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I.2.7 depends on I.2.6 so has H = 2 and V = 4; (I.2.6>I.1.10>I.1.6). I.2.8 can be solved with I.2.7 which would give it H = 2 and V = 5. I.2.9 uses I.1.2 so has H = V =2. I.2.10 also depends on I.2.6 so H = 2, V = 4.



  • part a) is a direct consequence of I.2.6
  • for part b) recall the notation Y_i, \overline{Y}_i is the intersection of Y, \overline{Y} with the standard affine U_i \subset \mathbb{P}^n<math>.
**the solution of I.2.6 shows <math>\dim Y_i = \dim Y when  Y_i \ne \emptyset
    • Argue  \dim \overline{Y} = \dim \overline{Y}_i which in turn is equal to  \dim Y_i = \dim Y


  • this can be proved without I.2.7 arguing similarly to how the analogous affine statement is proved, but with I.2.7 have:
\dim Y = n-1 iff \dim Y_i = n -1 iff Y_i = V(f)

where f is irreducible. Then homogenizing gives the result.


  • \alpha is the dehomoginization operator and \beta is the homogenization operator. Let  F \in I(\overline{Y}), then argue <math> \alpha(F) \in I(Y)<math> where <math>Y is considered as just an affine variety. Conclude the result from  F = \beta(\alpha(F)) \in \beta(I(Y))
  • for the next part, use the result of I.1.2:  I(Y) = (z - x^3, y - x^2). Then check (groebner basis, maybe) that  I(\overline{Y}) = (zw^2 - x^3, yw - x^2, xy - zw)


  • part a is straightforward, and part b is a consequence of part a
  • part c is a consequence of I.2.6 because  \dim C(Y) = \dim A(C(Y)) = \dim S(Y) = \dim Y + 1 where A(C(Y)) is the affine coordinate ring.

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