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?



Irreducible closed set birational to a hypersurface
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.


