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Gauss's lemma II
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(Theorem)
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Definition: A polynomial $p(x)=a_nx^n+\cdots+a_0$ over a UFD $R$ is said to be primitive if its coefficients are not all divisible by any element of $R$ other than a unit.
Proposition (Gauss): Let $R$ be a UFD and $F$ its field of fractions. If a polynomial $p\in R[x]$ is reducible in $F[x]$ then it is reducible in $R[x]$
Remark: The above statement is often used in its contrapositive form. For an example of this usage, see this entry.
Proof: We may assume that $p$ is primitive. Suppose $p=qr$ with $q,r\in F[x]$ There are unique elements $a,b\in F$ such that $q/a$ and $r/b$ are in $R[x]$ and are primitive. But $p/ab=(q/a)(r/b)$ Since $p$ is primitive, it follows from Gauss's lemma I that $ab$ is a unit, and therefore so are $a$ and $b$ This completes the proof.
Remark: Another result with the same name is Gauss' lemma on quadratic residues.
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"Gauss's lemma II" is owned by bshanks. [ full author list (9) ]
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Cross-references: quadratic residues, Gauss lemma, Gauss's lemma I, proof, contrapositive, reducible, field of fractions, Gauss, unit, divisible, coefficients, UFD, polynomial
There are 9 references to this entry.
This is version 11 of Gauss's lemma II, born on 2002-11-04, modified 2007-04-14.
Object id is 3567, canonical name is GausssLemmaII.
Accessed 13427 times total.
Classification:
| AMS MSC: | 12E05 (Field theory and polynomials :: General field theory :: Polynomials ) |
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Pending Errata and Addenda
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