Function field of an affine hypersurface I am reading Hulek' Elementary Algebraic Geometry, p.103
Let $V$ be an irreducible affine hypersurface, say $V=V(f)\subset\Bbb{A}^n$. Then the coordinate ring is by definition $k[V]=k[x_1,\ldots,x_n]/(f)$. Suppose $f$ contains the variable $x_1$. Then 
$$k(V)=k(x_2,\ldots,x_n)[x_1]/(f)$$
can somebody very kindly explain me this identification?
 A: By assumption $k[x_1,\dotsc,x_n]/(f) \cong k[x_2,\dotsc,x_n][x_1]/(f)$ is an integral domain. It follows that $f$ is irreducible over $k[x_2,\dotsc,x_n]$, as a polynomial in $x_1$. By assumption it is not constant. Hence, Gauss' Lemma tells us that it stays irreducible over $k(x_2,\dotsc,x_n)$. This means that $k(x_2,\dotsc,x_n)[x_1]/(f)$ is a field. Since it contains $k[x_1,\dotsc,x_n]/(f)$ as a subring and is generated by it, it must be its field of fractions.
A: The crucial point is that $f$ , which is irreducible in $k[x_2,\ldots,x_n][x_1]$, is still irreducible in $k(x_2,\ldots,x_n)[x_1]$: this is not trivial but follows from a result of Gauss on UFD's.
Hence $L:=k(x_2,\ldots,x_n)[x_1]/(f)$ is a field.    
The obvious $k$-morphism $k[V]=k[x_1,\ldots,x_n]/(f)\to L=k(x_2,\ldots,x_n)[x_1]/(f)$ extends to a morphism $$k(V)=\text {Frac} (k[x_1,\ldots,x_n]/(f))\to L=k(x_2,\ldots,x_n)[x_1]/(f)$$ Again this is not obvious but results from Gauss: if a polynomial  $g\in k[x_1,\ldots,x_n]$  is not a multiple of $f$ in  $k[x_2,\ldots,x_n][x_1]$, then $g$ will not be a multiple of $f$ in $k(x_2,\ldots,x_n)[x_1]$ either.   
Once we have this morphism of $k$-extensions $k(V)\to L$ the result follows easily: like all morphisms with source a field it is injective and   surjectivity is clear  once you realize that a polynomial  in $h\in k[x_2,...,x_n]$ is  invertible in $k(V)$, because $h$  cannot be a multiple of $f$ since $f$ contains $x_1$ and $h$ does not.  
Bibliography
A great reference on Gauss's results on UFDs is Artin's Algebra , Theorem (3.9) of Chapter 11, page 401.
Beware that the results in this theorem are elementary but quite subtle and that it is easy to make mistakes in their statements.
