Dimension of a ring

  • #1
disregardthat
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Let [tex]A = k[x,y,z][/tex] and [tex]Y = \{(t,t^2,t^3)|t \in k\}[/tex], which is irreducible. It corresponds to the prime ideal [tex]p=(y-x^2,z-x^3)[/tex].

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 [tex]x^{d-r-1}yz^r[/tex] for r < d, and [tex]x^{d-r}z^r[/tex] 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?
 
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Answers and Replies

  • #2
mathwonk
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i'm not very algebraic minded, but i forget how A(Y) is graded, when the ideal defining it is not homogeneous?
 
  • #3
disregardthat
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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 [tex]S_d \mod I(Y) * S_k \mod I(Y) \subseteq S_{d+k} \mod I(Y)[/tex], but I suppose that [tex]\oplus S_d \mod I(Y)[/tex] is not a grading of A(Y).

EDIT: Of course, I see it now. The [tex]S_d \mod I(Y)[/tex]'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.
 
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