Question concerning prime ideals of $\mathbb{C}[x,y]$ I know that $(0)$, $(x-a,y-b)$ for $(a,b)\in\mathbb{C}^2$ and $(f(x,y))$ for $f(x,y)$ irreducible in $\mathbb{C}[x,y]$ are all prime ideals in $\mathbb{C}[x,y]$.
What I'd like to understand is why they are the only prime ideals. In particular, I'd like to know why the following outline of a proof is valid (which comes from Vakil's algebraic geometry notes): Let $P$ be a prime ideal that is not principal. &quot;Show you can find $f(x,y),g(x,y)\in P$ with no common factor. By considering the Euclidean algorithm in the Euclidean domain $\mathbb{C}(x)[y]$, we can find a nonzero $h(x)\in (f(x,y),g(x,y))\subset P$.&quot;
I have two questions.

Why can we find such $f(x,y),g(x,y)$ without a common factor? If we were to pick to distinct generators of $P$, I don't see why they must be coprime. Further, I don't think we can just factor out a common factor either.
Why does the Euclidean algorithm imply we can find such a nonzero $h(x)$? Might this not still be of the form $h(x,y)$ since the Euclidean algorithm only guarantees the existence of $q(x,y)$ and $h(x,y)$ with $f(x,y)=g(x,y)q(x,y)+h(x,y)$ with the $y$ degree of $h(x,y)$ less than that of $g(x,y)$? Does this somehow follow from $f(x,y)$ and $g(x,y)$ having no common factors?
We know that $P$ can be generated by finitely many generators (because $\mathbb C[x, y]$ is Noetherian), so suppose that $f_1, \dots, f_n$ is a minimal set of generators for $P$. Since $P$ is not principal, $n \geq 2$.
Now recall that $\mathbb C[x, y]$ is a unique factorisation domain. So suppose that $h$ is a greatest common divisor for $f_1$ and $f_2$. Then $f_1 = hg_1$ and $f_2 = hg_2$, where $g_1, g_2$ have no non-trivial common factor.
Now $h$ can't be in $P$, otherwise $P$ would be generated by $h, f_3, \dots, f_n$, contradicting my assumption that the generators $f_1, f_2, \dots, f_n$ were minimal.
But $f_1 = hg_1 \in P$ and $P$ is prime, so either $h\in P$ or $g_1 \in P$. Since $h \notin P$, we have $g_1 \in P$. By a similar argument, $g_2 \in P$ too.
Thus we have constructed elements $g_1, g_2 \in P$ that have no non-trivial common factor.
For the second part, see this answer.
Ad 1: Pick $f$ of minimal (total) degree (but non-zero, of course). Then $P\ne(f)$, so there exists $g\notin (f)$ and again we pick it of minimal degree. What could you say about the degree of  a common factor?