Polynomials with integer coefficients - the GCD identity

Jan 2010
594
5
Hobart, Tasmania, Australia
I am reading Anderson and Feil on the Factorization of Polynomials - Section 5.3 Polynomials with Integer Coefficients

A&F point out that some of the therems they have proved for \(\displaystyle \mathbb{Q} [x] \) are false if we restrict ourselves to polynomials with integer coefficients.

For example they point out that the Divsion Theorem is false for \(\displaystyle \mathbb{Z} [x] \).

Also the GCD identity fails in \(\displaystyle \mathbb{Z} [x] \).

They then ask the reader the polynomials 2 and x in \(\displaystyle \mathbb{Z} [x] \) pointing out the 1 is the GCD for these polynomials. A&F then ask the reader to prove that we cannot write 1 as a linear combination of 2 and x

Can anyone help me with a rigorous and formal proof of this.

Peter
 

Deveno

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Mar 2011
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Tejas
suppose we COULD.

we then have 1 = 2f(x) + xg(x), for some polynomials f(x),g(x) in Z[x].

suppose that deg(f) = m, and deg(g) = n. let k = max(m,n). then we can write:

\(\displaystyle f(x) = a_0 + a_1x + \cdots + a_kx^k, g(x) = b_0 + b_1x + \cdots + b_kx^k\) (some of the coefficients may be 0).

then:

\(\displaystyle 1 = 2a_0 + (2a_1 + b_0)x + \cdots + (2a_k + b_{k-1})x^k + b_kx^{k+1}\).

since these are equal polynomials, we must have equal constant terms, so \(\displaystyle 1 = 2a_0\) for some integer \(\displaystyle a_0\), which cannot happen.
 
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Jan 2010
594
5
Hobart, Tasmania, Australia
Excellent ... yes, so obvious when you see how ... :)

Thanks so much ....

Peter