# Square root in Characteristic 2 Field

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.

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If the field is finite, yes. Then the Frobenius homomorphism is an isomorphism. If the field isn't finite, in general no, consider $\mathbb{F}_2(X)$, $X$ is not a square. – Daniel Fischer Aug 28 '13 at 19:24
This is not necessarily injective for all $K$ (check for $\mathbb R$). My point is, the characteristic is key here. – M Turgeon Aug 28 '13 at 19:25
If the square root does not exist in the field $K$, then it exists in an extension field of $K$. Here the "entire" curve consists of the point with coordinates in all the extension fields of $K$, and if there is a singular point among those points, then the curve is labelled singular. The solutions of the equation $(x,y)\in K^2$ are called $K$-rational points. They are, indeed, often of special interest, but the curve has other points also. For the purposes of deciding whether a curve is singular, we can always assume that $K$ is algebraically closed. – Jyrki Lahtonen Aug 28 '13 at 19:47
@JyrkiLahtonen :Thanks so much for the comment! I was so confused by the example since it seems to suggest that it always have some singular point at $K$. – Yong Hao Ng Aug 28 '13 at 20:09
@YongHaoNg The point I was trying to make was that there is a simple way of computing the square root of $\alpha \in \mathbb F_{2^n}$; just start with $\alpha$ and square $n-1$ times to get the square root. Furthermore, if $(a_1,a_2,\ldots, a_n)$ is a representation of $\alpha$ with respect to some basis of $\mathbb F_{2^n}$ over $\mathbb F_{2}$, then the representation of $\sqrt{\alpha}$ can be obtained by applying a linear transformation to $(a_1,a_2,\ldots, a_n)$. – Dilip Sarwate Aug 28 '13 at 20:55

## 1 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.

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Thanks for the example! That answers the main question. I guess I must be missing something for the part on Elliptic Discriminant. – Yong Hao Ng Aug 28 '13 at 19:38