Given $\mathbb{P}^k$ and $\mathbb{P}^{\ell}$, the Segre variety is projective subvariety of $\mathbb{P}^{(k+1)(\ell+1)-1}$ defined as the common vanishing set of quadratic polynomials which can be defined as the $2 \times 2$ minors of a $(k+1)\times (\ell +1)$ matrix.
For geometric reasons, I would expect this projective variety to have dimension $k + \ell$, and thus to be defined by $[(k+1)(\ell+1)-1]-k + \ell=k\ell$ polynomials.
However, combinatorially I would expect the number of $2 \times 2$ minors of a $(k+1)\times (\ell +1)$ matrix to be $\binom{(k+1)}{2}\cdot \binom{(\ell +1)}{2} = \frac{1}{4}(k^2+k)(\ell^2+\ell)$, thus the Segre variety is defined by that many polynomials.
But $\frac{1}{4}(k^2+k)(\ell^2+\ell) \not= k\ell $. (Take, e.g., $k=4, \ell=8$.)
Question: Where are the errors in my reasoning?
If there are no errors in my reasoning, then does this imply that some of the defining polynomials for the Segre variety are redundant? Why or why not? If so, which ones? ("Soft" answers would be sufficient.)
(Would this be a case where, for degeneracy reasons, we can't use the Implicit function theorem to easily use the minimum number of defining polynomials to compute the dimension of the variety at each point?)
There are almost certainly major mistakes in my reasoning, so I apologize in advance for the embarrassingly stupid question.
What you are noticing is that it is not, in general, true that the codimension of a projective (or affine) variety is equal to the number of equations used to define it (or even the minimum number of generators of it's ideal). When the codimension is equal to the minimum number of generators of the ideal of the variety, it is called a complete intersection and is quite special. See examples and non-examples here on the wikipedia page.