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Consider an algebraically closed field $k$, and define $\mathbb P^n_k:=\textrm{Proj}(k[T_0,\ldots,T_n])$. In some algebraic geometry books I see the notation ${(\mathbb P^n_k)}^\vee$ that is referred as "the dual projective space" without any other precisation.

Now I'm confused: $\mathbb P^n_k$ is a scheme and I don't understand what is the formal meaning of its "dual". Maybe hyperplane divisors are involved?

Many thanks in advance.

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    $\begingroup$ Just to add to Zhen Lin's precise answer: yes, hyperplane divisors are involved. The points of $(\mathbf P^n)^\vee$ are the hyperplanes in $\mathbf P^n$. $\endgroup$
    – user64687
    Commented Aug 15, 2014 at 13:39
  • $\begingroup$ This intuitively must be true thanks to the analogy with the classical projective space, but formally I don't understand the definition. $\endgroup$
    – Dubious
    Commented Aug 15, 2014 at 14:25
  • $\begingroup$ The definition of $\mathbb{P}^n$ is not 100% correct (use $n+1$ variables). $\endgroup$ Commented Aug 15, 2014 at 18:06
  • $\begingroup$ Yeah you're right $\endgroup$
    – Dubious
    Commented Aug 15, 2014 at 19:07

1 Answer 1

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If you have a $k$-vector space $V$ you can form the symmetric algebra on $V$, $$\operatorname{Sym} V = k \oplus V \oplus V^{\otimes 2} / S_2 \oplus V^{\otimes 3} / S_3 \oplus \cdots $$ and it is clearly a graded $k$-algebra. We define $\mathbb{P} (V) = \operatorname{Proj} \operatorname{Sym} V$. The dual of $\mathbb{P} (V)$ is just $\mathbb{P} (V^\vee)$.

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  • $\begingroup$ here $V=k[T_1,\ldots,T_n]$? $\endgroup$
    – Dubious
    Commented Aug 15, 2014 at 13:36
  • $\begingroup$ Typically $V$ is finite dimensional. If $\dim V = n + 1$ then $\mathbb{P} (V) \cong \mathbb{P}_k^n$. $\endgroup$
    – Zhen Lin
    Commented Aug 15, 2014 at 14:14
  • $\begingroup$ Ok, and which is the connection with the hyperplane divisors? $\endgroup$
    – Dubious
    Commented Aug 15, 2014 at 14:33
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    $\begingroup$ A point of the dual projective space is an equivalence class of linear forms on the original vector space, which defines a hyperplane in the usual way. $\endgroup$
    – Zhen Lin
    Commented Aug 15, 2014 at 15:38
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    $\begingroup$ @ZhenLin: The description of the symmetric algebra is not correct. $\endgroup$ Commented Aug 15, 2014 at 18:07

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