# Counting irreducible components

*Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).*

The logarithm rank lemma of the next lesson compares the number of irreducible components of a zero set in a torus with the number of its translates. It needs one geometric bound: a common zero set of polynomials of degree at most \(\delta\) in \(n\) variables has at most \(\delta^n\) irreducible components. This is a form of Bézout's inequality; the polynomials \(x_i(x_i-1)\cdots(x_i-\delta+1)\), \(1\le i\le n\), show that it is sharp. This lesson proves it from Hilbert polynomials of graded modules, following the weighted count indicated in [OpenAI-UC, proof of Lemma 3.1]: assign to each component its degree times \(\delta\) to the power of its dimension, and check that this weight does not increase when one more equation is imposed.

- Closed sets, components and dimension (Section 1).
- Homogeneous ideals and graded filtrations (Section 2).
- Hilbert polynomials, dimension and degree (Section 3).
- Cutting by a hypersurface (Section 4).
- The component bound (Section 5).

## 1. Closed sets, components and dimension

Let \(k\) be an algebraically closed field, \(R=k[x_1,\ldots,x_n]\), and give \(k^n\) the Zariski topology, whose closed sets are the zero sets \(V(I)\) of ideals \(I\subset R\). By the strong Nullstellensatz (Theorem 2.2 of [The Nullstellensatz and Jacobson rings](course:AG-CA/the-nullstellensatz-and-jacobson-rings)), the polynomials vanishing on \(V(I)\) form the radical \(\sqrt I\). Hence \(I\mapsto V(I)\) is an inclusion-reversing bijection between radical ideals and closed sets, under which prime ideals correspond to irreducible closed sets. A Noetherian ring has finitely many minimal primes over an ideal (Theorem 3.2 of [Associated primes and primary decomposition](course:AG-CA/associated-primes-and-primary-decomposition)), and \(V(I)\) is the union of the \(V(\mathfrak p)\) for the minimal primes \(\mathfrak p\) of \(I\). These are its *irreducible components*: the maximal irreducible closed subsets. So the number of irreducible components of \(V(I)\) is the number of minimal primes of \(I\).

The *dimension* of an irreducible closed set \(V(\mathfrak p)\) is the Krull dimension \(\dim R/\mathfrak p\), the supremum of lengths of chains of primes; equivalently, of chains of irreducible closed subsets (Proposition 1.2 of [Krull dimension and Noether normalization](course:AG-CA/krull-dimension-and-noether-normalization)). The dimension of a closed set is the largest dimension of its components. A proper irreducible closed subset of an irreducible closed set \(Y\) has smaller dimension than \(Y\), since a chain inside it can be extended by \(Y\).

The *torus* is the open subset \(\mathbb G_m^n=\{x\in k^n\mid x_1\cdots x_n\ne0\}\) with the subspace topology. Its closed subsets are the sets \(Z\cap\mathbb G_m^n\) with \(Z\) closed in \(k^n\). A nonempty open subset of an irreducible space is irreducible and dense. Consequently the irreducible components of \(Z\cap\mathbb G_m^n\) are the sets \(C\cap\mathbb G_m^n\) for the components \(C\) of \(Z\) meeting \(\mathbb G_m^n\), different \(C\) giving different sets, and the dimension of \(C\cap\mathbb G_m^n\) is defined as that of its closure \(C\). In particular \(Z\cap\mathbb G_m^n\) has at most as many components as \(Z\). Since \(k\) is infinite, a nonzero polynomial does not vanish on all of \(\mathbb G_m^n\), so \(\mathbb G_m^n\) is dense in \(k^n\) and has dimension \(n\).

## 2. Homogeneous ideals and graded filtrations

Let \(S=k[x_0,\ldots,x_n]\), graded by total degree, and \(\mathfrak m=(x_0,\ldots,x_n)\). An ideal is *homogeneous* if it contains the homogeneous components of its elements, equivalently if it is generated by homogeneous elements. Sums, products and intersections of homogeneous ideals are homogeneous, and so are radicals (Exercise 6.5). A graded \(S\)-module \(M=\bigoplus_tM_t\) is finitely generated by homogeneous elements in our applications, and \(M(c)\) denotes the shift with \(M(c)_t=M_{c+t}\).

**Lemma 2.1.**

1. A homogeneous ideal \(\mathfrak q\ne S\) is prime as soon as \(ab\in\mathfrak q\) implies \(a\in\mathfrak q\) or \(b\in\mathfrak q\) for homogeneous \(a,b\).
2. The annihilator of a homogeneous element of a graded module is a homogeneous ideal.

*Proof.* (1) Let \(ab\in\mathfrak q\) with \(a,b\notin\mathfrak q\). Removing from \(a\) and \(b\) their homogeneous components lying in \(\mathfrak q\) changes \(ab\) by an element of \(\mathfrak q\), so we may assume that no component of \(a\) or \(b\) lies in \(\mathfrak q\). Let \(a_i\), \(b_j\) be the components of highest degree. The component of degree \(i+j\) of \(ab\) is \(a_ib_j\), which lies in \(\mathfrak q\) because \(\mathfrak q\) is homogeneous; so \(a_i\in\mathfrak q\) or \(b_j\in\mathfrak q\), a contradiction. (2) If \(sm=0\) with \(m\) homogeneous of degree \(c\), the component of degree \(t+c\) of \(sm\) is \(s_tm\), so every \(s_t\) annihilates \(m\). \(\square\)

**Proposition 2.2** (graded prime filtration). Let \(J\subset S\) be a homogeneous ideal. The graded module \(S/J\) has a filtration by graded submodules
\[
\begin{gathered}
0=M_0\subsetneq M_1\subsetneq\cdots\subsetneq M_r=S/J,\\
M_i/M_{i-1}\cong(S/\mathfrak q_i)(-c_i),
\end{gathered}
\]
with homogeneous primes \(\mathfrak q_i\supseteq J\) and integers \(c_i\ge0\). Every minimal prime of \(J\) occurs among the \(\mathfrak q_i\); in particular the minimal primes of \(J\) are homogeneous.

*Proof.* Let \(N\ne0\) be a graded quotient \((S/J)/M_{i-1}\). Among the annihilators of nonzero homogeneous elements of \(N\) choose a maximal one, \(\mathfrak q=\operatorname{Ann}(m)\); it exists because \(S\) is Noetherian. It is homogeneous by Lemma 2.1(2), and proper. Let \(a,b\) be homogeneous with \(ab\in\mathfrak q\) and \(b\notin\mathfrak q\). Then \(bm\) is a nonzero homogeneous element with \(\operatorname{Ann}(bm)\supseteq\mathfrak q\), so equality holds by maximality, and \(a\in\operatorname{Ann}(bm)=\mathfrak q\). By Lemma 2.1(1), \(\mathfrak q\) is prime. The submodule \(Sm\) is graded and isomorphic to \((S/\mathfrak q)(-\deg m)\). Let \(M_i\) be the preimage of \(Sm\) in \(S/J\). The ascending chain of submodules of the Noetherian module \(S/J\) stops, which happens only when \(M_r=S/J\). Each \(\mathfrak q_i\) contains \(J\), which annihilates \(S/J\).

Let \(\mathfrak p\) be a minimal prime of \(J\). Localizing the filtration at \(\mathfrak p\) gives a filtration of \((S/J)_{\mathfrak p}\ne0\) with quotients \((S/\mathfrak q_i)_{\mathfrak p}\), so some \((S/\mathfrak q_i)_{\mathfrak p}\ne0\), that is, \(\mathfrak q_i\subseteq\mathfrak p\). As \(J\subseteq\mathfrak q_i\), minimality gives \(\mathfrak q_i=\mathfrak p\). \(\square\)

## 3. Hilbert polynomials, dimension and degree

For a finitely generated graded \(S\)-module \(M\) put \(H_M(t)=\dim_kM_t\). By the Hilbert–Serre theorem (Theorem 3.1 of [Graded modules and Hilbert–Samuel functions](course:AG-CA/graded-modules-and-hilbert-samuel-functions)), \(H_M(t)\) agrees for large \(t\) with a numerical polynomial, also written \(H_M\). By Lemma 2.1 there, a numerical polynomial of degree \(d\) is an integer combination of the binomial polynomials \(\binom ti\), \(i\le d\), so its leading coefficient times \(d!\) is an integer.

**Definition 3.1.** For a homogeneous ideal \(J\subsetneq S\) with \(H_{S/J}\) not eventually zero, let \(d(J)\) be the degree of \(H_{S/J}\) and \(e(J)\) the integer \(d(J)!\) times its leading coefficient. As \(H_{S/J}(t)\ge0\), \(e(J)\ge1\). If \(H_{S/J}\) is eventually zero, put \(d(J)=-1\) and \(e(J)=0\).

For example \(H_S(t)=\binom{t+n}n\), so \(d(0)=n\) and \(e(0)=1\).

**Proposition 3.2.** For a homogeneous prime \(\mathfrak q\subset S\),
\[
d(\mathfrak q)=\dim S/\mathfrak q-1 .
\]

*Proof.* Put \(B=S/\mathfrak q\) and let \(\bar{\mathfrak m}\) be the image of \(\mathfrak m\). Since \(B\) is generated over \(k=B_0\) by elements of degree one, \(\bar{\mathfrak m}^{\,t}=\bigoplus_{i\ge t}B_i\), so \(B/\bar{\mathfrak m}^{\,t}=\bigoplus_{i<t}B_i\). This ring is local with nilpotent maximal ideal, so it equals its localization at \(\bar{\mathfrak m}\), and
\[
\ell\bigl(B_{\bar{\mathfrak m}}/\bar{\mathfrak m}^{\,t}B_{\bar{\mathfrak m}}\bigr)=\sum_{i<t}H_B(i).
\]
For large \(t\) the right side is a polynomial of degree \(d(\mathfrak q)+1\): summing a numerical polynomial of degree \(d\ge0\) raises the degree by one, and an eventually zero function has eventually constant sums. By the local dimension theorem (Theorem 2.1 of [Dimension theory of Noetherian local rings](course:AG-CA/dimension-theory-of-noetherian-local-rings)), this degree is \(\dim B_{\bar{\mathfrak m}}\), the height of \(\bar{\mathfrak m}\) in \(B\). By the height formula (Theorem 4.3 of the Krull dimension lesson) for the finite-type domain \(B\), this height is \(\dim B-\dim B/\bar{\mathfrak m}=\dim B\), since \(B/\bar{\mathfrak m}=k\). \(\square\)

So \(\mathfrak q=\mathfrak m\) is the only homogeneous prime with \(d(\mathfrak q)=-1\), and \(d(\mathfrak q)\ge0\) for every other homogeneous prime.

**Corollary 3.3.** Let \(J\) be a homogeneous ideal, and let \(\mathfrak q_1,\ldots,\mathfrak q_r\) be as in Proposition 2.2. Then \(d(\mathfrak q_i)\le d(J)\) for all \(i\), and
\[
e(J)=\sum_{i:\,d(\mathfrak q_i)=d(J)}e(\mathfrak q_i).
\]
In particular \(e(J)\ge\sum e(\mathfrak p)\) over the minimal primes \(\mathfrak p\) of \(J\) with \(d(\mathfrak p)=d(J)\).

*Proof.* Taking graded pieces of the filtration and adding dimensions gives
\[
H_{S/J}(t)=\sum_{i=1}^rH_{S/\mathfrak q_i}(t-c_i)
\]
for all \(t\). All terms are nonnegative, so no term grows faster than the sum, and \(d(\mathfrak q_i)\le d(J)\). A shift \(t\mapsto t-c\) does not change the degree or the leading coefficient. Comparing the coefficients of \(t^{d(J)}\) gives the formula. Each minimal prime occurs among the \(\mathfrak q_i\) by Proposition 2.2. \(\square\)

## 4. Cutting by a hypersurface

**Proposition 4.1.** Let \(\mathfrak p\subset S\) be a homogeneous prime with \(\dim S/\mathfrak p\ge1\), and let \(F\notin\mathfrak p\) be homogeneous of degree \(\delta\ge1\). Then every minimal prime \(\mathfrak q\) of \(\mathfrak p+(F)\) satisfies \(\dim S/\mathfrak q=\dim S/\mathfrak p-1\), and
\[
\sum_{\mathfrak q}e(\mathfrak q)\le\delta\,e(\mathfrak p),
\]
the sum running over the minimal primes of \(\mathfrak p+(F)\) different from \(\mathfrak m\).

*Proof.* In the domain \(B=S/\mathfrak p\) the image \(\bar F\) is nonzero. A minimal prime \(\mathfrak q\) of \(\mathfrak p+(F)\) gives a prime \(\mathfrak q/\mathfrak p\) of \(B\) minimal over \((\bar F)\). Its height is at most \(1\) by Krull's principal ideal theorem (Theorem 3.1 of the dimension theory lesson), and at least \(1\) because \(\bar F\ne0\) in a domain. The height formula gives \(\dim S/\mathfrak q=\dim B-1\).

Multiplication by \(\bar F\) is injective on the domain \(B\), so there is an exact sequence of graded modules
\[
\begin{gathered}
0\longrightarrow B(-\delta)\xrightarrow{\ \bar F\ }B\\
\longrightarrow S/(\mathfrak p+(F))\longrightarrow0,
\end{gathered}
\]
and \(H_{S/(\mathfrak p+(F))}(t)=H_B(t)-H_B(t-\delta)\). Put \(d=d(\mathfrak p)=\dim B-1\ge0\). If \(d=0\), \(H_B\) is eventually constant, the difference is eventually zero, and every minimal prime \(\mathfrak q\) has \(\dim S/\mathfrak q=0\), so \(\mathfrak q=\mathfrak m\) and the sum is empty. If \(d\ge1\) and \(H_B(t)=e(\mathfrak p)t^d/d!+O(t^{d-1})\), then
\[
\begin{aligned}
H_B(t)-H_B(t-\delta)&=\frac{\delta\,e(\mathfrak p)}{(d-1)!}\,t^{d-1}\\
&\quad+O(t^{d-2}),
\end{aligned}
\]
so \(d(\mathfrak p+(F))=d-1\) and \(e(\mathfrak p+(F))=\delta e(\mathfrak p)\). Every minimal prime \(\mathfrak q\) has \(d(\mathfrak q)=\dim S/\mathfrak q-1=d-1\) by Proposition 3.2, and Corollary 3.3 gives the inequality. \(\square\)

## 5. The component bound

For a homogeneous ideal \(J\subset S\) and an integer \(\delta\ge1\), let \(\operatorname{Min}^+(J)\) be the set of minimal primes of \(J\) other than \(\mathfrak m\), and define the weight
\[
w_\delta(J)=\sum_{\mathfrak q\in\operatorname{Min}^+(J)}\delta^{\dim S/\mathfrak q-1}e(\mathfrak q).
\]
Every term is at least \(1\), so \(\#\operatorname{Min}^+(J)\le w_\delta(J)\).

**Lemma 5.1.** \(w_\delta(0)=\delta^n\), and \(w_\delta(J+(F))\le w_\delta(J)\) for every homogeneous ideal \(J\) and every homogeneous \(F\) of degree \(\delta\).

*Proof.* The only minimal prime of \(0\) is \(0\), with \(\dim S=n+1\) and \(e(0)=1\). Let \(\mathfrak q'\in\operatorname{Min}^+(J+(F))\). It contains a minimal prime \(\mathfrak p\) of \(J\). If \(F\in\mathfrak p\), then \(\mathfrak p\supseteq J+(F)\), so \(\mathfrak q'=\mathfrak p\) by minimality, and \(\mathfrak p\ne\mathfrak m\). If \(F\notin\mathfrak p\), then \(\mathfrak p\ne\mathfrak m\), and \(\mathfrak q'\) is minimal over \(\mathfrak p+(F)\): a prime between \(\mathfrak p+(F)\) and \(\mathfrak q'\) contains \(J+(F)\), hence equals \(\mathfrak q'\). Therefore
\[
\begin{aligned}
&\operatorname{Min}^+(J+(F))\\
&\quad\subseteq\{\mathfrak p\in\operatorname{Min}^+(J):F\in\mathfrak p\}\\
&\qquad\cup\bigcup_{\mathfrak p\in\operatorname{Min}^+(J),\,F\notin\mathfrak p}\operatorname{Min}^+(\mathfrak p+(F)).
\end{aligned}
\]
For \(\mathfrak p\) in the second family, Proposition 4.1 shows that the elements of \(\operatorname{Min}^+(\mathfrak p+(F))\) have \(\dim S/\mathfrak q=\dim S/\mathfrak p-1\) and total contribution
\[
\delta^{\dim S/\mathfrak p-2}\sum_{\mathfrak q}e(\mathfrak q)\le\delta^{\dim S/\mathfrak p-1}e(\mathfrak p),
\]
the contribution of \(\mathfrak p\) to \(w_\delta(J)\). (If \(\dim S/\mathfrak p=1\), the family \(\operatorname{Min}^+(\mathfrak p+(F))\) is empty.) Summing over \(\mathfrak p\) proves the inequality. \(\square\)

**Theorem 5.2** (component bound). Let \(\delta\ge1\) and let \(f_1,\ldots,f_s\in k[x_1,\ldots,x_n]\) have total degree at most \(\delta\). Then \(V(f_1,\ldots,f_s)\subset k^n\) has at most \(\delta^n\) irreducible components, and so does \(V(f_1,\ldots,f_s)\cap\mathbb G_m^n\).

*Proof.* Put \(F_i=x_0^{\delta}f_i(x_1/x_0,\ldots,x_n/x_0)\), homogeneous of degree \(\delta\), and \(J=(F_1,\ldots,F_s)\). By Lemma 5.1, applied \(s\) times, \(\#\operatorname{Min}^+(J)\le w_\delta(J)\le\delta^n\). By Section 1 it suffices to inject the set of minimal primes of \(I=(f_1,\ldots,f_s)\subset R\) into \(\operatorname{Min}^+(J)\).

Dehomogenization \(G\mapsto G_*=G(1,x_1,\ldots,x_n)\) is a surjective ring homomorphism \(S\to R\) with \(F_{i*}=f_i\). For an ideal \(\mathfrak p\subset R\) let \(\mathfrak p^h\) be the ideal generated by the homogeneous \(G\in S\) with \(G_*\in\mathfrak p\). Three facts are used.

(c) If \(A,B\in S\) are homogeneous with \(A_*=B_*\), then \(x_0^aA=x_0^bB\) for some \(a,b\ge0\). Write \(A=x_0^{\alpha}A'\) with \(x_0\nmid A'\). Then \(A'_*=A_*\) has the same degree as \(A'\), and \(A'=x_0^{\deg A'}A'_*(x_1/x_0,\ldots,x_n/x_0)\), so \(A'\) is determined by \(A_*\). The same holds for \(B=x_0^{\beta}B'\), so \(A'=B'\) and \(x_0^{\beta}A=x_0^{\alpha}B\). Moreover, every element \(Q'\in S\) has the same dehomogenization as the homogeneous element \(Q=\sum_tx_0^{T-t}Q'_t\), where the \(Q'_t\) are its homogeneous components and \(T\) is the largest degree; if \(Q'\) lies in a homogeneous ideal, so does \(Q\).

(a) If \(\mathfrak p\subset R\) is prime, \(\mathfrak p^h\) is a homogeneous prime not containing \(x_0\), and \((\mathfrak p^h)_*=\mathfrak p\). A homogeneous element of \(\mathfrak p^h\) is a combination of homogeneous generators with homogeneous coefficients, so a homogeneous \(G\) lies in \(\mathfrak p^h\) exactly when \(G_*\in\mathfrak p\). If \(G,H\) are homogeneous with \(GH\in\mathfrak p^h\), then \(G_*H_*\in\mathfrak p\), so one factor lies in \(\mathfrak p^h\); Lemma 2.1(1) gives primality. \(x_0\notin\mathfrak p^h\) because \(1\notin\mathfrak p\). Dehomogenization is a ring homomorphism, so \((\mathfrak p^h)_*\subseteq\mathfrak p\); every \(g\in\mathfrak p\) is \(G_*\) for the homogeneous \(G=x_0^{\deg g}g(x_1/x_0,\ldots,x_n/x_0)\in\mathfrak p^h\), so \((\mathfrak p^h)_*=\mathfrak p\).

(b) If \(\mathfrak q\subset S\) is a homogeneous prime with \(x_0\notin\mathfrak q\), then \(\mathfrak q_*\) is a prime of \(R\) and \((\mathfrak q_*)^h=\mathfrak q\). The image of an ideal under a surjective homomorphism is an ideal. By (c), every element of \(\mathfrak q_*\) is \(Q_*\) for a homogeneous \(Q\in\mathfrak q\). Let \(gh\in\mathfrak q_*\) with \(g,h\in R\); write \(g=G_*\), \(h=H_*\) with \(G,H\) homogeneous, and \(gh=Q_*\) with \(Q\in\mathfrak q\) homogeneous. By (c), \(x_0^aGH=x_0^bQ\in\mathfrak q\), and since \(x_0\notin\mathfrak q\), \(GH\in\mathfrak q\). So \(G\) or \(H\) lies in \(\mathfrak q\), and \(g\) or \(h\) lies in \(\mathfrak q_*\). Likewise \(1=Q_*\) with \(Q\in\mathfrak q\) homogeneous would give \(x_0^a=x_0^bQ\in\mathfrak q\), which is impossible; so \(\mathfrak q_*\ne R\). Finally a homogeneous \(G\) with \(G_*\in\mathfrak q_*\) satisfies \(x_0^aG=x_0^bQ\) with \(Q\in\mathfrak q\), so \(G\in\mathfrak q\); hence \((\mathfrak q_*)^h\subseteq\mathfrak q\), and the reverse inclusion is clear.

Let \(\mathfrak p\) be a minimal prime of \(I\). Then \(F_i\in\mathfrak p^h\), so \(\mathfrak p^h\supseteq J\) contains a minimal prime \(\mathfrak q\) of \(J\), homogeneous by Proposition 2.2, with \(x_0\notin\mathfrak q\). By (b), \(\mathfrak q_*\) is a prime containing \(f_1,\ldots,f_s\) and contained in \((\mathfrak p^h)_*=\mathfrak p\); minimality of \(\mathfrak p\) gives \(\mathfrak q_*=\mathfrak p\), and then \(\mathfrak q=(\mathfrak q_*)^h=\mathfrak p^h\). So \(\mathfrak p^h\in\operatorname{Min}^+(J)\), and \(\mathfrak p\mapsto\mathfrak p^h\) is injective because \((\mathfrak p^h)_*=\mathfrak p\). The statement for the torus follows from Section 1. \(\square\)

## 6. Exercises

**6.1.** Show that the bound of Theorem 5.2 is attained by \(f_i=x_i(x_i-1)\cdots(x_i-\delta+1)\), \(1\le i\le n\), when \(k\) has at least \(\delta\) elements, and by the analogous polynomials with \(x_i-a_j\), \(a_j\ne0\), in the torus.

**6.2.** For a homogeneous \(F\ne0\) of degree \(\delta\), show that \(d((F))=n-1\) and \(e((F))=\delta\).

**6.3.** Let \(\mathfrak q\) be generated by \(r\le n\) linearly independent linear forms. Show that \(d(\mathfrak q)=n-r\) and \(e(\mathfrak q)=1\).

**6.4.** Give an example of polynomials in \(k[x_1,x_2]\) for which \(V(f_1,f_2)\) has a component that does not meet \(\mathbb G_m^2\).

**6.5.** Show that the radical of a homogeneous ideal of \(S\) is homogeneous.

## 7. Solutions

**6.1.** The common zeros are the \(\delta^n\) points with coordinates in \(\{0,1,\ldots,\delta-1\}\), each an irreducible component (a point is irreducible and closed, and the set is finite). With distinct nonzero \(a_1,\ldots,a_\delta\) the zeros are the \(\delta^n\) points of \(\{a_1,\ldots,a_\delta\}^n\subset\mathbb G_m^n\).

**6.2.** Multiplication by \(F\) is injective on \(S\), so \(H_{S/(F)}(t)=\binom{t+n}n-\binom{t-\delta+n}n\). Its leading term is \(\delta t^{n-1}/(n-1)!\).

**6.3.** After a linear change of coordinates, which does not change Hilbert functions, \(\mathfrak q=(x_{n-r+1},\ldots,x_n)\) and \(S/\mathfrak q\cong k[x_0,\ldots,x_{n-r}]\), with Hilbert polynomial \(\binom{t+n-r}{n-r}\).

**6.4.** \(f_1=x_1\), \(f_2=x_1x_2\): the zero set is the line \(x_1=0\), outside the torus.

**6.5.** Let \(J\) be homogeneous and \(f\in\sqrt J\), with homogeneous components \(f_0,\ldots,f_T\), \(f_T\ne0\). If \(f^N\in J\), the component of degree \(NT\) of \(f^N\) is \(f_T^N\), which lies in \(J\); so \(f_T\in\sqrt J\), and \(f-f_T\in\sqrt J\) has fewer components. Induction on the number of components shows that all \(f_t\) lie in \(\sqrt J\).

## References

- [OpenAI-UC] OpenAI, *Open Homomorphisms of Global Solvably Closed Galois Groups*, OpenAI Math Release preprint, 5 October 2026, Section 3.1 (proof of Lemma 3.1). https://github.com/openai/math/tree/main/preprints/Open-Homomorphisms-of-Global-Solvably-Closed-Galois-Groups-October-5-2026
- S. Dasgupta, *Ranks of matrices of logarithms of algebraic numbers I: the theorems of Baker and Waldschmidt–Masser*, Essential Number Theory 2 (2023), 93–138, Section 6 (degrees of unmixed ideals). https://arxiv.org/abs/2303.02037