There is a theorem : In a field such that its Characteristic $n$ is greater than zero , the equality $(a+b)^n=a^n+b^n$ is true.
Does anyone know how to prove it ?
Thanks.
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1Try the binomial theorem – J. W. Tanner Mar 07 '19 at 18:23
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Yes, the binomial theorem and the fact that $n$ must be prime to be a non-zero field characteristic. – Thomas Andrews Mar 07 '19 at 18:24
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Related (duplicate?) When is $(a+b)^n \equiv a^n+b^n$? – Martin R Mar 07 '19 at 18:24
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https://en.wikipedia.org/wiki/Freshman's_dream – vadim123 Mar 07 '19 at 18:26
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The binomial theorem only applies when $a$ and $b$ commute. @orel You originally indicated that your question was about linear algebra, are $a$ and $b$ supposed to be matrices? – Ben Grossmann Mar 07 '19 at 18:26
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You might check out https://math.stackexchange.com/questions/3135077/x-mapsto-xp-with-operatornamecharf-p0-is-a-ring-homomorphism – Robert Lewis Mar 07 '19 at 18:27
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In a field the characteristic $n=p$ is prime and the result follows from the fact that $$ p\mid{p\choose k}=\frac{p!}{k!(p-k)!} $$ if $0<k<p$.
On the other hand in a ring that it is not generally true. For instance in a ring of characteristic $4$ we have $$ (a+b)^4=a^4+2a^2b^2+b^4 $$ identically.
Andrea Mori
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@J.W.Tanner, thanks for pointing out the typo. I edited the answer. – Andrea Mori Mar 09 '19 at 00:34