Lemma 10.28.9. Let $R$ be a ring. Let $f, g \in R[x]$ be polynomials. Assume the leading coefficient of $g$ is a unit of $R$. There exists elements $r_ i\in R$, $i = 1\ldots , n$ such that the image of $D(f) \cap V(g)$ in $\mathop{\mathrm{Spec}}(R)$ is $\bigcup _{i = 1, \ldots , n} D(r_ i)$.

**Proof.**
Write $g = ux^ d + a_{d-1}x^{d-1} + \ldots + a_0$, where $d$ is the degree of $g$, and hence $u \in R^*$. Consider the ring $A = R[x]/(g)$. It is, as an $R$-module, finite free with basis the images of $1, x, \ldots , x^{d-1}$. Consider multiplication by (the image of) $f$ on $A$. This is an $R$-module map. Hence we can let $P(T) \in R[T]$ be the characteristic polynomial of this map. Write $P(T) = T^ d + r_{d-1} T^{d-1} + \ldots + r_0$. We claim that $r_0, \ldots , r_{d-1}$ have the desired property. We will use below the property of characteristic polynomials that

This was proved in Lemma 10.28.8.

Suppose $\mathfrak q\in D(f) \cap V(g)$, and let $\mathfrak p = \mathfrak q \cap R$. Then there is a nonzero map $A \otimes _ R \kappa (\mathfrak p) \to \kappa (\mathfrak q)$ which is compatible with multiplication by $f$. And $f$ acts as a unit on $\kappa (\mathfrak q)$. Thus we conclude $\mathfrak p \not\in V(r_0, \ldots , r_{d-1})$.

On the other hand, suppose that $r_ i \not\in \mathfrak p$ for some prime $\mathfrak p$ of $R$ and some $0 \leq i \leq d - 1$. Then multiplication by $f$ is not nilpotent on the algebra $A \otimes _ R \kappa (\mathfrak p)$. Hence there exists a prime ideal $\overline{\mathfrak q} \subset A \otimes _ R \kappa (\mathfrak p)$ not containing the image of $f$. The inverse image of $\overline{\mathfrak q}$ in $R[x]$ is an element of $D(f) \cap V(g)$ mapping to $\mathfrak p$. $\square$

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## Comments (1)

Comment #1651 by Lucas Braune on

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