Scheme theoretic dual of $\mathbb P^n_k$

Question

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.

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 $\mathbf P^n$. — , Aug 15 '14 at 13:39
This intuitively must be true thanks to the analogy with the classical projective space, but formally I don't understand the definition. — Dubious, Aug 15 '14 at 14:25
The definition of $\mathbb{P}^n$ is not 100% correct (use $n+1$ variables). — Martin Brandenburg, Aug 15 '14 at 18:06

Zhen Lin · Answer 1 · 2014-08-15T18:32:26.410

4

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 $\mathbb{P} (V^\vee)$.

edited Aug 15 '14 at 18:32

answered Aug 15 '14 at 13:31

Zhen Lin

90,111

here $V=k[T_1,\ldots,T_n]$? – Dubious Aug 15 '14 at 13:36
Typically $V$ is finite dimensional. If $\dim V = n + 1$ then $\mathbb{P} (V) \cong \mathbb{P}_k^n$. – Zhen Lin Aug 15 '14 at 14:14
Ok, and which is the connection with the hyperplane divisors? – Dubious Aug 15 '14 at 14:33
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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. – Zhen Lin Aug 15 '14 at 15:38
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@ZhenLin: The description of the symmetric algebra is not correct. – Martin Brandenburg Aug 15 '14 at 18:07

Scheme theoretic dual of $\mathbb P^n_k$

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