## 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 étale 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 étale (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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