Prime ideals of the ring $\Bbb C[x,y]$, proof given in Vakil's book.

Question

I am reading Vakil's Algebraic Geometry lecture note, and the following is the proof of the statement that any non-principal prime ideal of $\Bbb C[x,y]$ is of the form $(x-a,y-b)$ for some $a,b\in \Bbb C$.

Suppose $\mathfrak{p}$ is a prime ideal that is not principal. Then we can find $f,g\in \mathfrak{p}$ with no common factor. By considering the Euclidean algorithm in $\Bbb C(x)[y]$, we can find a nonzero $h(x)\in (f,g)\subset \mathfrak{p}$. Since $\mathfrak{p}$ is prime, one of the linear factors of $h(x)$, say $x-a$, is in $\mathfrak{p}$. Similarly $y-b\in \mathfrak{p}$ for some $b\in \Bbb C$.

My question is: how can we find such $h(x)$ using Euclidean algorithm?

Answer 1

Note that $\Bbb{C}(x)[y]$ is a PID and gcd$(f,g)=1$, by Bezout identity (or Euclidean algorithm), there are $a',b'\in \Bbb{C}(x)[y]$ such that $a'f+b'g=1$. Now we can find some $h\in \Bbb{C}[x]$ such that $a=a'h$ and $b=b'h$ are in $\Bbb{C}[x][y]=\Bbb{C}[x,y]$. Therefore, $$af+bg=(a'h)f+(b'h)g=h(a'f+b'g)=h,$$ i.e., $h\in (f,g)$.

Answer 2

The Euclidean algorithm allows you to write $f=pg+h$ where $h$ is non-zero, so $h=f-pg$ is the desired item.