$$Z[x]$$ is not Euclidian but still a unique factorization domain

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!