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In Shafarevich Basic Algebraic Geometry I, there is a theorem (p.39 Theorem 5) that state:

Any Irreducible closed set is birational to a hypersurface of some affine space $\mathbb{A}^n$

I wonder why it should be irreducible? is the result false for a reducible closed set?

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    $\begingroup$ You have to define what you mean by "birational" for a reducible variety. One definition of a "birational" map is that it is an isomorphism on an open subset. In this sense a birational map from an irreducible component $Z$ of a reducible set $X$ is also birational with respect to $X$. $\endgroup$
    – pinaki
    Apr 23, 2011 at 6:40
  • $\begingroup$ thanks for the reply, yes i notice that in that book the functional field is only defined on the affine variety. so do rational map. If I consult Harsthorne's book, the birational map is like what you've said defined in any variety. assume we use Harsthorne's (and your definition), is the result still true for any variety? Edit: and i just found in Harsthorne that in case of any variety birational equivalence is constructed to the hypersurface of a projective variety (adding one in dimension).. .. i should have more careful. $\endgroup$
    – Ajat
    Apr 23, 2011 at 8:07
  • $\begingroup$ Ajat I'm confused. In Hartshorne, varieties are always irreducible (except for the curves in the surfaces chapter). Regardless, using auniket's definition, then the answer to your question is a clear yes. Choose any irreducible component and find a hypersurface birationally equivalent to it, with that definition of birational, you have obtained something birationally equivalent to your given variety as you wanted. $\endgroup$
    – Karl Schwede
    Apr 23, 2011 at 19:22

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I usually define birational, for reduced schemes of finite type, to mean that there exists an isomorphism on a dense open set. With this definition, the answer to this question is

  • If and only if the closed subset is equidimensional (in this context I mean all irreducible components have the same dimension).

Suppose for simplicity we are working over an algebraically closed field. We may as well assume our variety is projective, $X \subseteq \mathbb{P}^n$ of dimension $r$. Perform simultaneous generic projections for all components, until each component is projected to a hypersurface in $\mathbb{P}^{r+1}$.

For an explanation of generic projections, see for example:

  • Hartshorne, Chapter IV, Section 3, or
  • "Generic projections of algebraic varieties." by Joel Roberts, Section 5.

Of course, for non-equidimensional sets, there is no hope. If the set is the $z = 0$ plane in $\mathbb{A}^3$ unioned with the line $(x = 0, y = 0)$, then of course this set can never be a hypersurface up to birational equivalence, because each component would need to be hypersurfaces in different dimensional spaces.

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  • $\begingroup$ yes indeed I confused the affine and non-affine variety (some books i used, use affine variety for usual algebraic set). It is clear now. and thanks you very much. $\endgroup$
    – Ajat
    Apr 24, 2011 at 3:50

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