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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?

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Where exactly is that result stated ? – Georges Elencwajg Jan 21 '14 at 10:46
Oops sorry I'm wrong (I wrote that down on some notes and…) it is actually in Hulek' Elementary Algebraic Geometry p.103 last line. Let me edit the post. – Heitor Fontana Jan 21 '14 at 11:11
Thanks a lot for answering so quickly, dear Heitor. I can stop browsing Shafarevich now :-) And +1 for this nice question. – Georges Elencwajg Jan 21 '14 at 12:16
up vote 6 down vote accepted

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.

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I hadn't read your answer while I composed mine (and had lunch). In a way I'm glad to realize there isn't a clever shortcut. Anyway +1 for your answer. – Georges Elencwajg Jan 21 '14 at 12:09

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.

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.

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"Again this is not obvious but results from Gauss:" Isn't this just the fact that $A \to A[S^{-1}]$ is injective when $A$ is an integral domain and $0 \notin S$? – Martin Brandenburg Jan 21 '14 at 13:57
@Martin: the point is that a morphism $u:k[V]=k[x_1,\ldots,x_n]/(f)\to L=k(x_2,\ldots,x_n)[x_1]/(f)$ extends to a morphism $\hat {u}: k(V)\to L$ if and only if $u$ is injective. And injectivity amounts to the implication that a non-multiple of $f$ in $k[x_1,...,x_n]$ is sent to a non-multiple of $f$ in $k(x_2,\ldots,x_n)[x_1]$. – Georges Elencwajg Jan 21 '14 at 14:54
I know that. I wanted to indicate that injectivity is easy when you use localizations ... – Martin Brandenburg Jan 21 '14 at 15:42

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