Polynomials with integer coefficients - the GCD identity
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 ![\mathbb{Q} [x]](http://latex.codecogs.com/png.latex? \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 ![\mathbb{Z} [x]](http://latex.codecogs.com/png.latex? \mathbb{Z} [x] )
.
Also the GCD identity fails in .
They then ask the reader the polynomials 2 and x in 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
Re: Polynomials with integer coefficients - the GCD identity
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:
 = 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:
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 
for some integer 
, which cannot happen.
Re: Polynomials with integer coefficients - the GCD identity
Excellent ... yes, so obvious when you see how ... :-)
Thanks so much ....
Peter
