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Let $Y$ be a variety of dimension $r$ and degree $d>1$ in $\mathbf{P}^n$. Let $P\in Y$ be a nonsingular point. Define $X$ to be the closure of the union of all lines $PQ$, where $Q\in Y$, $Q\ne P$. Show that $\operatorname{deg} X<d$.

The comments below resolve this problem.

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    $\begingroup$ Have you tried going the other way? Choose a subspace L that is in generic position to X, and then see how the subspace spanned by L and P meets Y? $\endgroup$
    – t3suji
    Commented Oct 15, 2015 at 1:14
  • $\begingroup$ I think the existence of such a hyperplane would do it. Issue is that though I am familiar with proving statements about generic lines relative to a plane curve, I haven't encountered a result of the form 'through every nonsingular point of a projective variety $Y$ there exists a generic hyperplane section with $d=\operatorname{deg} Y$ irreducible components.' But of course I would expect that the set of such hyperplanes is in fact Zariski-dense in the appropriate sense. $\endgroup$
    – Tomo
    Commented Oct 15, 2015 at 1:38
  • $\begingroup$ Actually, $P$ needn't be nonsingular in $X$, so one needs instead the statement 'through every point of a projective variety $X$ there exists a hyperplane section with $d=\operatorname{deg} X$ irreducible components.' $\endgroup$
    – Tomo
    Commented Oct 15, 2015 at 2:56
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    $\begingroup$ What I mean is, don't draw a plane through P. Draw a subspace L of dimension complementary to dimension of X which meets X at finitely many points (different from P). Now look at the subspace spanned by L and P; this will intersect X at a bunch of lines. How does it meet Y? $\endgroup$
    – t3suji
    Commented Oct 15, 2015 at 3:11
  • $\begingroup$ At no fewer than $|L\cap X|+1$ points. Thank you for this idea. How can I pick $L$ such that $|L\cap X|$ has no points of multiplicity $>1$? $\endgroup$
    – Tomo
    Commented Oct 15, 2015 at 3:30

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