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?

Mathematics Asked by Nuntractatuses Amável on November 14, 2021

1 Answers

One Answer

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.)

Answered by Kenny Wong on November 14, 2021

Add your own answers!

Related Questions

Nimber multiplication

2  Asked on July 30, 2020 by yberman


Calculating distance when velocity is given

1  Asked on July 27, 2020 by aruha


Ask a Question

Get help from others!

© 2021 All rights reserved.