How can I find the rank of this algebraic vector bundle? Let $P(n,d)$ be the vector space of degree $d$ polynomials in $\mathbb{C}[x_0,\ldots,x_n]$. If I consider the variety
$$
Z = \{(x,f) \in \mathbb{P}^n\times P(n,d) : f(x) = 0 \}
$$
then the projection
$$
\pi_1: Z \to \mathbb{P}^n
$$
defines a vector bundle. This is because given two degree $d$ polynomials $f,g$ vanishing at $x$ and $\alpha,\beta \in \mathbb{C}$ we necessarily have that
$$
\alpha f(x) + \beta g(x) = 0
$$
My question is: how can I find the rank of this vector bundle?
 A: You are asking for the dimension of a fiber of the projection map $\pi_1$, that is, for $x \in \mathbb{P}^n$, what is the dimension of the set
$$
  \{ f \in P(n,d) \mid f(x) = 0 \}.
$$
This turns out to be a hyperplane in $P(n,d)$, so its dimension is $\dim P(n,d) - 1 = \binom{n+d}{n} - 1$, and that is the rank of the vector bundle.
One simple way to see that the indicated set is indeed a hyperplane in $P(n,d)$ is to take the coordinates on $P(n,d)$ to be the coefficients of monomials appearing in polynomials $f \in P(n,d)$. That is, write each $f = a_1 x_0^d + a_2 x_0^{d-1} x_1 + \dotsb$, and take the $a_i$ to be the coordinates on $P(n,d)$. Now for a fixed $x \in \mathbb{P}^n$, the condition $f(x) = 0$ is a single linear condition on the $a_i$. It is a nonzero condition because $x \in \mathbb{P}^n$ has at least one non-vanishing coordinate, so at least one of the $x_i^d$ is nonzero.
Finally a note of caution. Just because the fibers are vector spaces doesn't mean that $Z$ is a vector bundle. You have to have local trivializations. 
