Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 15 1 Solution Created 2026-10-03 Updated 2026-10-14
A monomial is invariant precisely when is divisible by : pulling back along the generator multiplies it by . Since distinct monomials are linearly independent, an invariant polynomial is a linear combination of exactly these invariant monomials. If , write and obtainIf , the analogous expression uses and . Thus the polynomial invariant ring is generated bywhich satisfy .
To check that this is the only relation, reduce each monomial in using until it has one of the formsTheir images are, respectively, , and . These are pairwise distinct monomials in , so no nontrivial linear combination of the reduced forms maps to zero. ThereforeThis identifies the determinant-one cyclic quotient of the affine plane with the affine hypersurface , and the quotient map is .
Apply the Jacobian criterion to . Its gradient isOver , with the printed , this vanishes only at . Every other point is smooth. This point is exactly the image of the origin, and its inverse image is only the origin, since forces . In fact the origin itself is singular: its tangent space has dimension three while the hypersurface has dimension two. Hence the quotient is smooth away from the image of the origin, with that point its unique singularity. Primality is not needed for this calculation; the same argument works for every integer .
Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 21 1 Solution Created 2026-10-03 Updated 2026-10-14
The determinant-one cyclic quotient of the affine plane has the closed embeddingHere denotes the cyclic group generated by the given diagonal matrix. We establish the presentation of the polynomial invariant ring and verify that its fibres really are the group orbits.
On the coordinate ring , the inverse-pullback action multiplies by . Since these monomials are a basis, a polynomial is invariant precisely when each of its nonzero monomials satisfies . If , such a monomial is ; if , it is . ThereforeThese generators give a surjective homomorphism , and lies in its kernel of a ring homomorphism. To prove there are no further relations, reduce any monomial using . The result is a linear combination ofTheir images are respectively , and . All have distinct exponent pairs, so they are linearly independent in . Any polynomial in the kernel of a ring homomorphism has zero reduced expression and hence belongs to . ConsequentlyThis proves the claimed closed subvariety embedding of the affine quotient.
For surjectivity of , if choose and set . Then . If , the relation forces ; choose and any th root of for . If two points have the same image and , equality of their th powers gives and equality of gives . If , both first coordinates vanish, and the same argument using gives the orbit relation; the case has only that one point in its fibre. Thus every fibre is exactly one group orbit.
The map is finite: and satisfy the monic equations and , and the finitely many monomials , , span as a module over its polynomial invariant ring. A finite morphism is closed when it is surjective as here and therefore gives the quotient topology. For a principal open subset defined by an invariant , its regular functions pull back to andIndeed, because is invariant, an invariant fraction with denominator has invariant numerator. These principal open subsets cover the affine hypersurface, proving the required local equality of invariant regular functions. Together with the orbit fibres, this verifies the geometric quotient, not merely a bijection of underlying sets.