biquadratic extension over a field of characteristic 2

Question

There are the following exercises in Dummit & Foote's book (Abstract Algebra, 3rd.):

Let $F$ be a field of characteristic $\neq2$. Let $D_{1}$ and $D_{2}$ be elements of $F$, neither of which is a square in $F$. Prove that $F(\sqrt{D_{1}},\sqrt{D_{2}})$ is of degree $4$ over $F$ if $D_{1}D_{2}$ is not a square in $F$ and is of degree $2$ over $F$ otherwise. When $F(\sqrt{D_{1}},\sqrt{D_{2}})$ is of degree $4$ over $F$ the field is called a biquadratic extension of $F$.

Is there a counterexample in case of the field of characteristic $=2$?

It does not come up well. Give some advice! Thank you!

Answer 1

Consider $F_2[X]$ and take $D_1=X$ : $D_1$ is not a square.

Then $D_2=X^2+X$ : $D_2$ is not a square.

$D_1D_2=X^3+X^2$ is not a square either.

But $\sqrt{D_2}=\sqrt{X^2+X}=\sqrt{X^2+\sqrt{D_1}^2}=X+\sqrt{D_1}$. So the extension will only be of degree 2, not 4.

It comes from the fact that in a characteristic 2 field : $(a+b)^2=a^2+2ab+b^2=a^2+b^2$