
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 10.19.1) 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 10.37.4 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 10.81.11). The map $A \to A'$ flat because it is étale (Algebra, Lemma 10.141.3). On the other hand, the induced map on spectra is open (see Algebra, Proposition 10.40.8 and use previous lemma referenced) and the image contains $V(I)$. Since $I$ is contained in the Jacobson radical of $A$ we conclude. $\square$

Comment #3634 by Brian Conrad on

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