Ideal of Cubic curve I solved the following questions, but it is not correct, because it contradicts the last question. Where is wrong?
Let $Y \subset \mathbb{A}^3$ be the set $Y=\{(t,t^2,t^3); t \in k\}$.
Q: Show that dim $Y=1$. A: Because any irreducible closed subset of $\mathbb{A}^3$  is a point, 
 any irreducible closed subset of $Y$ is also a point. Thus the maximal length of a sequence of distinct irreducible closed  subsets $Z_0 \subset Z_1$ is one which is given by  $\phi \subset\{a\}$ for any point $a \in \mathbb{A}^3$ .
Q: Find generators for the ideal $I(Y)$.A: Because the polynomial ring is the principle ideal domain, it is sufficient to find only one generator. Intuitively I found $xy-z$, $y^2-xz$ but I am not sure how to confirm this .　
Q: Show that A(Y) is isomorphic to polynomial ring in one variable over algebraically closed field $k$.A: 
If we take the equation $xy-z$  as the generator of ideal $I(Y)$, the quotinet ring $k[x,y,z]/(xy-z) = k[x,y]$ and not isomorphic in one variable??? Contradiction ... why??
A: Q1. It is not true that the only irreducible subsets of $\mathbb{A}^3$ are points. In fact there are many irreducible subsets of dimension $1$ and $2$ (for example, the zero locus of an irreducible polynomial will be an irreducible subset of dimension 2).
Q2 and Q3. It is not true that $k[x, y, z]$ is a principal ideal domain. I think this resolves your contradiction. You will need more than one generator for the ideal $I$.
