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Let $F$ be a field of characteristic 2 with more than 2 elements. Show that there are elements $a$ and $b$ in $F$ such that $(a+b)^3 \not= a^3 + b^3$.

$F$ couldn't possibly have less than 2 elements, and if it had exactly 2 — that is, $F = \mathbb Z_2$ —, $(a+b)^3$ would actually always be $a^3+b^3$. With $\#F>2$, how do I find $a$ and $b$ to violate that?

All I could do so far is simplify the inequation, using the characteristic and the commutative prroperty:

$$\begin{align} (a+b)^3 &\not= a^3 + b^3\\ a^3 + 3a^2b + 3ab^2 + b^3 &\not= a^3 + b^3\\ a^3 + a^2b + ab^2 + b^3 &\not= a^3 + b^3\\ a^2b + ab^2 &\not= 0\\ ab(a + b) &\not= 0\quad. \end{align}$$

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How do you ensure a product is not zero? – Jack Schmidt Jul 25 '12 at 4:03
@JackSchmidt I had an insight as soon as I read your comment. A field is a domain, and this one has at least two non-zero distinct elements. Pretty easy, I should've thought of that by myself. Thanks. – Luke Jul 25 '12 at 21:01
up vote 5 down vote accepted

In a field, product is $0$ iff atleast one of the element in product is $0$. Here, if $a\neq0,b\neq 0$ and $a\neq -b$ then , $ab(a+b)\neq 0$, so choose non-zero elements of field $\Bbb F$ such that one is not the additive inverse of other one.

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It might help to know the addition and multiplication tables of $F_4$, the smallest extension of the simplest Galois field of characteristic 2.

Suppose $F_4 = \{0, 1, \alpha, \alpha+1\}$, then addition works as you might expect, and multiplication is such that $\alpha^2 = \alpha + 1$. This lets you work out the full addition and multiplication tables, and now it's child's play to pick $a, b$ such that $ab(a+b) \neq 0$.

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I read the question to ask for a proof for all fields with more than 2 elements, not a particular one. True, an example can be helpful in the search for a proof. – Ross Millikan Jul 25 '12 at 4:11
whoops, you're right. still, this example is probably enlightening in some way. – Kris Jul 25 '12 at 4:20

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