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Gauss's lemma II (Theorem)

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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See Also: Gauss's lemma I, Eisenstein criterion, proof of Eisenstein criterion, prime factors of $x^n-1$, alternative proof that $\sqrt{2}$ is irrational

Other names:  Gauss' lemma II
Also defines:  primitive polynomial
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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 13449 times total.

Classification:
AMS MSC12E05 (Field theory and polynomials :: General field theory :: Polynomials )

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