# I.2.7-I.2.10

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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.

### HAPPY

#### 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 $U_i \subset \mathbb{P}^n[itex]. **the solution of I.2.6 shows [itex]\dim Y_i = \dim Y$ 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

• 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.

#### I.2.9

• $\alpha$ is the dehomoginization operator and $\beta$ is the homogenization operator. Let $F \in I(\overline{Y}), then argue [itex] \alpha(F) \in I(Y)[itex] where [itex]Y$ 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.