Dimension of A(Y) and Height of p in Ring Let A = k[x,y,z]

[disregardthat]

Let $$A = k[x,y,z]$$ and $$Y = \{(t,t^2,t^3)|t \in k\}$$, which is irreducible. It corresponds to the prime ideal $$p=(y-x^2,z-x^3)$$.

A(Y) is generated by x,y,z of degree 1 as a k-algebra in its graded ring structure. Each group corresponding to the degree d is spanned by the linearly independent monomials $$x^{d-r-1}yz^r$$ for r < d, and $$x^{d-r}z^r$$ for r <= d.

For each group there are 2d+1 such monomials. This polynomial has degree 1, so doesn't this imply the dimension of A(Y) is 2 and hence the height of p is 1? But the height of p is obviously 2, so what is wrong here?

 

[mathwonk]

i'm not very algebraic minded, but i forget how A(Y) is graded, when the ideal defining it is not homogeneous?

 

[disregardthat]

i'm not very algebraic minded, but i forget how A(Y) is graded, when the ideal defining it is not homogeneous?

Oh, maybe you are right, I may have been too quick in verifying that A(Y) is graded. Basically I assumed S_d mod I(Y) would constitute the groups. Certainly $$S_d \mod I(Y) * S_k \mod I(Y) \subseteq S_{d+k} \mod I(Y)$$, but I suppose that $$\oplus S_d \mod I(Y)$$ is not a grading of A(Y).

EDIT: Of course, I see it now. The $$S_d \mod I(Y)$$'s are intersecting non-trivially, so the direct sum is not equal to A(Y). Reading on homogeneous ideals it makes sense now that only homogeneous ideals have a grading in this sense.