I.2.7-I.2.10

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.

I.2.7

 * 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.
 * the solution of I.2.6 shows $$\dim Y_i = \dim Y$$ when $$ Y_i \ne \emptyset$$
 * Argue $$ \dim \overline{Y} = \dim \overline{Y}_i $$ which in turn is equal to $$ \dim Y_i = \dim Y$$

I.2.8
$$\dim Y = n-1$$ iff $$\dim Y_i = n -1$$ iff $$Y_i = V(f)$$ where $$f$$ is irreducible. Then homogenizing gives the result.
 * this can be proved without I.2.7 arguing similarly to how the analogous affine statement is proved, but with I.2.7 have:

I.2.9

 * $$\alpha$$ is the dehomoginization operator and $$\beta$$ is the homogenization operator. Let  F \in I(\overline{Y}), then argue  \alpha(F) \in I(Y) where $$Y$$ 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 $$ I(\overline{Y}) = (zw^2 - x^3, yw - x^2, xy - zw)$$

I.2.10

 * 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.