Gauss's lemma (polynomial)

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This article is about Gauss's lemma for polynomials. See also Gauss's lemma (number theory).

In the theory of polynomials, Gauss's lemma, named after Carl Friedrich Gauss, refers to two related statements:

This second statement is a consequence of the first (see proof).

There is another result, a consequence of a generalization of the second, also sometimes referred to as Gauss's lemma; namely that if an integral domain is a unique factorization domain), then its ring of polynomials is also a unique factorization domain.

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[edit] Formal statement

These statements can be expanded to any unique factorization domain:

  • If R is a UFD, and f(x) and g(x) are both primitive polynomials in R[x], then so is f(x).g(x).
  • If R is a UFD and F is its field of fractions, then if a polynomial f(x) in R[x] is irreducible over R[x], then it is also irreducible over F[x].

Both these results can be generalised to several variables.

[edit] Proofs of the primitivity property

Firstly a proof from first principles.

Theorem: The product of two primitive polynomials is itself primitive.

Proof:

Clearly the product f(x)\cdot g(x) of two primitive polynomials has integral coefficients. Therefore, if it is not primitive, there must be a common divisor d of all its coefficients, that can not divide all the coefficients of either f(x) or g(x) (otherwise they would not be primitive). Let arxr be the first coefficient of f(x) not divisible by d and let bsxs be the first coefficient of g(x) not divisible by d. Now consider the term xr + s in the product. Its coefficient must take the form:

a_r b_s + a_{r+1} b_{s-1} + a_{r+2} b_{s-2} + \ldots + a_{r-1} b_{s+1} + a_{r-2} b_{s+2} + \ldots

The first term is not divisible by d, yet all the remaining ones are, so the entire sum cannot be divisible by d. We assumed that all coefficients in the product were divisible by d, leading to a contradiction. Therefore, the coefficients of the product can have no common divisor and are thus primitive. This completes the proof.

An outline of an abstract algebra formulation.

  1. Assume f(x).g(x) is not primitive, and let p be a prime factor of the coefficients of f(x).g(x).
  2. This implies f(x).g(x)=0 over the ring R/pR (R modulo p).
  3. Since R/pR is an integral domain, then either f(x) or g(x) =0 over R/pR, and hence p must divide the coefficients of either f(x) or g(x).

[edit] Proof of the lemma

For the second statement, we will prove the contrapositive:

  1. Without loss of generality we may assume f(x) is primitive.
  2. Assume f(x) is reducible over F[x]. Then there exist g(x) and h(x) in F[x] such that f(x) = g(x).h(x).
  3. There exist a,b in F such that both a.g(x) and b.h(x) are in R[x] and are primitive.
  4. By the first result (a.g(x)).(b.h(x)) = (ab).f(x) is also primitive, and hence ab = ±1

This implies f(x) is reducible over R[x].

[edit] Implications

The first result implies that the GCD of the coefficients of the product of two polynomials will be the product of the GCD of the coefficients of each polynomial. This is very useful, as a limit to the common divisors of the coefficients of the product.

The second result implies that if a polynomial can be factorised over the rationals, then there exists a factorisation over the integers. This property is also useful when combined with properties such as Eisenstein's criterion.

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