Prove that, if $f$ is an irreducible element in $\mathbb{C}[x,y]$, then there exist $(a,b)\in\mathbb{C}^2$ such that $(f)\cap\mathbb{Q}[x,y]=(x-a,y-b)\cap\mathbb{Q}[x,y]$.
The original question is, if $\pi:\mathbb{C}^2\to \mathrm{prime\ ideals\ of\ }\mathbb{Q}[x,y]$ is defined to be mapping $(a,b)$ to $\{f\in\mathbb{Q}[x,y]:f(a,b)=0\}$, prove that this map is surjective, which is an exercies in the Rising Sea. I ended up with the statement above.
My thought is, assume that $f\neq0$ (in which case we can find two elements $a,b$ that are transcendential independent, and $(a,b)$ will satisfy), and degree of $x$ in $f$ is not zero. Then we can find a root $a$ of $f(x,\pi+q)=0$, where $q\in\mathbb{Q}$ such that it is not a zero function. Then I guess the point $(a,\pi+q)$ will work, but I don't know how to prove it.