Lemma 15.9.9. Let $I$ be an ideal of a ring $A$. Let $A \to B$ be an integral ring map. Let $b \in B$ map to an idempotent in $B/IB$. Then there exists a monic $f \in A[x]$ with $f(b) = 0$ and $f \bmod I = x^ d(x - 1)^ d$ for some $d \geq 1$.

Proof. Observe that $z = b^2 - b$ is an element of $IB$. By Algebra, Lemma 10.38.4 there exist a monic polynomial $g(x) = x^ d + \sum a_ j x^ j$ of degree $d$ with $a_ j \in I$ such that $g(z) = 0$ in $B$. Hence $f(x) = g(x^2 - x) \in A[x]$ is a monic polynomial such that $f(x) \equiv x^ d(x - 1)^ d \bmod I$ and such that $f(b) = 0$ in $B$. $\square$

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