Let \(Y \subseteq \AA^3\) be the set \(Y = \ts{ (t, t^2, t^3) \st t \in k }\).
Show that \(Y\) is an affine variety of dimension \(1\).
Find generators for the ideal \(I(Y)\).
Show that \(A(Y)\) is isomorphic to a polynomial ring in one variable over \(k\).
We say \(Y\) is given by the parametric representation \(x = t,\ y = t^2,\ z = t^3\).
Solution
The coordinate ring, and the dimension. \[ A(Y) = k[x,y,z]/\gens{y - x^2,\ z - x^3} \cong k[t, t^2, t^3] \cong k[t] , \] since \(t^2\) and \(t^3\) are already polynomials in \(t\). This settles the third part, and with \(\dim Y = \krulldim A(Y)\) and \(\krulldim k[t] = 1\) it settles the first.
That \(Y\) is a variety. It suffices that \(Y\) is an irreducible closed subset of \(\AA^3\). It is closed because \(Y = V(y - x^2,\ z - x^3)\). It is irreducible exactly when \(I(Y)\) is prime, and \(I(Y) = \gens{y-x^2,\ z-x^3}\) is prime because the quotient \(k[t]\) is a domain.
That the ideal is the full \(I(Y)\) rather than a smaller one uses radicality: in a UFD a principal ideal generated by irreducibles is radical, and an ideal is radical exactly when its quotient is reduced.
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P-AGH29PROJCLOSUREProjective closure of an affine variety, and the twisted cubic curveuses