Hyperplanes and linear forms

Question

For a number field $k$, let $H$ be a hyperplane on $\mathbb{P}^n(k)$ be given by $$H = \{\mathbf{x} = [x_0:...:x_n]|L(\mathbf{x})=0\},$$ where $L = a_0X_0+...+a_nX_n$ is the linear form (of $n+1$ variables) defining $H$.

Question 1. What does it mean when $t$ linear forms $L_1,L_2,...,L_t$ are linearly independent, where $2 \leq t \leq n+1$?

My guess is that we write every linear form $L_i$ as a row vector with $n+1$ entries $a_0,...,a_n$ and then the $L_i$'s are linearly independent when these vectors are. Just wanted to check this, I don't see how it can be any other way.

Now each hyperplane $H_i$ corresponding to the linear form $L_i$ can be viewed as an effective Cartier divisor of $\mathbb{P}^n(k)$.

Question 2. Is it true that $t$ hyperplanes intersect properly if and only if the corresponding linear forms are linearly independent?

The definition of proper intersection is as follows:

We say that the effective Cartier divisors $D_1,...,D_q$ intersect properly if for all subsets $I \subset \{1,...,q\}$ and for all $x \in \cap_{i \in I}\,\mathrm{Supp}(D_i)$, the sequence $(\phi_i)_{i \in I}$ is regular in $\mathcal{O}_{X,x}$, where the $\phi_i$'s are the defining equations of $D_i$ around $x$.

I'm not sure how to relate this concept to linear independence, some hints would be appreciated.

Answer 1

Points in $\Bbb{P}^n$ are described in homogeneous coordinates $[x_0:\cdots:x_n]$. A linear form in $x_0,\ldots, x_n$ is just an expression $$ f(x)=\sum_{i=0}^n a_ix_i =a^tx $$ for the $a_i \in k$, $a=(a_0,\ldots, a_n)$, and $x=(x_0,\ldots, x_n)$ regarded as column vectors. You might already know this, but such an $f(x)$ does not define a function on $\Bbb{P}^n$ when evaluated in this fashion, as the value depends on the choice of representative of $[x_0:\cdots:x_n]$. However, it is well-defined to say that $f$ vanishes at $[x_0:\cdots:x_n]$.

However, when we discuss linear independence of linear forms, we are regarding them as elements of $\mathrm{Hom}(k^{n+1},k) = (k^{n+1})^\vee$. So, linear independence here is as elements of the dual space.

As far as intersecting "properly" is concerned: yes, this is equivalent to linear independence of the forms in question. There are a few ways to see this. Heuristically, $f_1,\ldots, f_r$ defining a regular sequence in $\mathcal{O}_{\Bbb{P}^n,x}$ for $x\in Z(f_1,\ldots, f_r)$ and $f_i$ homogeneous polynomials means that each $f_i$ cuts down the dimension by $1$. More rigorously, you can consult the definition here where it is stated that this is equivalent to the commutative algebra notion of $Z(f_1,\ldots, f_r)$ being a (local) complete intersection. The complete intersection property can be verified with a Jacobian criterion; namely that the matrix of partials $(\frac{\partial f_i}{\partial x_j})$ should have full rank. If we return to the linear forms case, consider forms $$ f_j(x) = \sum_{i=0}^n a_{ij}x_i. $$ The matrix of partials becomes $(a_{ij})$ and we see that the $f_i$ define a regular sequence at every point in their intersection if and only if the rank of the matrix $(a_{ij})$ is maximal. This is equivalent to the functionals themselves being linearly independent.