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Let $k$ be a field. Let $P=(0:0...:0:1)\in \mathbb{P}_k^n$. Show that the set of lines $L_P$ in $\mathbb{P}_k^n$ passing through $P$ could be idenitified with a projective space $\mathbb{P}_k^{n-1}$

To start this proof, a line in projective space, $\mathbb{P}^1\rightarrow \mathbb{P}^n$: $(u:v)\mapsto (a_0u+b_0v, ..., a_nu+b_nv)$ is a line in projective space. What is the equation of $L_P$ for points goes through $P$? Any hints on the problem?

Moreover, I m not understanding the notion of lines in projective space. What do they actually look like in the affine space? Thanks for the help

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  • $\begingroup$ You could choose a "copy" $H$ of $\mathbb P^{n-1}$ inside of $\mathbb P^n$ and identify a line through $P$ with its intersection with $H$. You have to choose $H$ correctly, of course. $\endgroup$ – Hoot Oct 27 '15 at 2:57
  • $\begingroup$ So you mean a point $H$ with one coordinate 0? $\endgroup$ – nerd Oct 27 '15 at 3:26
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A point in projective $n$-space $\mathbb P^n$ corresponds to a line through the origin in $\mathbb A^{n+1}$.

A line in projective $n$-space $\mathbb P^n$ corresponds to a plane through the origin in $\mathbb A^{n+1}$.

Thus the equation of a line in $\mathbb P^n$ is given by $n-1$ linearly independent linear forms.

Now for your problem: without loss of generality you can assume that $P=(1:0:0:\ldots:0)$ (there is always an automorphism sending $P$ to this point). Thus lines through $P$ corresponds to $n-1$ linear forms $\sum_{i=0}^n a_i x_i=0$ with $a_0=0$.

But this is just the condition to get a line through the origin in $\mathbb A^n$. That is, a point in $\mathbb P^{n-1}$.

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  • $\begingroup$ I wrote very similar to your argument. However, according to the instructor, the definition of line is $ (u:v)↦(a_0u+b_0v,...,a_nu+b_nv)$ $\endgroup$ – nerd Oct 27 '15 at 16:04
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    $\begingroup$ @nerd Well, it's just a matter of definition. What you describe is the image of an injective linear map $\mathbb A^2 \to \mathbb A^{n+1}$. But taking duals and kernels gives that this is the same as a linear map $0 \to \mathbb A^{n-1} \to \mathbb A^{n+1}$, as above. $\endgroup$ – Fredrik Meyer Oct 27 '15 at 18:54

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