Field of order 8, $a^2+ab+b^2=0$ implies $a=0$ and $b=0$.

Question

I was able to come up with a proof for this problem however, it seems like my argument can work for any field of even order and not just odd powers of 2 so I'm convinced there is something wrong here. Can someone verify or see where the error in reasoning is?

Problem: Let $F$ be a field with $2^n$ elements, with $n$ odd. Show that for $a,b \in F$ that $a^2+ab+b^2=0$ implies that $a=0$ and $b=0$.

Proof: Suppose $a,b \in F$ and $a^2+ab+b^2=0$.

$\implies a^2+2ab+b^2 = ab$

$\implies \frac{2^n}{2}(a^2+2ab+b^2) = \frac{2^n}{2}ab$

$\implies \frac{2^n}{2}a^2+ 2^nab+\frac{2^n}{2}b^2 = \frac{2^n}{2}ab$

$\implies \frac{2^n}{2}a^2+\frac{2^n}{2}b^2 = \frac{2^n}{2}ab$ (since F is a group under addition, then every element to the $|F|$ multiple is the identity thus $2^n(ab) = 0$)

$\implies \frac{2^n}{2}(a^2+b^2) = \frac{2^n}{n}ab$

$\implies a^2+b^2 = ab$

$\implies a^2-ab+b^2 = 0 = a^2+ab+b^2$

$\implies -ab = ab \implies 2ab=0 \implies ab=0$.

Thus, $a=0$ or $b=0$. However, if just one of them is zero, then so is the other ($a=0 \implies a^2+ab+b^2 = 0 \implies b^2 = 0 \implies b=0$). Thus, $a=0$ and $b=0$.

QED

Anyways, if there is something wrong with this proof, could someone give me a subtle hint perhaps? I've been stuck on this seemingly simple problem for awhile now.

Answer 1

The mistake you make is:

$$\frac{2^n}{2}(a^2+b^2) = \frac{2^n}{2}ab \Rightarrow a^2+b^2 =ab $$

Note that your field has characteristic $2$, which means that $\frac{2^n}{2}=0$! You divide again by $0$ in the last line.

Hint $$a^3-b^3=(a-b)(a^2+ab+b^2)=0$$

Thus $a^3=b^3$, and you also know what $a^7, b^7$ are....

You asked for a subtle hint, I didn't include more details, let me know if it is helpful, or you want more details.

Answer 2

Hint $\ $ You cannot divide by $\,2\,$ since $\,2 = 0\,$ in $\,\Bbb F_{2^n}.\,$ Instead, notice that $\, 0 = (a-b)(a^2+ab+b^2) = a^3-b^3.\,$ If $\,a\,$ or $\,b\neq 0,\,$ wlog. $\,b\ne 0,\,$ then $\,c^3 = 1\,$ for $\,c = a/b.\,$ If $\,c \ne 1\,$ then $\,c\,$ has order $= 3,\,$ so Lagrange's Theorem $\Rightarrow 3\,$ divides $\,|\Bbb F_{2^n}^{*}| = 2^n-1,\,$ contra $\,{\rm mod}\ 3\!:\ 2^n\! =(-1)^{2k+1}\!\equiv -1.\,$ So $\,c = 1,\,$ so $\,a\!=\!b,\,$ so $\,a^2\!+\!ab\!+\!b^2\!=3b^2 = 0,\,$ so $\,b = 0.$