The Stacks project

Lemma 86.14.1. Let $A$ be a Noetherian adic topological ring. Let $\mathfrak q \subset A$ be a prime ideal. The following are equivalent

  1. for some ideal of definition $I \subset A$ we have $I \not\subset \mathfrak q$ and $\mathfrak q$ is maximal with respect to this property,

  2. for some ideal of definition $I \subset A$ the prime $\mathfrak q$ defines a closed point of $\mathop{\mathrm{Spec}}(A) \setminus V(I)$,

  3. for any ideal of definition $I \subset A$ we have $I \not\subset \mathfrak q$ and $\mathfrak q$ is maximal with respect to this property,

  4. for any ideal of definition $I \subset A$ the prime $\mathfrak q$ defines a closed point of $\mathop{\mathrm{Spec}}(A) \setminus V(I)$,

  5. $\dim (A/\mathfrak q) = 1$ and for some ideal of definition $I \subset A$ we have $I \not\subset \mathfrak q$,

  6. $\dim (A/\mathfrak q) = 1$ and for any ideal of definition $I \subset A$ we have $I \not\subset \mathfrak q$,

  7. $\dim (A/\mathfrak q) = 1$ and the induced topology on $A/\mathfrak q$ is nontrivial,

  8. $A/\mathfrak q$ is a $1$-dimensional Noetherian complete local domain whose maximal ideal is the radical of the image of any ideal of definition of $A$, and

  9. add more here.

Proof. It is clear that (1) and (2) are equivalent and for the same reason that (3) and (4) are equivalent. Since $V(I)$ is independent of the choice of the ideal of definition $I$ of $A$, we see that (2) and (4) are equivalent.

Assume the equivalent conditions (1) – (4) hold. If $\dim (A/\mathfrak q) > 1$ we can choose a maximal ideal $\mathfrak q \subset \mathfrak m \subset A$ such that $\dim ((A/\mathfrak q)_\mathfrak m) > 1$. Then $\mathop{\mathrm{Spec}}((A/\mathfrak q)_\mathfrak m) - V(I(A/\mathfrak q)_\mathfrak m)$ would be infinite by Algebra, Lemma 10.61.1. This contradicts the fact that $\mathfrak q$ is closed in $\mathop{\mathrm{Spec}}(A) \setminus V(I)$. Hence we see that (6) holds. Trivially (6) implies (5).

Conversely, assume (5) holds. Let $I \subset A$ be an ideal of definition. Since $A/\mathfrak q$ is complete with respect to $I(A/\mathfrak q)$ (for example by Algebra, Lemma 10.97.1) we see that all closed points of $\mathop{\mathrm{Spec}}(A/\mathfrak q)$ are contained in $V(IA/\mathfrak q)$ by Algebra, Lemma 10.96.6. Since $\dim (A/\mathfrak q) = 1$ and since $I \not\subset \mathfrak q$ we conclude two things: (a) $V(IA/\mathfrak q)$ must contain all points distinct from the generic point of $\mathop{\mathrm{Spec}}(A/\mathfrak q)$, and (b) $V(IA/\mathfrak q)$ must be a (finite) discrete set. From (a) we see that $\mathfrak q$ is a closed point of $\mathop{\mathrm{Spec}}(A) \setminus V(I)$ and we conclude that (2) holds.

Continuing to assume (5) we see that the finite discrete space $V(IA/\mathfrak q)$ must be a singleton by More on Algebra, Lemma 15.11.16 for example (and the fact that complete pairs are henselian pairs, see More on Algebra, Lemma 15.11.4). Hence we see that (8) is true. Conversely, it is clear that (8) implies (5).

At this point we know that (1) – (6) and (8) are equivalent. We omit the verification that these are also equivalent to (7). $\square$


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