About a field of order $2^{n}$ with $n$ an odd integer and an additional property

Question

I'm new in the world of fields (so I don't have any strong theorem at my disposal) and I've got stuck in this problem:

Given a field of order $2^{n}$ with $n$ an odd integer and $a,b$ elements in the field such that $a^2+ab+b^2=0$, prove that $a=b=0$.

I've found this answer but in the book i'm studying from (Gallian's book) we've not touched that stuff, the only thing I can use is that the characteristic is two and the fact that this fields forms a group under multiplication. I'm looking for something like this in the general case.

Please, if someone can help me that would be great.

Answer 1

The proof that introduces degrees, irreducible polynomials, etc. can be "written down" to an explicit calculation.

If either $a,b$ is zero, then $a^2 + ab + b^2 = 0$ implies both are. So assume neither $a,b$ is zero, and in particular $x = a/b$ is nonzero.

As before $x^2 + x + 1 = 0$, or as we are in characteristic 2, $x^2 = x+1$ and $x(x+1) = 1$. That is, in the multiplicative group of the field, $x$ has order 3:

$$ x^3 = x(x+1) = 1 $$

Therefore 3 must divide the order of the multiplicative group $2^n-1$. But this is only possible when $n$ is even by a simple congruence calculation, $2^2 \equiv 1 \pmod{3}$.

Answer 2

If the field $\Bbb F$ has $2^n$ elements, it is an extension of degree $n$ of the field $\Bbb F_2$ with $2$ elements.

If $a$ and $b$ are in $\Bbb F$ and non zero, their ratio $x=a/b\in\Bbb F$ is a root of the polynomial $X^2+X+1\in\Bbb F_2[X]$ which is irreducible over $\Bbb F_2$. Thus $x$ generates the (unique) quadratic extension $\Bbb F_4$ of $\Bbb F_2$. But since $n$ is odd $\Bbb F$ cannot contain $\Bbb F_4$.

Thus the only possibility left is that $a=b=0$.

Answer 3

Or you could do it entirely with group theory: Suppose that $a \neq b$ . Notice that $a^{3}-b^{3} = (a-b)(a^{2}+ab+b^{2})=0,$ so $a^{3} = b^{3}.$ We claim that $m = 2^{n }-1$ must be divisible by $3.$ This is the order of the order of the multiplicative group of the field containing $a$ and $b,$ so we have $a^{m} = b^{m} = 1.$ Now if $3$ does not divide $m,$ then we have $3x + my = 1$ for integers $x$ and $y.$ Then $a = a^{3x+my} = a^{3x} = b^{3x} = b^{3x+my} = b^{1} = b,$ a contradiction. Hence $3$ divides $2^{n}-1,$ so as noted by others $n$ must be even.