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Dimension of an algebraic variety
- P(F,G) = 0.
That implies that F and G are constrained to take related values (up to some finite freedom of choice): they cannot be truly independent.
For the function field even to be defined, V here must be an irreducible algebraic set; in which case the function field (for an affine variety) is just the field of fractions of the co-ordinate ring of V. It is easy to define by polynomials sets that have 'mixed dimension': a union of a curve and a plane in space, for example. These fail to be irreducible.
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