A counterexample when $\text{char}(\mathbb{F})=2$

Question

When trying to prove this:

$$\det\begin{bmatrix} A & B \\ B & A \end{bmatrix}=\det(A+B)\det(A-B),$$

for $A,B\in M_n(\mathbb{F})$ when $\mathbb{F}$ is a field, I used the fact that $\text{char}(\mathbb{F})\neq 2$ because at the end of the proof, after a few calculations both side of que equation above have an extra factor $2^n$ (if you want, I can post it). This fact made me look for a counterexample when the characteristic of the field is $=2$. There is such counterexample? If not, how to prove this statement for any field? Thanks in advance.

Answer 1

Adding the first $n$ rows to the last $n$ rows and then adding the negative of the last $n$ columns to the first $n$ columns we obtain via the formula for a triangular matrix per blocks:

$$\det\begin{bmatrix} A & B \\ B & A \end{bmatrix}=\det\begin{bmatrix} A & B \\ A+B & A+B \end{bmatrix}=\det\begin{bmatrix} A-B & B \\ 0 & A+B \end{bmatrix}=\det(A+B)\det(A-B).$$ So the hypothesis of $F$ be a field can be reduced to merely a commutative ring.

Answer 2

Both sides of the equation above are polynomials in $\mathbb{Z}[A_{ij}, B_{ij}]$, where $A_{ij}$ and $B_{ij}$ are the entries of $A$ and $B$, respectively. If they agree over characteristic $p\not = 2$, then they also agree in characteristic $2$.