Exercise A0.V.2.04

Proposition

In $\mathbb{Z}[x,y]$, $\gcd(x,y) = 1,$ yet $1$ is not a linear combination of $x$ and $y$.

Proof

Let $(d)$ be a principal ideal such that $(x,y) \subseteq (d).$ If $\deg_x(d) \gt 0$ then $y \notin (d),$ since $\deg_x(y) = 0.$ Symmetrically, $\deg_y(d) \gt 0$ cannot hold either. So $d \in \mathbb{Z}.$ If (up to units) $d \neq 1$ then $x \notin (d),$ since $x$ has coefficient of $1.$ So it must be that (up to units) $d = 1.$

Suppose $f, g \in \mathbb{Z}[x,y]$ are such that $fx + gy = 1.$ Let $\varphi : \mathbb{Z}[x,y] \to \mathbb{Z}$ be the evaluation map given by $x \mapsto 0, y \mapsto 0.$ In this case, $\varphi(fx + gy) = 0$ yet $\varphi(1) = 1.$ This is a contradiction.

$\blacksquare$