As I'm reading through Vakil's notes on algebraic geometry, I'm stumbling over some of the details in his proof of Bertini's theorem. The proof in question is discussed already on this site (1, 2 and doubtless others).
Summary of Vakil's proof for context.
(Please see page 346 of the above-linked notes for the source material.)
Fix an algebraically closed field $k$ for the remainder of the discussion. Further fix a smooth closed irreducible subvariety $X \subseteq \mathbb{P}^n$. We wish to show that there is a dense open subset $U \subseteq (\mathbb{P}^n)^\vee$ such that for each closed point $H \in U$, $H \cap X$ is smooth and $X \not\subseteq H$.
Essentially, Vakil proceeds by defining a closed subvariety of $Z \subseteq \mathbb{P}^n \times (\mathbb{P}^n)^\vee$ spanned by points $(p \in X, H)$ such that $p \in H$ and either $p$ is singular in $X \cap H$ or $X \subseteq H$. He accomplishes this by cutting out the equations for $X$, adding one equation enforcing $p \in H$ and then requiring the corank of the resultant Jacobian to be $\mathsf{dim}(X)$. The details are in the following quote:
12.4.3. We first define $Z$ more precisely, in terms of equations on $\Bbb P^n\times\Bbb P^{n\vee}$, where the coordinates on $\Bbb P^n$ are $x_0,\dots,x_n$, and the dual coordinates on $\Bbb P^{n\vee}$ are $a_0,\dots,a_n$. Suppose $X$ is cut out by $f_1,\dots,f_r$. Then we take these equations as the first of the defining equations of $Z$. (So far we have defined the subscheme $X\times\Bbb P^{n\vee}$.) We also add the equation $a_0x_0+\cdots+a_nx_n=0$. (So far we have described the subscheme of $\Bbb P^n\times\Bbb P^{n\vee}$ corresponding to the points $(p,H)$ where $p\in X$ and $p\in H$.) Note that the Jacobian matrix ((12.1.6.1), except with variables starting with $x_0$ rather than $x_1$) $$\begin{pmatrix} \frac{\partial f_1}{\partial x_0}(p) & \cdots & \frac{\partial f_r}{\partial x_0}(p) \\ \vdots & \ddots & \vdots \\ \frac{\partial f_1}{\partial x_n}(p) & \cdots & \frac{\partial f_r}{\partial x_n}(p) \end{pmatrix}$$ has corank equal to $\dim X$ at all closed points of $X$ — this is precisely the Jacobian criterion for regularity (Exercise 12.2.D). We then require that the Jacobian matrix with a new colum $$\begin{pmatrix} a_0 \\ \vdots \\ a_n\end{pmatrix}$$ appended has corank $\geq \dim X$ (hence $=\dim X$). This is cut out by equations (the determinants of certain minors). By the Jacobian description of the Zariski tangent space, this condition encodes the requirement that the Zariski tangent space of $H\cap X$ at $p$ has dimension precisely $\dim X$, which is $\dim H\cap X+1$ (i.e., $H\cap X$ is singular at $p$) if $H$ does not contain $X$, or if $H$ contains $X$. This is precisely the notion we wished to capture. (Remark 12.4.4 works through an example, which may help clarify how this works.)
There appears to be a small typo here: since we are considering the affine cone of $X$ in this, we should be careful to potentially replace $\mathsf{dim}(X)$ with $\mathsf{dim}(X + 1)$.
Question
I am stuck attempting to understand the next part of the proof. Having defined $Z$, Vakil claims the fiber of $Z$ over $\{p\} \times (\mathbb{P}^k)^\vee$ is cut out of $(\mathbb{P}^k)^\vee$ by $\mathsf{dim}(X) + 1$ linear conditions, but I'm afraid I don't see where these come from at all. The second linked question I've linked to above chases down the obvious answer: it's the requirement that certain minors vanish, but this doesn't seem to lead to the correct number of equations, but no clear alternatives present themselves. The set is clearly closed, but the requirement that it be determined by $\mathsf{dim}(X) + 1$ linear equations specifically is important and elusive.
For completeness, I've included the direct quote from Vakil below:
We next show that $\dim Z\leq n-1$. For each closed point $p\in X$, let $W_p$ be the locus of hyperplanes containing $p$, such that $H\cap X$ is singular at $p$, or else contains all of $X$; what is the dimension of $W_p$? Suppose $\dim X = d$. Then the restrictions on the hyperplanes in definition of $W_p$ correspond to $d+1$ linear conditions. (Do you see why?) This means that $W_p$ is a codimension $d+1$, or dimension $n-d-1$, …
Best Answer
You're correct that each occurrence of $\dim X$ in the summary should be $\dim X+1$.
Let's put algebraic geometry aside for a moment and consider the following linear algebra problem: given an $m\times n$ matrix $A$ of rank $r$ and a vector $v$, what are the condition we need to put on $v$ so that the augmented matrix $(A\mid v)$ has the same rank as $A$? Well, the rank of a matrix is the dimension of the image, the image of a matrix is the span of the columns, and therefore we need $v$ to belong to the span of the columns of $A$. Choosing a basis $e_1,\cdots,e_r$ for the span of the columns of $A$ and then extending to a basis $e_1,\cdots,e_m$ for the whole space, we see that we get exactly $m-r$ independent linear conditions: the coefficients on $e_{r+1},e_{r+2},\cdots,e_m$ must all be zero, and the coefficients on the other basis vectors can be whatever we want.
Okay, now let's translate this linear algebra result to the situation at hand. We're looking at the Jacobian matrix which is $(n+1)\times r$ and of corank $\dim X+1$ or rank $(n+1)-(\dim X+1)=n-\dim X$, and we're curious about when appending a column with the $a_i$ will keep this rank. Just apply the above result! From the work in the first paragraph we get $(n+1)-(n-\dim X)=\dim X+1$ independent linear conditions.