# 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

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

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