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Strong transcendence descends to reduced minimal-prime quotients

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $R \subseteq S$ be an inclusion of reduced commutative rings (The nilradical and reduced rings), let $x \in S$ be strongly transcendental over $R$ (Strong transcendence over a subring), let $q \subseteq S$ be a minimal prime and let $p = R \cap q$ be its contraction to $R$ . Then the image of $x$ in the domain $S / q$ is strongly transcendental over the subring $R / p \subseteq S / q$ (The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ ).

The minimality of $q$ is essential: the local ring $S_{q}$ is then a field, which is what lets one clear a denominator outside $q$ , and the annihilator-sensitive form of strong transcendence is what makes the clearing argument work without assuming that $S$ is a domain.

Facts & Assumptions

Given: The Axiom of Choice; an inclusion of reduced commutative rings $R \subseteq S$ , an element $x \in S$ strongly transcendental over $R$ , a minimal prime $q \subseteq S$ and $p = R \cap q$ .

[L1]

For $R \subseteq S$ and $x \in S$ , strong transcendence of $x$ over $R$ means that $u (a_{0} + a_{1} x + \dots + a_{k} x^{k}) = 0$ with $u \in S$ and $a_{i} \in R$ implies $u a_{i} = 0$ for all $i$ (Strong transcendence over a subring).

[L2]

The nilradical $Nil (R)$ of a commutative ring is the radical of the zero ideal, so $x \in Nil (R)$ means $x^{n} = 0$ for some $n \geq 1$ , and $R$ is reduced when $Nil (R) = (0)$ (The nilradical and reduced rings).

[L3]

The nilradical of a commutative ring is the intersection of all of its prime ideals; this is the point where the Axiom of Choice is used, through the existence of primes avoiding a given element (The nilradical is the intersection of all prime ideals).

[L4]

For a commutative ring $R$ , a multiplicative subset $S \subseteq R$ and an ideal $I ⊴ R$ one has $S^{- 1} \sqrt{I} = \sqrt{S^{- 1} I}$ as ideals of $S^{- 1} R$ (Radicals commute with localization).

[L5]

For a commutative ring $R$ and a prime $p \in Spec (R)$ , contraction along $R \to R_{p}$ is an inclusion-preserving bijection from $Spec (R_{p})$ onto the set of primes $q \subseteq p$ of $R$ ; the inverse sends $q$ to $q R_{p}$ (Primes of a localization at a prime).

[L6]

For a multiplicative subset $S$ of a commutative ring $R$ and $r \in R$ , the image of $r$ in $S^{- 1} R$ is zero if and only if $u r = 0$ for some $u \in S$ (Equality, vanishing, and the kernel of the localisation map).

[L7]

For a prime $p \subseteq R$ the localisation $R_{p}$ is $(R ∖ p)^{- 1} R$ , with elements fractions $r / s$ for $s \notin p$ (Localisation at a prime ideal: $R_{p} = (R ∖ p)^{- 1} R$ ).

[L8]

For an ideal $I$ of a commutative ring $R$ the quotient ring $R / I$ has elements the cosets $r + I$ , and the canonical projection $R \to R / I$ , $π (r) = r + I$ , is a surjective ring homomorphism with kernel $I$ (The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ , The canonical projection $R \to R / I$ is a surjective ring homomorphism with kernel $I$ ).

[L9]

The Axiom of Choice is the assertion that every family of nonempty sets has a choice function; it is assumed in this item and used exactly through [L3] (The Axiom of Choice).

Proof

technique · direct

1.1

We work under [L9]. Set $\overline{R} = R / p$ and $\overline{S} = S / q$ , and let $\overline{x}$ denote the image of $x$ in $\overline{S}$ by [L8]. Since $p = R \cap q$ is the kernel of the composite $R \to S \to S / q$ , the map $\overline{R} \to \overline{S}$ induced by the inclusion is injective, so $\overline{R}$ is a subring of $\overline{S}$ ; both are domains, hence reduced by [L2].

givenL2L8

1.2

The localisation $S_{q}$ is reduced. Indeed, applying [L4] to the ideal $I = (0)$ of $S$ gives $S^{- 1} \sqrt{(0)} = \sqrt{S^{- 1} (0)}$ for the multiplicative subset $S ∖ q$ of [L7], that is, the extension of $Nil (S) = (0)$ is $Nil (S_{q})$ ; the extension of the zero ideal is zero, so $Nil (S_{q}) = (0)$ by [L2].

givenL2L4L7

1.3

By [L5] the primes of $S_{q}$ correspond bijectively to the primes of $S$ contained in $q$ . Since $q$ is a minimal prime, the only prime of $S$ contained in $q$ is $q$ itself, so $q S_{q}$ is the unique prime ideal of $S_{q}$ .

givenL5L7

2.1

By [L3] applied to the ring $S_{q}$ , its nilradical is the intersection of its prime ideals; by step 1.3 that intersection is the single ideal $q S_{q}$ , so $Nil (S_{q}) = q S_{q}$ . By step 1.2 the nilradical is zero, hence $q S_{q} = 0$ . In particular the image in $S_{q}$ of every element of $q$ is a member of $q S_{q} = 0$ , hence is zero.

step 1.2step 1.3L3

2.2

Now let $\overline{u} \in \overline{S}$ and ${\overline{a}}_{0}, \dots, {\overline{a}}_{k} \in \overline{R}$ satisfy $\overline{u} ({\overline{a}}_{0} + {\overline{a}}_{1} \overline{x} + \dots + {\overline{a}}_{k} {\overline{x}}^{k}) = 0$ in $\overline{S}$ . Choose preimages $u \in S$ and $a_{i} \in R$ under the quotient maps of [L8]; then $u (a_{0} + a_{1} x + \dots + a_{k} x^{k}) \in q$ , because its image in $\overline{S}$ is the left-hand side of the assumed relation.

givenstep 1.1L8algebra

3.1

By step 2.1 the image of the element $u (a_{0} + a_{1} x + \dots + a_{k} x^{k}) \in q$ in $S_{q}$ is zero, so the kernel criterion [L6] applied to the localisation $S \to S_{q}$ at the multiplicative subset $S ∖ q$ of [L7] provides $u^{'} \in S ∖ q$ with $u^{'} u (a_{0} + a_{1} x + \dots + a_{k} x^{k}) = 0$ in $S$ .

step 2.1step 2.2L6L7

4.1

The element $x$ is strongly transcendental over $R$ by hypothesis, so [L1] applied to the vanishing product of step 3.1 with multiplier $u^{'} u \in S$ gives $u^{'} u a_{i} = 0$ in $S$ for every $i$ .

givenstep 3.1L1

5.1

Since $u^{'} u a_{i} = 0 \in q$ and $u^{'} \notin q$ , the primality of $q$ gives $u a_{i} \in q$ for every $i$ . Passing to $\overline{S}$ by [L8], this says $\overline{u} {\overline{a}}_{i} = 0$ in $\overline{S}$ for every $i$ .

step 4.1L8algebra

6.1

Steps 2.2, 5.1 and the discussion of step 1.1 show: for every $k \geq 0$ , every multiplier $\overline{u} \in \overline{S}$ and all ${\overline{a}}_{i} \in \overline{R}$ , the vanishing $\overline{u} ({\overline{a}}_{0} + \dots + {\overline{a}}_{k} {\overline{x}}^{k}) = 0$ implies $\overline{u} {\overline{a}}_{i} = 0$ for all $i$ . By [L1] the image $\overline{x}$ is strongly transcendental over $\overline{R} = R / p$ inside $\overline{S} = S / q$ , which is the assertion; the Axiom of Choice was assumed in step 1.1 and used only through [L3] in step 2.1. ∎

step 1.1step 2.2step 5.1L1

Depends on

Used by

Finite algebras over a strongly transcendental variable are nowhere quasi-finite Lemma

Dependency tree · two levels

32 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Stacks Project, Commutative Algebra, Section 10.123, Lemma 10.123.8 (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, version 4.03, Section 17 (standard reference, not scraped)