Let $K$ be a field of characteristic 2.
For each $a\in K$, can we always find some $x$ such that $x^2=a$?
I came upon this question while reading "Arithmetic of Elliptic Curves".
The original motivation is to prove that $\Delta = 0 \Leftrightarrow E$ (elliptic curve) is singular.
The proof given for characteristic 2 is as follows:
Start with a general Weierstrass form:
$$E: y^2+a_1 xy+a_3y=x^3+a_2x^2+a_4x+a6$$
If $a_1=0$, then setting $x\mapsto x+a_2$ we get a new curve:
$$E:y^2+a_3y=x^3+a_4x+a_6$$ where $\Delta=a_3^4$ (Note that there was an errata).
Suppose $\Delta=0\implies a_3=0$. Then we want to show that it is singular.
So we look at
$$f(x,y)=y^2+a_3y-x^3-a_4x-a_6$$
And we need $\dfrac{\partial f}{\partial x}(x,y)=0\implies -3x^2-a_4=x^2-a_4=0$ as part of the requirement for singularity. This leads to requiring a solution $x^2=a_4$ should $a_4\neq 0$, which implies that we can always find a square root. Is this true? Or is my reasoning wrong somewhere? (If this is true then we can easily fulfill the other requirements)
If we define a map $\phi:K\to K$ such that $\phi(x)=x^2$, we see that $\phi$ is an injective ring endomorphism. So I suppose the question is sort of equivalent to whether $\phi$ must be surjective, which means it is an isomorphism. I don't see why it must be true though.
Best Answer
If the field is finite, the Frobenius homomorphism must be surjective, since it is injective. So in a finite field of characteristic $2$, every element is a square.
If the field is infinite, in general not every element is a square, consider the field of rational functions $\mathbb{F}_2(X)$. In that field, $X$ is not a square.