The Stacks project

15.10 Zariski pairs

In this section and the next a pair is a pair $(A, I)$ where $A$ is a ring and $I \subset A$ is an ideal. A morphism of pairs $(A, I) \to (B, J)$ is a ring map $\varphi : A \to B$ with $\varphi (I) \subset J$.

Definition 15.10.1. A Zariski pair is a pair $(A, I)$ such that $I$ is contained in the Jacobson radical of $A$.

Lemma 15.10.2. Let $(A, I)$ be a Zariski pair. Then the map from idempotents of $A$ to idempotents of $A/I$ is injective.

Proof. An idempotent of a local ring is either $0$ or $1$. Thus an idempotent is determined by the set of maximal ideals where it vanishes, by Algebra, Lemma 10.23.1. $\square$

Lemma 15.10.3. Let $(A, I)$ be a Zariski pair. Let $A \to B$ be a flat, integral, finitely presented ring map such that $A/I \to B/IB$ is an isomorphism. Then $A \to B$ is an isomorphism.

Proof. The ring map $A \to B$ is finite by Algebra, Lemma 10.36.5. Hence $B$ is finitely presented as an $A$-module by Algebra, Lemma 10.36.23. Hence $B$ is a finite locally free $A$-module by Algebra, Lemma 10.78.2. Since the module $B$ has rank $1$ along $V(I)$ (see rank function described in Algebra, Lemma 10.78.2), and as $(A, I)$ is a Zariski pair, we conclude that the rank is $1$ everywhere. It follows that $A \to B$ is an isomorphism: it is a pleasant exercise to show that a ring map $R \to S$ such that $S$ is a locally free $R$-module of rank $1$ is an isomorphism (hint: look at local rings). $\square$

Lemma 15.10.4. Let $(A, I)$ be a Zariski pair. Let $A \to B$ be a finite ring map. Assume

  1. $B/IB = B_1 \times B_2$ is a product of $A/I$-algebras

  2. $A/I \to B_1/IB_1$ is surjective,

  3. $b \in B$ maps to $(1, 0)$ in the product.

Then there exists a monic $f \in A[x]$ with $f(b) = 0$ and $f \bmod I = (x - 1)x^ d$ for some $d \geq 1$.

Proof. By Lemma 15.9.10 we can find an ├ętale ring map $A \to A'$ inducing an isomorphism $A/I \to A'/IA'$ such that $B' = B \otimes _ A A'$ contains an idempotent $e'$ lifting the image of $b$ in $B'/IB'$. Consider the corresponding $A'$-algebra decomposition

\[ B' = B'_1 \times B'_2 \]

which is compatible with the one given in the lemma upon reduction modulo $I$. The map $A' \to B'_1$ is surjective modulo $IA'$. By Nakayama's lemma (Algebra, Lemma 10.20.1) we can find $i \in IA'$ such that after replacing $A'$ by $A'_{1 + i}$ the map $A' \to B'_1$ is surjective. Observe that the image $b'_1 \in B'_1$ of $b$ satisfies $b'_1 - 1 \in IB'_1$. Thus we may pick $a' \in IA'$ mapping to $b'_1 - 1$. On the other hand, the image $b'_2 \in B'_2$ of $b$ is in $IB'_2$. By Algebra, Lemma 10.38.4 there exist a monic polynomial $g(x) = x^ d + \sum a'_ j x^ j$ of degree $d$ with $a'_ j \in IA'$ such that $g(b'_2) = 0$ in $B'_2$. Thus the image $b' = (b'_1, b'_2) \in B'$ of $b$ is a root of the polynomial $(x - 1 - a')g(x)$. We conclude that

\[ (b' - 1)(b')^ d \in \sum \nolimits _{j = 0, \ldots , d} IA' \cdot (b')^ j \]

We claim that this implies

\[ (b - 1)b^ d \in \sum \nolimits _{j = 0, \ldots , d} I \cdot b^ j \]

in $B$. For this it is enough to see that the ring map $A \to A'$ is faithfully flat, because the condition is that the image of $(b - 1)b^ d$ is zero in $B/\sum _{j = 0, \ldots , d} Ib^ j$ (use Algebra, Lemma 10.82.11). The map $A \to A'$ flat because it is ├ętale (Algebra, Lemma 10.143.3). On the other hand, the induced map on spectra is open (see Algebra, Proposition 10.41.8 and use previous lemma referenced) and the image contains $V(I)$. Since $I$ is contained in the Jacobson radical of $A$ we conclude. $\square$

Lemma 15.10.5. Let $(A, I)$ be a Zariski pair with $A$ Noetherian. Let $f \in I$. Then $A_ f$ is a Jacobson ring.

Proof. We will use the criterion of Algebra, Lemma 10.61.4. Let $\mathfrak p \subset A$ be a prime ideal such that $\mathfrak p_ f = \mathfrak p A_ f$ is prime and not maximal. We have to show that $A_ f/\mathfrak p_ f = (A/\mathfrak p)_ f$ has infinitely many prime ideals. After replacing $A$ by $A/\mathfrak p$ we may assume $A$ is a domain, $\dim A_ f > 0$, and our goal is to show that $\mathop{\mathrm{Spec}}(A_ f)$ is infinite. Since $\dim A_ f > 0$ we can find a nonzero prime ideal $\mathfrak q \subset A$ not containing $f$. Choose a maximal ideal $\mathfrak m \subset A$ containing $\mathfrak q$. Since $(A, I)$ is a Zariski pair, we see $I \subset \mathfrak m$. Hence $\mathfrak m \not= \mathfrak q$ and $\dim (A_\mathfrak m) > 1$. Hence $\mathop{\mathrm{Spec}}((A_\mathfrak m)_ f) \subset \mathop{\mathrm{Spec}}(A_ f)$ is infinite by Algebra, Lemma 10.61.1 and we win. $\square$

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