Prime ideals of $Bbb C[x, y]$

You're not supposed to divide $g(x, y)$ by $f(x, y)$. You're suppose to use the Euclidean algorithm in $mathbb C(x)[y]$ to find a greatest common divisor for $f(x, y)$ and $g(x, y)$. (Remember, $mathbb C(x)[y]$ is a Euclidean domain, so it makes sense to use the Euclidean algorithm.)

The greatest common divisor for $f(x,y)$ and $g(x,y)$ in $mathbb C(x)[y]$ can be written in the form $frac{a(x)c(x,y)}{b(x)}$, where $c(x,y)$ has no non-trivial factors in $mathbb C[x,y]$ that are purely polynomials in $x$. (Remember, $mathbb C[x,y]$ is a UFD, so this statement makes sense.)

Then for some $p(x,y)$, $q(x)$, $r(x, y)$ and $s(x)$, we have $$ frac{a(x)c(x,y)}{b(x)} frac{p(x, y)}{q(x)} = f(x, y), frac{a(x)c(x,y)}{b(x)} frac{r(x, y)}{s(x)} = g(x, y), $$ which is to say that $$ a(x)c(x,y)p(x,y) = b(x)q(x)f(x, y), a(x)c(x,y)r(x,y) = b(x) s(x) g(x, y). $$

But $c(x,y)$ doesn't have any factors in common with $b(x)$, $q(x)$ or $s(x)$ in $mathbb C[x, y]$, since $c(x, y)$ has no factors in $mathbb C[x, y]$ that are purely polynomials in $x$. So $c(x, y)$ must divide both $f(x, y)$ and $g(x, y)$ in $mathbb C[x, y]$.

And now, we use the fact that $f(x, y)$ and $g(x, y)$ have no common factor in $mathbb C[x, y]$ to conclude that $c(x, y)$ is a constant.

Hence the greatest common divisor for $f(x,y)$ and $g(x,y)$ in $mathbb C(x)[y]$ can be written in the form $a(x) / b(x)$. (We absorb the constant into $a(x)$.)

Using the Euclidean algorithm (i.e. the Bezout identity), it must be possible to write the greatest common divisor $a(x) / b(x)$ of $f(x, y)$ and $g(x, y)$ in $mathbb C(x)[y]$ as a linear combination, $$ frac{a(x)}{b(x)} = frac{u(x, y)}{t(x)}f(x, y) + frac{v(x, y)}{w(x)}g(x, y),$$ or, clearing denominators, $$ a(x) t(x) w(x) = u(x, y) b(x) w(x) f(x, y) + v(x, y) b(x) t(x) g(x, y).$$

Defining $$h(x) := a(x) t(x) w(x),$$ we have a non-zero polynomial in $x$ which is contained in $(f(x, y), g(x, y))$, and hence is contained in $mathfrak p$ too.

Using the fact that $(f(x, y), g(x, y))$ is prime (by assumption), we see that some linear factor $x - a$ of $h(x)$ must be in $mathfrak p$.

(Note that $h(x)$ really does have linear factors. If $h(x)$ were a constant, then $mathfrak p$ would be the whole of $mathbb C[x, y]$, contradicting the fact that $mathfrak p$ is prime.)