# Scheme theoretic dual of $\mathbb P^n_k$

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 $\mathbf P^n$.
– user64687
Commented Aug 15, 2014 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. Commented Aug 15, 2014 at 14:25
• The definition of $\mathbb{P}^n$ is not 100% correct (use $n+1$ variables). Commented Aug 15, 2014 at 18:06
• Yeah you're right Commented Aug 15, 2014 at 19:07

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 $\mathbb{P} (V^\vee)$.
• here $V=k[T_1,\ldots,T_n]$? Commented Aug 15, 2014 at 13:36
• Typically $V$ is finite dimensional. If $\dim V = n + 1$ then $\mathbb{P} (V) \cong \mathbb{P}_k^n$. Commented Aug 15, 2014 at 14:14