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The units of over an integral domain are exactly the constant polynomials whose values are units of
Statement
Let be an integral domain. A polynomial is a unit if and only if it is a constant polynomial whose constant value is a unit of .
Facts & Assumptions
Given: An integral domain and a polynomial .
For nonzero polynomials over a domain, (Over an integral domain, degrees add under multiplication of nonzero polynomials).
The constant-polynomial map is an injective unital ring homomorphism (Polynomial convolution makes a commutative ring containing as its constant subring).
Proof
If is a unit, choose with ; neither factor is zero and [L1] gives , so both degrees are , and comparison of constant coefficients shows that the constant value of is a unit of .
Conversely, if is a unit with inverse , then [L2] gives , so the constant polynomial is a unit of .
Depends on
Used by
- A cusp family defeats the missing source hypothesis Counterexample
- A nondegenerate projective quadric Example
- A projective cone with a smooth conic base Example
- Different singularities can share a tangent cone Example
- The cusp y²=x³ has Puiseux parameter (t²,t³) Example
- The plane branch y²=x⁵ has Puiseux parameter (t²,t⁵) Example
- The rank-one 2 by 2 determinantal cone Example
- Every nonzero nonunit polynomial over a field factors into irreducible polynomials Lemma
- Fibres of the orbit map and the scheme-theoretic stabilizer as a closed subgroup scheme Lemma
- The twist index on the projective line is an isomorphism invariant Lemma
- Units, powers and the domain property of the Laurent polynomial ring Lemma
- A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field Theorem
Dependency tree · two levels
8 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
- James McKernan, MIT 18.703 Lecture 21, Lemma 21.1 (standard reference, not scraped)