For a field $F$, the ideal $(x,y)$ in $F[x,y]$ is not principal

In a commutative ring the multiples of $h$ form an ideal containing $h$ and lie in every ideal containing $h$, so [L3] identifies $(h)$ with the set of multiples of $h$. Suppose for contradiction that $I=(h)$; since $x,y\in I$, the element $h$ divides both. Treating the ring as $F[x][y]$ by [L1], degree in $y$ and the domain property [L2] show from $h\mid x$ that $h\in F[x]$.

From $h\mid y$, comparison of the leading coefficient in $y$ gives $1=hc$ for some $c\in F[x]$, so $h$ is a unit and $(h)$ is the whole ring.

Evaluation at $x=0,y=0$ exists by [L4] and has kernel an ideal containing $x,y$, so [L3] gives $I$ inside that proper kernel; thus $I$ is proper, contradicting step 2.1, and $I$ is not principal.