# I.2.12

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This has H = 1.

### HAPPY

• for part a) take $f \in a$ and break it into homogeneous parts $f = \sum_i f_i$ and argue (using $f(M_0, ..., M_N) = 0$) that in fact each $f_i \in a$; mod out by $a$ and show you get an inclusion into a domain.
• for part b), the easy direction is $im \rho_d \subset V(a)$.

For the other direction, let $Q = (Q_0,..., Q_N) \in V(a)$, the task is to fine $(a_0, ..., a_n)$ such that $Q = (M_0(a_i),...,M_N(a_i))$; use induction on $n$; the base case $n =1$ require that k is algebraically closed, so you can take dth roots.

• Let $m_j$ ($j = 0,1, ..., n$) be the index such that $M_{m_j} = x_j^d$; argue for some j, $Q_{m_j} \ne 0$.
• if some $Q_{m_j}$ is zero then can set $a_j = 0$ i.e. can reduce to n-1 variables and apply induction hypothesis.
• thus reduce to case where all $Q_{m_j}$ are all nonzero; for fixed j get a set of equations (indexed by i)
$M_i (x_0/x_j, .... , x_n/x_j ) = Q_i$
• there are consistent! since ($Q \in Z(a)$).
• argue all $Q_i$ are nonzero (not just the $Q_{m_j}$)
• then argue that by continuing to divide by certain equations can reduce the number of variables and solve (similar to how base case is proved).
• for part c), show for a general closed set $Z(b) \subset \mathbb{P}^n$ that $\rho_d(Z(b)) = Z(\theta^{-1}(b)) \cap Z(a)$ hence $\rho_d$ is a closed map and consequently a homeomorhpism (i.e. had bijective and continuous, now showed the inverse was continuous).
• part d) is straightforward.