$k$ algebraically closed field $\Rightarrow$ $V(f)\subset k^2$ infinite Let $k$ be an algebraically closed field, and $f\in k[X,Y]$ a non-constant polynomial.
Show that $V(f)\subset k^2$ is infinite.

We solved this exercise in my tutorial class, but I have some questions to the solution.

Firse we proved that if $k=\overset{-}k$ we have $|k|=\infty$.
Then he wrote: 
We fix $x\in k$. Let $f_x=f(x,y)\in k[Y]$ has $deg(f_x)$ many roots. Then we change $x$ to obtain infinitely many roots.
We only have to take care if $deg(f_x)<def(f)$ in $Y$. (Why?)
$f$ is non-constant, so $f$ depends on $X$ or $Y$. WLOG $Y$. So $f = \sum\limits_{i=0}^n f_i(X)Y^i$, $n>0$.
$deg(f_x)<deg(f)$ in $Y$ if and only if $f_n(X)=0$.
But $f_n$ hat finitely many roots (Why can't hold the above then?)
Hence $V(f)$ is infinite. (I don't see why we are already finished here and what happened in these last steps.)
I hope someone can explain me the questions.
All the best! Luca
 A: You only need $\deg(f_x) > 0$ for infinitely many $x\in k$, so showing $\deg(f_x)=\deg(f)$ for all but finitely many $x$ is a bit overkill, but gets the job done without much extra effort.  
A: The basic sketch of the proof is that in order to show that f(x,y) has infinitely many zeroes (so infinitely many points of $k^2$) you will show that for infinitely many values of $x \in k$, there is a $y \in k$ such that $f(x, y) = 0$, i.e. so that $(x, y)$ is in $V(f)$.  The strategy here is to deal with one variable polynomials, which are much easier to understand, and the way we go about doing this is to plug in a number for $x$ so that we just have a polynomial in $y$ left.
Now, in general, when we think of $f(x, y)$ as a polynomial in $y$ with coefficients in $x$, it has some degree, $\deg (f)$, and so most of the time when we plug in some $x$ value, say $x_0 \in k$ we get $\deg (f)$ solutions to $f(x_0, y)$.  (Here we are relying on $k$ being algebraically closed.)  However, we might have a problem if plugging in $x_0$ makes the coefficient of $y^{\deg(f)}$ zero.  In this case, $f(x_0, y)$ has degree smaller than $\deg(f)$ (which you have called $\deg(f_{x_0})$ and thus fewer roots.  However, $f_n$ has finitely many roots, so while it can happen that $\deg(f_x) < \deg (f)$, it only happens for finitely many $x$.  Thus we can see that for all but finitely many values of $x$, we can find a $y$ such that $(x,y) \in V(f)$.  Since $k$ is infinite, we see that "all but finitely many" in this case means infinitely many, and so we get infinitely many points in $V(f)$.
