Prime ideals of $\Bbb C[x, y]$ In the exercise 3.2.E of Vakil's "Foundations of Algebraic Geometry", it is asked to prove that all the prime ideals of $\Bbb C[x, y]$ are of the form $(0)$, $(x-a, y-b)$ or $(f(x, y))$, where $f$ is an irreducible polynomial. In order to do so, it is suggested to consider a non-principal prime ideal $\mathcal{p}$ along with $f, g \in \mathcal{p}$ with no common factor. Then, dividing $g$ by $f$ in $\Bbb C(x)[y]$, one is supposed to find $h(x) \in (f, g)$ - so some factor of the form $x-a$ is in $\mathcal{p}$. But I can't seem to figure out how to find such $h(x)$, because the expression of the division is something of the form:
$$g(x, y) = f(x,y) \left (\frac{p_0(x)}{q_0(x)} + \cdots + \frac{p_n(x)}{q_n(x)}y^n \right) + \left(\frac{r_0(x)}{a_0(x)} + \cdots + \frac{r_m(x)}{a_m(x)}y^m \right)$$
And I don't know how to relate the fact that $f, g$ have no common factor with this expression. My guess is that it has something to do with the term $\frac{r_0(x)}{a_0(x)}$, but I don't know what it is. Can anyone shed some light?
 A: 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.)
