# Rational Varieties

1. Feb 26, 2007

### Diophantus

I am trying to show that some given simple affine algebraic varieties are rational (i.e. birationally equivalent to some A^k).

Are there any tricks or even nice algorithms for finding the birational maps and their inverses? Examples are the curve x^2 + y^2 = 1 and the surface x^2 + y^2 + z^2 = 1?

I have tried to tackle the first one by assuming that the map from A^1 to the curve takes the form of a pair of quotients of linear polynomials, and then try to work out suitable coeffiecients of these polynomials but it gets very messy very quickly and offers no insight into making the thing invertible.

Any help would be greatly appreciated. Thanks.

Last edited: Feb 26, 2007
2. Feb 26, 2007

### mathwonk

there are standard tricks for examples including the ones you name.

e.g. if a hypersurface (e.g. a plane curve or surface in space) of degree n has a point of multiplicity n-1, then projection from that point defines a generically one to one map to a coordinate space.

but I think you need characteristic zero to conclude the map is birational. maybe not.

in general it is very hard to prove varieties are rational.

3. Feb 28, 2007

### matt grime

You might also need some condition on algebraic closure as well.

4. Mar 1, 2007

### Diophantus

Yes the field is assumed to be closed, sorry. I have now learned of some tricks for the above cases which involve a generalisation of projecting a line from a real circle/sphere to another point on the circle/plane to an algebraically closed field.

If anything I now have a feel for how hard this type of problem is in general which I suppose is a good lesson learned.

I may have some more algebraic geometry questions to follow soon!

5. Mar 1, 2007

### mathwonk

e.g. it was known for decades that a cubic hypersurface in P^4 is the image of P^3 under a map of degree 2, but whether there is a map of degree one was unknown for deacades more. finally it was proved there can be no such degree one map.