Are linear polynomials like this irreducible?

Question

Let $K$ be a field and let $f,g\in K[y]$ such that $f,g \notin K$ (hence not constant), and $(f,g)=1$ (relatively prime). Is it true that $\phi(z)=zg(y)-f(y) \in K[y][z]$ is irreducible? I know that linear polynomials with units as leading coefficients are irreducible when the coefficients are in an integral domain, but I do not know if $g(y)$ is a unit in $K[y]$ since I only know that $K[y]$ is a UFD.

Answer 1

I think I get this, so I'm going to post my own answer. We can show that $\phi$ is irreducible in $K(y)[z]$ iff $\phi$ is irreducible in $K[y][z]$ (this is because the content of $\phi$ in $K[y][z]$ is a unit, in particular, $1$, because $(f,g)=1$) (This is Lemma III.6.13 in Hungerford's Algebra). Since $K(y)$ is a field then $g(y)$ is a unit in it, therefore $\phi$ is irreducible in $K(y)[z]$, and hence in $K[y][z]$.

Answer 2

Units in $R[y]$ ($R$ an integral domain) are the units in $R$. hence $zg(y)-f(y)$ is irreducible in $K[y][z]$, unless $g(y)=0$ and $f(y)$ is a non-zero constant.

Answer 3

By considering the degree in $z$ a factorisation will have to be of the form $$ zg(y)-f(y)=(z g_1(y)-f_1(y))h(y). $$ Then equating coefficients means that $h|f, h|g$ so that $h|1$ and hence must be degree $0$ in $y$, and so in $K$. That means our factorisation is not proper after all.