Taking vanishing ideals of $\mathbb{C}^2$ in $\mathbb{Q}[x,y]$ is surjective to $\operatorname{Spec}\mathbb{Q}[x,y]$ A friend of mine and me are currently reading through Vakil's draft
and are having a hard time solving following exercise (s. 3.2.I in the version of 29.december 2015):

Show, that the map
  $$\phi: \mathbb{C}^2 \to \operatorname{Spec}\mathbb{Q}[x,y]\quad, \quad (x,y) \mapsto I(x,y):=\{p \in \mathbb{Q}[x,y] \ |\ p(x,y)=0\}$$
  is surjective, where $\operatorname{Spec}\mathbb{Q}[x,y]$ is the set of prime ideals in $\mathbb{Q}[x,y]$.

We have managed so far, that it hits every maximal ideal of $\mathbb{Q}[x,y]:$
Let $m \subseteq \mathbb{Q}[x,y]$ be a maximal ideal, so $\mathbb{Q}[x,y]/m$ is an algebraic field extension of $\mathbb{Q}$ and so embedds into $\mathbb{C}$. Especially we have images $a,b$ of $x$ and $y$ in $\mathbb{C}$. Furthermore we have (because of the evaluation morphism) $m \subseteq \phi(a,b) \neq \mathbb{Q}[x,y]$ and since $m$ is maximal equality holds.
Also the case $(0)=\phi(e,\pi)$ is shown ($e,\pi$ as any transcendental numbers).
On the other hand for prime ideals generated by a irreducible polynomial $p$ we have the hunch, that taking $a=\pi, b=\text{any solution of }p(a,b)=0$ should do the trick (once again $\pi$ as the transcendental number of your choice). Once again one can easily see, that $p \subseteq \phi(a,b)$. The problems we encountered here, are:


*

*Is also $(p) \supseteq \phi(a,b)$ true?

*Are these all prime ideals in $\mathbb{Q}[x,y]$ or are there non-maximal prime ideals with more than one generators?

*Is there a sensefull way to extend the answer to the polynomial ring with n variables (the maximal ideals don't care about that)?


I would be really grateful, if somebody can enlighten us.
Edit: The second question is partly answered here for the asked case (polynomials over 2 variables). It remains to show, that a prime ideal with 2 generators is maximal.
 A: Following anon's comment we solved the remaining case like that (PS: The first paragraph is mainly about transcendence degree 1):
Let $0 \neq p \subseteq \mathbb Q[x,y]$ be a prime ideal, which is neither maximal nor zero. We consider now the quotient field $K=Quot(\mathbb Q[x,y]/p)$ ($p$ is prime, hence we can do that). $K$ is a field extension of $\mathbb Q$ and because $p$ is not maximal it has at least transcendence degree 1 (else $\mathbb Q[x,y]/p$ would be already a field). Also since $p \neq 0$ we have an algebraic relation between $x$ and $y$, hence the field extension has transcendence degree at most 1. W.l.o.g we can hence assume the image of $x$ in $K$ to be transcendental over $\mathbb Q$. Hence we get the chain of field extensions (first purely transcendental, second algebraic):
$$ \mathbb Q \subset M:=Quot(\mathbb Q[x]/(p \cap \mathbb Q[x])) \subseteq K$$
We now identify $M \cong \mathbb Q(\pi) \subseteq \mathbb C$ for $\pi$ the transcendental number of your choice (and hence we also can consider $K \subseteq \mathbb C$). Since $M \subseteq K$ is a finite algebraic extension (and separabel), we have a primitive element $\zeta \in K \subseteq \mathbb C$ with $M[\zeta]=K$. So altogether we have a field isomorphism
$$\mathbb Q (\pi, \zeta) \to K \ , \ (\pi, \zeta) \mapsto (x,y) $$
Concatenating now the evaluation morphism for $\mathbb Q[x,y]$ at $(\pi, \zeta)$ with this isomorphism yields, that the vanishing ideal is exactly $p$, hence $I(\pi, \zeta)=p$.

Similarly (maybe nicer than the special case before):
Let $p \subseteq \mathbb Q[x_1, \dots, x_n]$ be a prime ideal and $K=Quot(\mathbb Q[x_1, \dots, x_n]/p)$, then $K$ is a field extension of finite transcendence degree over $\mathbb Q$. Hence there exist $t_1, \dots, t_k, \zeta_{k+1}, \dots, \zeta_n \in \mathbb C$ with $\mathbb Q(t_1, \dots, t_k, \zeta_{k+1}, \dots, \zeta_n) \cong K$ and as before the evaluation morphism now yields the desired kernel.
