rightarrow$ $\forall m$ maximal , no prime ideal of $A$ are strictly between $m$ and $(0)$

$\Rightarrow$ $\forall m$ maximal , the only prime of $A$ contained in $m$ is $(0)$

$\Rightarrow$ $\forall m$ maximal , the only prime of $A_m$ is $(0)$

$\Rightarrow$ $\forall m$ maximal , $A_m$ is a field

this proof is obviously wrong , it doesn't even use the fact that Nil(A)=0 , I tired to show this myself but I get stuck at some point when I need to prove that $A_m$ is an integral domain . any clear proof would be appreciated .

Laassila souhayl
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2 Answers2

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I'd tend to think of this as a consequence of the fact that the localization of reduced ring at a minimal prime is always a field. When you apply this to a reduced ring in which every prime is maximal (hence minimal too), you get the desired result.

Lemma Let $R$ be a ring. The following are equivalent:

(1) $R$ has exactly one prime ideal.

(2) Every element of $R$ is either nilpotent or a unit.

Proof: $(1) \implies (2)$ Follows immediately from the fact that the nilradical of a ring is the intersection of its prime ideals. $(2) \implies (1)$ The non-units of $R$ are by assumption closed under addition, so $R$ is local with maximal ideal $\mathfrak{m}$. Since $\mathfrak{m}$ consists of nilpotents, we have $\mathfrak{m} \subseteq \operatorname{Nil}(R) = \bigcap\limits_{\mathfrak{p} \in \operatorname{Spec}(R)} \mathfrak{p} \subseteq \mathfrak{p}$ for every prime ideal $\mathfrak{p}$.

Corollary Let $R$ be a reduced ring (i.e. $\operatorname{Nil}(R) = 0)$ and let $\mathfrak{p}$ be a minimal prime of $R$. Then $R_\mathfrak{p}$ is a field.

Proof: The primes of $R_\mathfrak{p}$ are in one-to-one correspondence with the primes of $R$ contained in $\mathfrak{p}$, so by the previous fact every element of $R_\mathfrak{p}$ is either a unit or nilpotent. Note that a localization of a reduced ring is still reduced, so $R_\mathfrak{p}$ contains only units, i.e. it's a field.

Community
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Badam Baplan
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$A_m$ is local since $m$ is a prime ideal. $A_m$ is also trivially absolutely flat. (Just use the element characterization.)

But a local absolutely flat ring is a field, since idempotents are trivial in local rings.

rschwieb
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