Irreducible polynomial in the extension k(x)/k(u).

Question

If I consider $k$ a field of prime characteristic (maybe this is not important here...). Consider the field of fractions $k(x)$ and $u \in k(x)$ where $u=\frac{f(x)}{g(x)}$ where $f$ and $g$ are relatively prime. If we look at the extension of fields k(x)/k(u), why is the polynomial $f(z)-ug(z) \in k(u)[z]$ irreductible (with $f,g \in k[z]$)?

Answer 1

As $P(u) = f(x)-u g(x)$ is of degree one in $u$, it is irreducible in $k(x)[u]$. As $f$ and $g$ are relatively prime, $P$ is primitive in $k(x)[u]$. According to Gauss's lemma (polynomial), $P$ is irreducible in $k[x][u] = k[x,u]$. As the fraction field of $k(x)[u]$ is $k(x)(u)$ and $P \notin k[x]$ (because $u \notin k[x]$) the same Gauss's lemma implies that $f(x)-u g(x)$ is irreducible in $k(u)[x]$.