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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?

Heitor Fontana
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2 Answers2

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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.

Martin Brandenburg
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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
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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.

Georges Elencwajg
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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