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One-variable integral correction after leading-coefficient localization

Statement

Let $R$ be a commutative ring, let $φ : R [x] \to S$ be a unital ring map of commutative rings, and let $t \in S$ be integral over the image subring $φ (R [x]) \subseteq S$ (Integral elements over a commutative ring and algebraic integers, Integral ring maps and integral extensions). Let $p = a_{0} + a_{1} x + \dots + a_{k} x^{k} \in R [x]$ be a polynomial with

$t φ (p) \in Im (φ) .$

If $p$ is monic, then there exists $q \in R [x]$ such that $t - φ (q)$ is integral over $R$ .
In general there exist $q \in R [x]$ and an integer $n \geq 0$ such that $φ (a_{k})^{n} t - φ (q)$ is integral over $R$ .

No injectivity of $φ$ , no regularity of the leading coefficient $a_{k}$ , and no reducedness or domain hypothesis is imposed: $a_{k}$ may be $0$ , a zero divisor or a nilpotent, $t$ may be $0$ or a zero divisor, and the polynomial ring and all localisations are taken over the possibly nonreduced ring $R$ . Part 1 is the monic case of the one-variable integral correction; part 2 obtains it in general by inverting $a_{k}$ .

Facts & Assumptions

Given: A unital ring map $φ : R [x] \to S$ of commutative rings, an element $t \in S$ integral over the subring $φ (R [x]) \subseteq S$ , and a polynomial $p = a_{0} + a_{1} x + \dots + a_{k} x^{k} \in R [x]$ with $t φ (p) \in Im (φ)$ .

[L1]

An element $b$ of a commutative ring $B$ is integral over a subring $A$ exactly when it is a root of a monic polynomial in $A [X]$ (Integral elements over a commutative ring and algebraic integers).

[L2]

Let $R$ be a commutative ring and let $g \in R [x]$ be monic. For every $f \in R [x]$ there are unique $q, r \in R [x]$ with $f = q g + r$ and $r = 0$ or $\deg r < \deg g$ (Division by a monic polynomial over a commutative ring).

[L3]

Let $A \to B$ be a homomorphism of commutative rings, $S \subseteq A$ multiplicative and $b \in B$ . If $b$ is integral over $A$ then $b / 1$ is integral over $S^{- 1} A$ in $S^{- 1} B$ ; and if $b / 1$ is integral over $S^{- 1} A$ in $S^{- 1} B$ then some $s \in S$ makes $s b$ integral over $A$ (Integrality and integral closure commute with localisation).

[L4]

If $A \to B$ and $B \to C$ are integral ring maps of commutative rings then the composite $A \to C$ is integral (Integral extensions are transitive).

[L5]

Let $A \subseteq B$ be commutative rings with $A \neq 0$ and $b \in B$ . Then $b$ is integral over $A$ if and only if $A [b]$ is finitely generated as an $A$ -module (Integrality and finite-module characterizations for one element).

[L6]

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

[L7]

For a commutative ring $R$ and $f \in R$ the powers of $f$ form a multiplicative subset and the principal localisation $R_{f}$ has elements $r / f^{n}$ ; $R_{0}$ is the zero ring (Principal localisation $R_{f} = {1, f, f^{2}, \dots}^{- 1} R$ ).

[L8]

Let $A \subseteq B$ be commutative rings with $A \neq 0$ . The elements of $B$ integral over $A$ form a subring of $B$ (Integral elements over a nonzero base ring form a subring).

Proof

technique · direct

1.1

We first dispose of the degenerate cases. If $R = 0$ then $R [x] = 0$ and $1_{S} = φ (0) = 0$ , so $S = 0$ , $t = 0$ and $q = 0$ satisfies both claims. If $S = 0$ then also $t = 0$ and $q = 0$ satisfies both claims. If $t$ is nilpotent, say $t^{N} = 0$ with $N \geq 1$ , then $t$ is a root of the monic polynomial $X^{N}$ , hence integral over $R$ by [L1], so $q = 0$ satisfies claim 1 and $(n, q) = (0, 0)$ satisfies claim 2. If $a_{k}$ is nilpotent, say $a_{k}^{N} = 0$ with $N \geq 1$ , then $φ (a_{k})^{N} t - φ (0) = 0$ is integral over $R$ by [L1], so $(n, q) = (N, 0)$ satisfies claim 2. We therefore assume $R \neq 0$ , $S \neq 0$ , $t$ not nilpotent, and in the treatment of claim 2 the element $a_{k}$ not nilpotent.

givenL1

1.2

Assume that $p$ is monic, so that claim 1 is at stake. Since $t φ (p) \in Im (φ)$ , choose $r \in R [x]$ with $t φ (p) = φ (r)$ . By [L2] applied to the monic divisor $p$ there are $q, r^{'} \in R [x]$ with $r = q p + r^{'}$ and $r^{'} = 0$ or $\deg r^{'} < \deg p$ . Set $t^{'} = t - φ (q) \in S$ . Then $t^{'} φ (p) = t φ (p) - φ (q) φ (p) = φ (r) - φ (q p) = φ (r^{'})$ . If $t^{'}$ is nilpotent, say $t^{' N} = 0$ for some $N \geq 1$ , then $t^{'}$ is integral over $R$ by the monic equation $X^{N} = 0$ , proving claim 1; below assume $t^{'}$ is not nilpotent.

givenL1L2construct

2.1

The element $t^{'} = t - φ (q)$ is integral over $φ (R [x])$ : the given element $t$ is integral over $φ (R [x])$ and $φ (q) \in φ (R [x])$ , while $1_{S} \neq 0$ makes the subring $φ (R [x]) \subseteq S$ nonzero, so [L8] applies.

step 1.2L8algebra

2.2

Write $p = x^{d} + p_{d - 1} x^{d - 1} + \dots + p_{0}$ with $d \geq 0$ (a monic constant is the case $d = 0$ ) and write $r^{'} = r_{d - 1}^{'} x^{d - 1} + \dots + r_{0}^{'}$ with $r_{i}^{'} = 0$ for $i \geq \deg r^{'}$ . In the principal localisation $S_{t^{'}}$ of [L7] the identity $t^{'} φ (p) = φ (r^{'})$ of step 1.2 becomes $φ (p) = t^{' - 1} φ (r^{'})$ , that is, the identity $φ (x)^{d} + \sum_{i = 0}^{d - 1} (φ (p_{i}) - t^{' - 1} φ (r_{i}^{'})) φ (x)^{i} = 0$ in $S_{t^{'}}$ . Its coefficients $φ (p_{i}) - t^{' - 1} φ (r_{i}^{'})$ lie in the subring $φ (R) [1 / t^{'}] \subseteq S_{t^{'}}$ and its leading coefficient is $1$ ; by [L1] the element $φ (x)$ is integral over $φ (R) [1 / t^{'}]$ .

step 1.2L1L7algebra

3.1

By [L5] applied to the nonzero ring $φ (R) [1 / t^{'}]$ and the integral element $φ (x)$ of step 2.2, the subalgebra $φ (R) [1 / t^{'}] [φ (x)]$ is a finitely generated $φ (R) [1 / t^{'}]$ -module, so every one of its elements is integral over $φ (R) [1 / t^{'}]$ by [L8]; equivalently the ring map $φ (R) [1 / t^{'}] \to φ (R) [1 / t^{'}] [φ (x)]$ is integral.

step 2.2L5L8algebra

4.1

Put $A = φ (R) [1 / t^{'}]$ and $B = A [φ (x)]$ inside $S_{t^{'}}$ . The element $t^{'}$ is integral over $B$ : its monic equation over $φ (R [x])$ from step 2.1 remains a monic equation over the larger subring $B$ . By step 3.1, $A \to B$ is integral. Since the integral elements over $B$ form a subring by [L8], the ring $B [t^{'}]$ is integral over $B$ ; transitivity [L4] makes $A \to B [t^{'}]$ integral. Therefore $t^{'} \in B [t^{'}]$ is integral over $A$ by [L1].

step 2.1step 3.1L1L4L8

5.1

By [L1] there are an integer $e \geq 1$ and coefficients $c_{0}, \dots, c_{e - 1} \in A = φ (R) [1 / t^{'}]$ with $t^{' e} + \sum_{i = 0}^{e - 1} c_{i} t^{' i} = 0$ . Each $c_{i}$ is a finite sum $\sum_{j = 1}^{m_{i}} φ (b_{i j}) / t^{' ℓ_{i j}}$ with $b_{i j} \in R$ and $ℓ_{i j} \geq 0$ , since $A$ is generated as a subring by $φ (R)$ and $t^{' - 1}$ . Choose $N \geq 0$ at least every exponent $ℓ_{i j}$ (take $N = 0$ if there are no summands). Multiplying the relation by $t^{' N}$ in $S_{t^{'}}$ gives $t^{' e + N} + \sum_{i = 0}^{e - 1} \sum_{j = 1}^{m_{i}} φ (b_{i j}) t^{' i + N - ℓ_{i j}} = 0,$ where every exponent $i + N - ℓ_{i j}$ is nonnegative.

step 4.1L1L7algebra

6.1

The identity of step 5.1 holds in $S_{t^{'}}$ , so by the kernel criterion [L6] applied to the localisation map $S \to S_{t^{'}}$ and the element $b = t^{' e + N} + \sum_{i = 0}^{e - 1} \sum_{j = 1}^{m_{i}} φ (b_{i j}) t^{' i + N - ℓ_{i j}} \in S$ there is $M \geq 0$ with $t^{' M} b = 0$ in $S$ . Hence $t^{' e + N + M} + \sum_{i = 0}^{e - 1} \sum_{j = 1}^{m_{i}} φ (b_{i j}) t^{' i + N - ℓ_{i j} + M} = 0$ in $S$ : a monic polynomial relation for $t^{'} = t - φ (q)$ with coefficients in $φ (R) \subseteq S$ . By [L1] the element $t - φ (q)$ is integral over $R$ , which is claim 1.

step 5.1L1L6algebra

7.1

Now let $p = \sum_{i = 0}^{k} a_{i} x^{i}$ be arbitrary with $t φ (p) \in Im (φ)$ , and assume first that $a_{k}$ is not nilpotent. Then the principal localisation $R^{'} = R_{a_{k}}$ of [L7] is nonzero by [L6], since its multiplicative set contains $1 = a_{k}^{0}$ while no power of $a_{k}$ is $0$ ; and $p^{'} = a_{k}^{- 1} p = x^{k} + \sum_{i = 0}^{k - 1} (a_{i} / a_{k}) x^{i} \in R^{'} [x]$ is monic. Let $S^{'} = S_{a_{k}}$ , let $φ^{'} : R^{'} [x] \to S^{'}$ be the localisation of $φ$ , and let $t^{'} \in S^{'}$ be the image of $t$ , so that $R^{'} [x] = R [x]_{a_{k}} = R_{a_{k}} [x]$ . The element $t^{'}$ is integral over $φ^{'} (R^{'} [x])$ , because the monic equation for $t$ over $φ (R [x])$ transports along the ring map induced by $φ$ ; and $t^{'} φ^{'} (p^{'}) = t φ (p) / a_{k}$ lies in $Im (φ^{'})$ , because $t φ (p) = φ (r)$ for some $r \in R [x]$ gives $t^{'} φ^{'} (p^{'}) = φ^{'} (r / a_{k})$ . Applying claim 1, already proved in steps 1.2-6.1 for the data $(R^{'}, S^{'}, φ^{'}, t^{'}, p^{'})$ , yields $q^{'} \in R^{'} [x]$ with $t^{'} - φ^{'} (q^{'})$ integral over $R^{'}$ .

step 6.1L1L6L7algebra

8.1

Write $q^{'}$ as a finite sum of terms $c / a_{k}^{m}$ with $c \in R [x]$ and $m \geq 0$ , and let $n$ be the maximum of the finitely many exponents $m$ occurring, with $n = 0$ when $q^{'} = 0$ . Then $a_{k}^{n} q^{'}$ is the image of some $q \in R [x]$ under $R [x] \to R^{'} [x]$ , namely $q = \sum_{j} c_{j} a_{k}^{n - m_{j}}$ for the finitely many summands $c_{j} / a_{k}^{m_{j}}$ of $q^{'}$ . With this $n$ and $q$ , the image in $S^{'}$ of $φ (a_{k})^{n} t - φ (q)$ is $a_{k}^{n} t^{'} - a_{k}^{n} φ^{'} (q^{'}) = a_{k}^{n} (t^{'} - φ^{'} (q^{'}))$ , and this is integral over $R^{'}$ because $t^{'} - φ^{'} (q^{'})$ is (step 7.1) and $a_{k}^{n} \in R^{'}$ .

step 7.1L1L7construct

9.1

Apply clause 2 of [L3] to the ring map $R \to S$ , the multiplicative subset ${a_{k}^{m} : m \geq 0} \subseteq R$ and the element $b = φ (a_{k})^{n} t - φ (q) \in S$ : its image in $S_{a_{k}} = S^{'}$ is integral over $R_{a_{k}} = R^{'}$ by step 8.1, so there exists $m \geq 0$ such that $a_{k}^{m} b$ is integral over $R$ . With $n^{'} = m + n \geq 0$ and $q^{''} = a_{k}^{m} q \in R [x]$ one has $φ (a_{k})^{n^{'}} t - φ (q^{''}) = a_{k}^{m} (φ (a_{k})^{n} t - φ (q))$ , which is integral over $R$ . This is claim 2 for arbitrary $p$ with $t φ (p) \in Im (φ)$ .

step 8.1L3algebra

10.1

Together, step 1.1 (the degenerate cases, including nilpotent $a_{k}$ and nilpotent $t$ ), step 6.1 (claim 1) and step 9.1 (claim 2) prove both assertions of the statement for every commutative ring $R$ , every unital $φ$ , every $t$ integral over $φ (R [x])$ and every $p$ with $t φ (p) \in Im (φ)$ ; this includes $p = 0$ , monic constants $p$ , the zero element $t = 0$ , and zero divisors or nilpotents among the $a_{i}$ . ∎

step 1.1step 6.1step 9.1given

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Conductor radical detects every polynomial coefficient Lemma

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Sources

The Stacks Project, Commutative Algebra, Section 10.123, Lemmas 10.123.2 and 10.123.3 (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, version 4.03, Section 17 (standard reference, not scraped)