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Gauss's lemma I
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(Theorem)
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There are a few different things that are sometimes called ``Gauss's Lemma''. See also Gauss's Lemma II.
Gauss's Lemma I: If $R$ is a UFD and $f(x)$ and $g(x)$ are both primitive polynomials in $R[x]$ , so is $f(x) g(x)$ .
Proof: Suppose $f(x) g(x)$ not primitive. We will show either $f(x)$ or $g(x)$ isn't as well. $f(x) g(x)$ not primitive means the gcd of the coefficients of $f(x) g(x)$ is not a unit. Let $p$ be a prime factor of that gcd. We consider the image of $R$ mod $p$ - i.e. under the natural ring homomorphism $\theta: R \to R/pR$ - and extend to the polynomial ring.
Since $R$ is an integral domain, $R/pR$ is an integral domain, so $(R/pR)[x]$ is an integral domain. And we have
where $\overline{f(x)}$ is the image of $f(x)$ in $(R/pR)[x]$ , similarly $\overline{g(x)}$ . So $\overline{f(x)} = 0$ or $\overline{g(x)} = 0$ . So $f(x)$ or $g(x)$ is divisible by $p$ , so one of them is not primitive.
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"Gauss's lemma I" is owned by bshanks. [ full author list (6) ]
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Cross-references: divisible, integral domain, polynomial ring, ring homomorphism, image, prime factor, unit, coefficients, gcd, proof, primitive polynomials, UFD, Gauss's lemma II
There are 2 references to this entry.
This is version 13 of Gauss's lemma I, born on 2002-11-04, modified 2008-05-01.
Object id is 3566, canonical name is GausssLemmaI.
Accessed 5442 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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