Exercise 5, Ch. 2, Sec. 2 - Fundamentals of Number Theory (William J. LeVeque):

If \(D\) is a Euclidean domain, and \(a\) and \(b\) are relatively prime elements of \(D\), then there are \(m,n \in D\) such that \(ma + nb = 1\).

• Show that 2 and \(x\) are relatively prime elements of \(Z[x]\).
• Conclude that \(Z[x]\) is not a euclidean domain (nevertheless it is still a unique factorization domain).

Proof:

If 2 and \(x\) are relatively prime elements of \(Z[x]\) then \(gcd(2,x) = 1\) so let's start by assuming otherwise and say

\[\begin{aligned}
gcd(2,x) = d(x)
\end{aligned}\]

With \(d(x)\) not equal to 1. Then \(d(x)\) must divide both x and 2 or

\[\begin{aligned}
2 = d(x)u(x) \\
x = d(x)v(x)
\end{aligned}\]

where \(v(x)\) and \(u(x)\) are polynomials in \(Z[x]\). Taking the degrees of the last two equations into account we must have

\[\begin{aligned}
0 = d + u \\
1 = d + v
\end{aligned}\]

where \(d,u,v\) are the degrees of their corresponding polynomials in \(Z[x]\). The first says \(d\) must be a constant (integer) and together with the second says the degree of \(v(x)\) is one so \(v(x) = Cx\) where \(C)\) is an integer. Thus

\[\begin{aligned}
x = dCx
\end{aligned}\]

and so the only way this can be true is if \(d = 1\) and \(C = 1\) contradicting our initial assumption that \(d(x)\) is not 1.

Now to prove the second part start by assuming \(Z[x]\) is Euclidean so we can write,

\[\begin{aligned}
2m(x) + xn(x) = 1.
\end{aligned}\]

Both terms on the left can have degree at most 1 and so either they are both constants leading to

\[\begin{aligned}
2m + xn = 1.
\end{aligned}\]

which is not possible or \(m(x) = Cx\) and \(n(x) = n\) where \(n,C\) are constant integers so that

\[\begin{aligned}
2Cx + xn = 1.
\end{aligned}\]

or by factoring,

\[\begin{aligned}
(2C + n)x = 1.
\end{aligned}\]

for which there are no such integers \(n,C\) which can make this true. Thus we conclude there are no \(m(x), n(x)\) for which

\[\begin{aligned}
2m(x) + xn(x) = 1.
\end{aligned}\]

so \(Z[x]\) can not be Euclidean.

If anyone sees an error in this proof please post a comment and let me know where the argument is incorrect. I'm trying to learn some algebra and I'm looking to learn -- not just to be right!