Proving Gauss's Lemma Extension in Polynomial Factoring

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SUMMARY

The discussion centers on proving the extension of Gauss's Lemma in the context of polynomial factoring. The lemma states that if a monic polynomial with integral coefficients factors into two monic polynomials with rational coefficients, then those polynomials also have integer coefficients. The exercise extends this to non-monic polynomials, asserting that if a polynomial with integral coefficients factors into rational polynomials, it can still be factored into polynomials with integer coefficients. The key insight is that by manipulating the coefficients, one can apply Gauss's Lemma to derive the necessary integer factorizations.

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  • Understanding of Gauss's Lemma in polynomial theory
  • Familiarity with monic and non-monic polynomials
  • Knowledge of polynomial factorization techniques
  • Basic concepts of rational and integral coefficients in polynomials
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  • Explore examples of polynomial factorization in both monic and non-monic cases
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Mathematicians, algebra students, and educators interested in polynomial theory and its applications in number theory will benefit from this discussion.

math_grl
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Ok, the book I'm reading states Gauss's lemma as such:
If f(x) is a monic polynomial with integral coefficients that factors into two monic polynomials with coefficients that are rational, f(x) = g(x)h(x), then g(x), h(x) \in \mathbb{Z}[x].

Now one of the exercises says to prove that:
If f(x) is a polynomial with integral coefficients that factors into two polynomials with coefficients that are rational, f(x) = g(x)h(x), then there is a factoring p(x), q(x) \in \mathbb{Z}[x] such that f(x) = p(x)q(x).

Which seem like almost the same problem except the exercise is more general in the case that they do not have to be monic. Yet, with g(x) and h(x) having rational coefficients, can't they factored into polynomials from \mathbb{Z}[x] and, moreover, monic by pulling out a rational factor?
I guess what I'm asking is thru what means can you go from Gauss's lemma to the exercise in the simplest way...
 
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Well first if f(x) is a monic polynomial, then if g(x)=bx^n+...+b_0 and h(x)=ax^m+...+a_0, then if we factor out ab=1, so we get two monic polynomials with rational parameters and you can use the lemma.

If it's not a monic polynomial, just divide both sides of the equation by this parameter (of x^(n+m)) to get a polynomial for which the conditions of the lemma are met, then multiply the equation by this factor to get the conclusion in your exercise.
 

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