Showing that a morphism is not finite I'va started algebraic geometry this year, and yesterday we corrected a few exercicses at school, but there's one little point that I didn't understand. 
The question is : 
in $\mathbb{A}^{2}$ let $W=V(xy-1)$. Show that $y : W \mapsto \mathbb{A}^{1}$ is not finite.
We defined a finite morphism as follow : a morphism $f : X \mapsto Y$, where X and $Y$ are affine varieties, is finite if $f^{*} : k[Y] \mapsto k[X]$ makes $k[X]$ a finitely generated module over $k[Y]$ ($k$ is an algebraically closed field).
The teacher explained saying that it is not finite because $k[x^{-1}]$ is not integral over $k[x,x^{-1}]$ (which I totally see).
However, if we take into consideration the definition, the $k[Y]$ is $k[x]$, and $k[X]$ is $k[x,x^{-1}]$. So don't we have to check if $k[x,x^{-1}]$ is finitely generated over $k[x]$ ? There is definitely something I don't understand about the definition of a finite morphim..
Thank you in advance for your help :)
 A: The statement "$k[x^{-1}]$ is not integral over $k[x,x^{-1}]$" does not make sense and the most obvious interpretation of it is false: to say $B$ is integral over $A$, we normally need $B$ to be an $A$-algebra, and $k[x^{-1}]$ is not a $k[x,x^{-1}]$-algebra.  And if you just mean that there exists an element of $k[x^{-1}]$ which is not integral over $k[x,x^{-1}]$, then that is false: every element of $k[x^{-1}]$ is an element of $k[x,x^{-1}]$, and thus integral over $k[x,x^{-1}]$.
So it seems your teacher misspoke (or you misinterpreted what they said).  What you need to check instead is that $k[x,x^{-1}]$ is not finitely generated as a $k[x^{-1}]$-module.  Note that you want $k[x^{-1}]$, not $k[x]$, as the second ring here because the map $W\to\mathbb{A}^1$ is $y$, not $x$.  That is, it is the morphism corresponding to the inclusion map $k[y]\to k[x,y]/(xy-1)$, and identifying $k[x,y]/(xy-1)$ with $k[x,x^{-1}]$, $y$ maps to $x^{-1}$, so we have the inclusion of the subring $k[x^{-1}]$, not $k[x]$.
