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Gauss's lemma I (Theorem)

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

$\displaystyle \overline{f(x)} \ \overline{g(x)} = 0$    

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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See Also: Gauss's lemma II
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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 4762 times total.

Classification:
AMS MSC: 12E05 (Field theory and polynomials :: General field theory :: Polynomials )
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