Lemma 33.39.1. Let $(A, \mathfrak m)$ be a Noetherian $1$-dimensional local ring. Let $f \in \mathfrak m$. The following are equivalent
$\mathfrak m = \sqrt{(f)}$,
$f$ is not contained in any minimal prime of $A$, and
$A_ f = \prod _{\mathfrak p\text{ minimal}} A_\mathfrak p$ as $A$-algebras.
Such an $f \in \mathfrak m$ exists. If $\text{depth}(A) = 1$ (for example $A$ is reduced), then (1) – (3) are also equivalent to
$f$ is a nonzerodivisor,
$A_ f$ is the total ring of fractions of $A$.
If $A$ is reduced, then (1) – (5) are also equivalent to
$A_ f$ is the product of the residue fields at the minimal primes of $A$.
Proof.
The spectrum of $A$ has finitely many primes $\mathfrak p_1, \ldots , \mathfrak p_ n$ besides $\mathfrak m$ and these are all minimal, see Algebra, Lemma 10.31.6. Then the equivalence of (1) and (2) follows from Algebra, Lemma 10.17.2. Clearly, (3) implies (2). Conversely, if (2) is true, then the spectrum of $A_ f$ is the subset $\{ \mathfrak p_1, \ldots , \mathfrak p_ n\} $ of $\mathop{\mathrm{Spec}}(A)$ with induced topology, see Algebra, Lemma 10.17.5. This is a finite discrete topological space. Hence $A_ f = \prod _{\mathfrak p\text{ minimal}} A_\mathfrak p$ by Algebra, Proposition 10.60.7. The existence of an $f$ is asserted in Algebra, Lemma 10.60.8.
Assume $A$ has depth $1$. (This is the maximum by Algebra, Lemma 10.72.3 and holds if $A$ is reduced by Algebra, Lemma 10.157.3.) Then $\mathfrak m$ is not an associated prime of $A$. Every minimal prime of $A$ is an associated prime (Algebra, Proposition 10.63.6). Hence the set of nonzerodivisors of $A$ is exactly the set of elements not contained in any of the minimal primes by Algebra, Lemma 10.63.9. Thus (4) is equivalent to (2). Part (5) is equivalent to (3) by Algebra, Lemma 10.25.4.
Then $A_\mathfrak p$ is a field for $\mathfrak p \subset A$ minimal, see Algebra, Lemma 10.25.1. Hence (3) is equivalent to (6).
$\square$
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