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In Hartshorne's proof of Bertini's theorem, given a linear system $|H|$, he defines the locus of "bad" hyperplanes $B_x$ for each point $x\in X\subset \Bbb P^n$ a projective variety, shows that this is a proper linear subset of $\{x\}\times |H|$, defines $B$ to be the union of all pairs $(x,H)$ so that $H\in B_x$, and then claims that "clearly $B$ is the set of closed points of a closed subset of $X\times|H|$".

I don't understand this. Clearly the statement "if a subset has proper closed intersection with all fibers of a map over closed points, then it's a proper closed subset" is false - consider the inclusion of a copy of $(\Bbb P^1\setminus\{0\})\times\{p\} \to \Bbb P^1\times\Bbb P^1$ followed by a projection. How can I rigorously see Hartshorne's claim?

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    $\begingroup$ Yes, topologically speaking it’s clearly wrong (you could simply imagine a reunion of random closed subsets of each fiber). The only way I can think of to salvage the claim is if the condition $H \in B_x$ is closed wrt $(x,H)$. $\endgroup$
    – Mindlack
    Nov 24 '20 at 22:38
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    $\begingroup$ @Mindlack Thanks for the comment. What do you mean "the condition $H\in B_x$ is closed wrt $(x,H)$"? How could I check that? $\endgroup$ Nov 24 '20 at 22:39
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    $\begingroup$ @Mindlack Let $f$ be the homogeneous linear form cutting out $H$. Fix $f_0$ a homogeneous linear form not vanishing at $x$. $H\in B_x$ if $f$ is in the kernel of the map from all homogeneous linear forms to $O_{X,x}/m_x^2$ given by sending $f\mapsto f/f_0$. Would still love to know what you mean by "closed wrt $(x,H)$"!. $\endgroup$ Nov 24 '20 at 22:47
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    $\begingroup$ (Before I read your latest comment) Is the condition “$H \in B_x$” equivalent to the vanishing of some rational functions of $(x,H)$? $\endgroup$
    – Mindlack
    Nov 24 '20 at 22:47
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    $\begingroup$ Right. So find a finite cover of $X$ by finitely many open subsets (cut out as the nonzero loci of homogenous linear forms) – topology should ensure the global result from the result on each of these open subsets. So let $f_0$ be a linear form not vanishing on $X$. Then, if $H$ is represented by the linear form $f$, $H \in B_x$ iff $f/f_0 \in m_x^2$ iff $f/f_0 \in m_x$ and $(f/f_0)’ \in m_x$ (I think – as $X$ is smooth). And these conditions are polynomial in affine coordinates of $X$. (Disclaimer: you’d better double check that yourself, because I’m not sure I didn’t make any mistakes here). $\endgroup$
    – Mindlack
    Nov 24 '20 at 23:06
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I pieced this together after the discussion with Mindlack in the comments. Define $B$ to be the subset of $X\times\Bbb P^n$ cut out by the minors of size $n-\dim X+1$ of the matrix $$\begin{pmatrix} \frac{\partial f_1}{\partial x_0} & \cdots & \frac{\partial f_1}{x_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial f_m}{\partial x_0} & \cdots & \frac{\partial f_m}{x_n} \\ a_0 & \cdots & a_n \end{pmatrix}$$

where $f_i$ are a generating set for $X$ and $a_j$ are coefficients for the hyperplane $V(\sum a_jx_j)$ where the $x_j$ are coordinates on $\Bbb P^n$. It's not so hard to see that the vanishing of all of these minors at a point $x\in X$ means that the dehomogenization of $\sum a_jx_j$ is in the square of the maximal ideal there via the Jacobian criterion. This exactly gives $B_x$ as the fiber of $B$ over $x$.

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