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Suppose $X\subset \Bbb P^n$ is a projective variety over an algebraically closed field (in the sense of Hartshorne Chapter I) and $P$ is a fixed point in $\Bbb P^n\setminus X$. Why should it be the case that for generic line $L$ through $P$ meeting $X$, we have that the line $L$ is not in the tangent space $T_QX$ for any point $Q\in L\cap X$? It's "geometrically obvious" to me that it must be true, but I cannot quite put together a full proof. Here are some of the things I've tried:

  • Restricting to the copy of $\Bbb A^1$ given by $L\setminus\{P\}$, this condition should be expressible as "the principal ideal cutting out our closed subset is generated by a polynomial with no multiple roots", but I'm not sure how to make this something I can actually analyze.

  • Using an incidence correspondence. There should be some way to embed $X^{sm}$ in to something times the Grassmanian parameterizing the tangent planes (?), take the closure, and then things pop out. This idea is inspired by this post, and I am really flailing at this one.

I would like to avoid using big theorems not present in some form in Hartshorne for this, though if you have a particularly nice solution which involves such a thing, please do go ahead and post it anyways. (In particular, if you want to use the characterization of degree as "number of intersection points with a generic linear subspace of complimentary dimension", I would like to see a proof or reference to a proof that this is the same as Hartshorne's characterization of degree via Hilbert polynomials.)

Hank Scorpio
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  • The inclusion of a tangent space $T_Q X \subset \mathbb P(V)$ induces an inclusion of Grassmannians of lines, exhibiting $\mathbb G(1,T_Q X)$ as a closed subset of $\mathbb G(1,\mathbb P (V))$. Is this not enough? – Tabes Bridges Nov 20 '20 at 03:39
  • @TabesBridges How do you know that this generically misses lines through $P$? – Hank Scorpio Nov 20 '20 at 04:31
  • By reasons of dimension? Assuming that $X$ is not so singular at $Q$ that its tangent space is all of $\mathbb P^n$, let $\dim T_Q X = k < n$. Then the dimensions of these Grassmannians are $2((k+1)-2) = 2k - 2$ and $2((n+1)-2) = 2n - 2$, and $2k - 2 < 2n - 2$. – Tabes Bridges Nov 20 '20 at 04:49
  • @TabesBridges Sorry, I don't understand - you showed "lines in $T_QX$" is of smaller dimension than "lines through $P$", and I don't see how this implies that they're generically disjoint. I also don't see how your dimension counts are right - shouldn't lines through $P$ be of dimension $n-1$, corresponding to "points on a $\Bbb P^{n-1}$ disjoint from $P$"? – Hank Scorpio Nov 20 '20 at 05:00
  • Sorry, I misread your question. I will try to think about it further tomorrow. – Tabes Bridges Nov 20 '20 at 05:31
  • Not up to your standards (in terms of references), but here's at least a translation to something familiar: consider linear projection from $P$ to a complimentary $\mathbb P^{n-1}$; let $\pi:X\to\mathbb P^{n-1}$ be the restriction. The locus of points in $X$ with a tangent line passing through $P$ is the ramification locus of $\pi$, which is codimension $1$ by a theorem of Nagata-Zariski ("purity of the branch locus") for which I don't know a good reference off hand. Since the generic line through $P$ meeting $X$ does not hit the branch locus, the line must be transverse to each $T_Q X$. – Tabes Bridges Nov 20 '20 at 18:07
  • @TabesBridges I'm at least glad to know that my intuition is correct! Thanks for your help, though I do wonder if there is any simpler method of solution. – Hank Scorpio Nov 21 '20 at 01:22
  • I assume you are in characteristic zero, otherwise things may go wrong. Project from the point to a lower dimensional projective space. Then the branch locus is a proper subscheme of the image and these are the points corresponding to tangents. – Mohan Jul 03 '23 at 21:43
  • @Mohan I am not necessarily in characteristic zero. What may go wrong? Is there an explicit example of an integral projective variety where this property fails? – Hank Scorpio Jul 04 '23 at 15:16
  • Take the plane curve $x^{p-1}y-z^p=0$ in characteristic $p>0$ and $P=(0,0,1)$. – Mohan Jul 04 '23 at 17:15

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I didn't took the effort to formalize this properly, but can't you argue in the following way:

We can assume that $n\ge 2$ (otherwise the result is trivial). We may also assume that $X$ does not contain a plane containing $L$. We then take a plane containing the line through $P$, this plane intersects $X$ at a curve. The claim is that $L$ is not a tangent line of that curve. This reduces the case to a planar curve $X$. We can assume that we are on the affine plane so that our planar curves are defined as the zeros of bivariate polynomials. If we take one of the variables as a parameter and compute the discriminant, we will get a non-trivial polynomial in that parameter. Saying that most lines $L$ are tangent to the curve would imply that the polynomial is equivalent to $0$.

There might be a problem though if the base field has characteristic $2$.

quantum
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  • Thank you for your answer. This makes intuitive sense - I'd need to spend a bit of time on formalizing it before accepting. What would the problem in characteristic two be? – Hank Scorpio Jul 04 '23 at 15:17
  • No worries maybe a better answer will come along. I'm not sure about characteristic 2. But the discriminant involves the derivative and bad things happen when you are at characteristic $p>0$. I thought like the case of strange curves in Samuel's Theorem (you can find it in Hartshorne Theorem IV.3.9), characteristic 2 will give you an exception. – quantum Jul 06 '23 at 19:56