How to algebraically express the condition for a tensor to be of rank $1$? Let $W$ be a finite-dimensional complex vector space and let $V = W\otimes \cdots \otimes W$ ($m$ times). A non-zero element $v \in V$ is said to be of rank $1$ if $v$ can be written as
$$v = w_1 \otimes \cdots \otimes w_m,$$
for some $w_i \in W$, for $i = i, \ldots, m$. More generally, the rank of a non-zero element $v \in V$ is the smallest number of such tensor products for which $v$ is the sum.
My question can now be formulated. How can one formulate the condition of being of rank $1$ algebraically, meaning with polynomial equations? If $m=2$, this is standard linear algebra. The condition is that all $2$ by $2$ minors of $v$, thought of as a square matrix, should vanish. What is the generalization of this condition for $m \geq 2$?
Here is what I suspect. I suspect it is $v \wedge v = 0$ where $\wedge$ denotes the bilinear map from $V \times V$ to $V \otimes V$ which maps
$(w_1 \otimes \cdots \otimes w_m, w'_1 \otimes \cdots \otimes w'_m)$ to
$$(w_1 \wedge w'_1) \otimes \cdots \otimes (w_m \otimes w'_m) + \cdots +
(w_1 \otimes w'_1) \otimes \cdots \otimes (w_m \wedge w'_m).$$
This seems like a possible generalization, but I am not sure if it is correct! Could someone confirm or correct please? I checked that it is correct for $m=2$ for instance.
Edit: I realize that I am essentially asking about the equations defining the product of $m$ copies of $P(W)$ embedded inside $P(V)$ via the Segre embedding. Ok, now I know where to look this up in the literature!
Edit 2: What I wrote above is incorrect. Please go to my answer below for a corrected formula. Ultimately, the equations are that the $2$-by-$2$ minors of $v$ must vanish (this is explained in my answer below).
 A: Define, for $i = 1, \ldots, m$, the bilinear map $\wedge_i$ from $V \times V$ to $(W \otimes W) \otimes \cdots \otimes (\Lambda^2 W) \otimes \cdots (W \otimes W)$ where $\Lambda^2 W$ is right after the first $i-1$ th pair of $W \otimes W$, which maps
$(w_1 \otimes \cdots \otimes w_m, w'_1 \otimes \cdots \otimes w'_m)$ to
$$(w_1 \otimes w'_1) \otimes \cdots \otimes (w_i \wedge w'_i) \otimes \cdots \otimes (w_m \otimes w'_m).$$
Then the equations defining the Segre embedding of $P(W) \times \cdots P(W)$ ($m$ copies of $P(W)$) inside $P(W \otimes \cdots \otimes W)$ ($m$ times) can be written using our notation as:
$v \wedge_i v = 0,$ for $i = 1, \ldots, m$.
What I had written in my post was actually incorrect (though somewhat close to a correct answer).
These conditions are redundant though. A less redundant way of writing these conditions is that the $2$-by-$2$ minors of $v$ should vanish. To simplify the discussion and notation, suppose that $W = (\mathbb{C}^n)^*$. Let $S_1$ and $S_2$ be two subsets of $[n] = \{1, \ldots, n\}$ such that $|S_1| = 2$ and $|S_2| = 2$. We do not require $S_1$ and $S_2$ to be distinct (they may or may not be distinct). Let us say that $S_1 = \{a_1,b_1\}$ and $S_2 = \{a_2,b_2\}$ where $1 \leq a_i < b_i \leq n$, for $i = 1, 2$. Also, let $i_1, i_2$ be such that $1 \leq i_1 < i_2 \leq m$.
We can think of $v \in V$ as a tensor having $m$ indices. Then, for each fixed values of the indices at locations other than $i_1$ and $i_2$ (and we suppress these indices from our notation), we get the following condition:
$$v_{a_1a_2} v_{b_1b_2} - v_{a_1b_2} v_{b_1a_2} = 0.$$
The left-hand side is what one may call a $2$-by-$2$ minor.
