# Projecting Projective Curves

I've been stuck for quite a while on what is probably a trivial problem. Let $X\subset\mathbb{P}^n$ be a smooth projective curve, and let $$\mathcal{I}=\{(p,q,r):p,q\in X,p\neq q,r\in\overline{pq}\}$$ (where $\overline{pq}$ is the line that joins $p$ and $q$) and let $$\mathcal{J}=\{(p,r):p\in X,r\mbox{ lies on the tangent line to }X\mbox{ at }p\}.$$ It is easy to see that $\mathcal{I}$ is a complex 3-manifold and $\mathcal{J}$ is a complex 2-manifold. Let $\alpha:\mathcal{I}\to\mathbb{P}^n$ so that $(p,q,r)\mapsto r$, and let $\beta:\mathcal{J}\to\mathbb{P}^n$ so that $(p,r)\mapsto r$.

Why is it that if $n\geq 4$ then there is a point in $\mathbb{P}^n$ that is not in the image of either function? I can see that the image of $\alpha$ has complex dimension at most 3 and the image of $\beta$ at most 2, but I can't see why their images can't cover all $\mathbb{P}^n$.

For reference, this was taken from Algebraic Curves and Surfaces by Rick Miranda, page 101.

Thanks.

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Robert, you need to change the letter "e" just after the second time you write $\overline{pq}$ for the letter "i". It seems that your Spanish tried to come out, that sometimes happens to me also ;) – Adrián Barquero Jun 21 '11 at 20:58
Thanks, I'm used to writing in Spanish! – rfauffar Jun 21 '11 at 21:23
If you can prove that the image of $\alpha$ has dimension at most $3$, then you know that image is not all of $P^n$ because $P^n$ has dimension at least $4$. – Mariano Suárez-Alvarez Jun 21 '11 at 21:25
Yeah, but I want to find an element that is neither in the image of $\alpha$ nor $\beta$... – rfauffar Jun 21 '11 at 22:30