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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.

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