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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.

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Note that $2=0$ in $F$, so $2ab=0$ does not imply $ab=0$. –  Dietrich Burde Feb 1 at 19:13

2 Answers 2

up vote 7 down vote accepted

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.

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Oh thank you, I didn't realize the characteristic was 2. I looked in my book and saw the theorem that all integral domains have characteristic 0 or prime (and all fields are integral domains). And there is an earlier exercise where we show the characteristic divides the order of the field. So the only prime that divides 8 is 2 so we have characteristic 2. Thanks, I'll see if I can get anywhere with that! –  benguin Feb 1 at 19:19
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So in the case where we have 8 elements, then $a^7=1=b^7$. Thus, we have $a^3 = b^3 \implies a^3a^7 = b^3a^7 \implies a^3a^7 = b^3b^7 \implies a^{10} = b^{10} \implies (a^3)^3a = (b^3)^3b \implies (a^3)^3a = (a^3)^3b \implies a=b$ –  benguin Feb 1 at 19:50
    
And then if $a=b$, then $0 = a^2+ab+b^2 = a^2+b^2+b^2 = a^2+2b^2 = a^2$ then $a=0$ or $a=0$ so $a=0$ (and I already showed if one was zero then so is the other). Is that sound? –  benguin Feb 1 at 19:58
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@user121903 Yes. And be carefull, when you divide by $(a^3)^3$ your argument is that if this is zero you are done, otherwise you can divide... –  N. S. Feb 1 at 20:30
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@user121903 A faster solution is $b^7=a^7=(a^3)^2a=(b^3)^2a$. –  N. S. Feb 1 at 20:31

Hint $\ $ You cannot divide by $\,2\,$ since $2 = 0\,$ in $\,\Bbb F_{2^n}.\,$ Notice $\, 0 = (a-b)(a^2+ab+b^2) = a^3-b^3.\,$ If one, say $\,b\ne 0,\,$ then $\,c^3 = 1,\,\ c = a/b.\,$ If $\,c \ne 1\,$ then $\,c\,$ has order $= 3,\,$ so, by Lagrange, $\,3\,$ divides $\,|\Bbb F_{2^n}^{*}| = 2^n-1,\,$ contra $\,{\rm mod}\ 3\!:\ 2^{2k+1}\!\equiv 2(4)^k\equiv 2.\,$ So $\,c = 1,\,$ so $\,a=b,\,$ so $\,3b^2 = 0,\,$ so $b = 0.$

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