Let \(Y\) be the plane curve \(y = x^2\), the zero set of \(f = y - x^2\). Show that \(A(Y)\) is isomorphic to a polynomial ring in one variable over \(k\).
Let \(Z\) be the plane curve \(xy = 1\). Show that \(A(Z)\) is not isomorphic to a polynomial ring in one variable over \(k\).
Let \(f\) be any irreducible quadratic polynomial in \(k[x,y]\), and let \(W\) be the conic it defines. Show that \(A(W)\) is isomorphic to \(A(Y)\) or to \(A(Z)\). Which one, and when?
Solution
(a) \(A(Y) = k[x,y]/\gens{y-x^2} \cong k[t,t^2] \cong k[t]\).
(b) \(A(Z) = k[x,y]/\gens{xy-1} \cong k[x^{\pm 1}]\). This is not a polynomial ring: it contains the unit \(\inverseof{x} \notin k\), while the units of \(k[t]\) are exactly \(k^*\).
(c) An affine change of coordinates reduces \(F\) to one of the two cases above. Write the conic as \[ F(x,y) = \tv{x, y, 1}^t \begin{pmatrix} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{pmatrix} \tv{x, y, 1} . \] The matrix is symmetric, so it diagonalises, and the corresponding affine change of coordinates puts \(F\) in the form \(\lambda_1 x^2 + \lambda_2 y^2 + \lambda_3\).
Which case appears is decided by whether the conic is degenerate at infinity: the parabola \(y = x^2\) meets the line at infinity in one point, the hyperbola \(xy = 1\) in two. So \(A(W) \cong A(Y)\) when the conic is tangent to the line at infinity, and \(A(W) \cong A(Z)\) otherwise.
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PR-7OT2ZThe ideal-variety correspondenceuses