Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Gauss's lemma says If the primitive polynomial $f(x)$ can be factored as product of two polynomials having rational coefficients, it can be factored as the product of two polynomials having integer coefficient.

My doubt is why is the condition that $f(x)$ is primitive necessary? Isn't true for all integer polynomials?

share|improve this question
    
en.wikipedia.org/wiki/… The irreducibility statement there might help understand, non? –  awllower Nov 13 '11 at 13:42
4  
Note: "doubt" here is used in the sense of Indian English ... meta.math.stackexchange.com/questions/3200 –  GEdgar Nov 13 '11 at 14:25
    
Amazingly, this meaning confirms to my naive concept of this word!! Although I am certainly not an Indian... –  awllower Nov 13 '11 at 14:48
add comment

2 Answers

Your statement is somewhat imprecise. Your expression "can be factored as the product of two polynomials" should be replaced by "is reducible" (over $\mathbb Z$ or $\mathbb Q$), which excludes the possibility of the factors being invertible elements in their respective rings. Otherwise the statement would be trivially true, since any polynomial can be written as $1$ times itself.

The primitivity condition in the statement thus corrected is indeed superfluous. In fact, a polynomial is reducible over $\mathbb Z$ if and only if it is reducible over $\mathbb Q$ or it is not primitive. (This is the contrapositive of the statement of Gauss's lemma in the Wikipedia article others have linked to.) Thus, in a sense, the polynomial being primitive is not a condition for applying the lemma, but for needing the lemma, since for a non-primitive polynomial you know that it's reducible over $\mathbb Z$ without knowing whether it's reducible over $\mathbb Q$.

share|improve this answer
    
No, it would not be false, if you allow constants, then the constant 1 is also allowed. Just not very interesting. –  Phira Nov 13 '11 at 15:47
    
@Phira: You're right; I'll edit accordingly. –  joriki Nov 13 '11 at 16:05
add comment

Yes, it is true for all integer polynomial, because the additional global integer factor can just be multiplied to one of the factors.

If the polynomial is primitive, one can actually choose the two polynomials to be primitive integer polynomials as well.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.