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Lift a Bezout identity for coprime residue factors

Statement

Let $A$ be a commutative ring, let $I \subseteq A$ be an ideal, and let $g_{0}, h_{0} \in (A / I) [T]$ generate the unit ideal. If $g, h \in A [T]$ lift $g_{0}, h_{0}$ , then there exist polynomials $a, b \in A [T]$ such that $a g + b h \equiv 1 (m o d I) .$

Facts & Assumptions

Given: A commutative ring $A$ , an ideal $I$ , residue polynomials $g_{0}, h_{0} \in (A / I) [T]$ with $(g_{0}, h_{0}) = (1)$ , and lifts $g, h \in A [T]$ .

[L1]

The quotient ring $A / I$ and the polynomial ring over a commutative ring are again commutative rings, so Bezout identities and coefficientwise lifting make sense in $(A / I) [T]$ and $A [T]$ (The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ , The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

Proof

technique · lift one residue Bezout identity coefficientwise

1.1

Because $(g_{0}, h_{0}) = (1)$ in $(A / I) [T]$ , there exist $a_{0}, b_{0} \in (A / I) [T]$ such that $a_{0} g_{0} + b_{0} h_{0} = 1.$

L1givenchoose

2.1

Lift the coefficients of $a_{0}$ and $b_{0}$ to polynomials $a, b \in A [T]$ . Reducing coefficientwise modulo $I$ gives $\overline{a g + b h} = \overline{a} \overline{g} + \overline{b} \overline{h} = a_{0} g_{0} + b_{0} h_{0} = 1.$ Hence $a g + b h \equiv 1 (m o d I)$ .

step 1.1L1choose

3.1

Thus a coprime residue factorization always admits a lifted Bezout relation modulo $I$ .

step 2.1∎

Depends on

Used by

Dependency tree · two levels

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Sources

Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., Chapter 22 (standard reference, not scraped)
The Stacks Project, Section 15.11: Henselian pairs (standard reference, not scraped)