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If $p$ is prime, then $(x+y)^p=x^p+y^p$ holds in any field of characteristic $p$. However all the proofs I have seen use induction and some relatively nasty algebra despite how fundamental this fact seems.

What is the nicest, "highest level proof" you know?

symplectomorphic
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Asvin
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  • This section may at least be of some interest: http://en.wikipedia.org/wiki/Freshman%27s_dream#Prime_characteristic – nathan.j.mcdougall Apr 17 '15 at 04:17
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    It holds in any commutative ring of characteristic $p$. Many would consider the theory of finite fields - determining their existence and uniqueness and thus their classification, and knowing how to construct them - to be nasty algebra despite how fundamental the topic seems. Moreover, what kind of high-level math do you expect to have something to say about finite fields while simultaneously not assuming any knowledge of them? Certainly I would count the binomial theorem in that category! Do you expect any other "high-level math" to be supported by less nasty algebra than that? – anon Apr 17 '15 at 13:08
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    for $p \neq 0$, right? In $\mathbb R$, $(x + y)^0 = 1$, but $x^0 + y^0 = 1 + 1 = 2$ (for $x, y, x + y \neq 0$). – wchargin Apr 17 '15 at 20:25
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    @wchargin $0$ is not prime... – Alex Kruckman Jan 08 '21 at 18:28

2 Answers2

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The binomial coefficient $\binom p i$ is divisible by $p$ for $1 \leq i \leq p-1$

One way of seeing this is Legendre's formula on the power of a prime dividing some factorial, http://www.cut-the-knot.org/blue/LegendresTheorem.shtml

and http://en.wikipedia.org/wiki/Factorial#Number_theory

From the formula, $p$ divides $p!$ with exponent exactly $1,$ but $p$ does not divide $i!$ or $(p-i)!$ when $1 \leq i \leq p-1.$

rschwieb
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Will Jagy
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  • Do we also need to prove this? Or is this considered well-known? – justhalf Apr 17 '15 at 06:12
  • Well, this is the inductive proof I was referring to in the text. It seems very inelegant to me. Induction is used twice, in proving the binomial theorem and then if we want to generalize it to more than two terms( say $(x+y+z)^p)$. This is unsatisfying basically because we are using something from outside the domain of finite fields to prove it if that makes sense. The $ \Bbb{F} _p $ do not have a principle of induction, one would expect there to be some proof that does not use it either (or atleast hides it very well). Maybe something group theoretic though I don't see how? – Asvin Apr 17 '15 at 07:05
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    @Asvin: The point is that you are using the action of the integers on the elements of the finite field. There is no reason why we should or could expect not to use the integers in the proof, since the very statement of field characteristic requires "using something from outside the finite field". – user21820 Apr 17 '15 at 07:32
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    @justhalf: the proof is trivial once you believe that $\binom{p}{i} = \frac{p!}{i!(p-i)!}$, and also believe that the integers have unique prime factorisations, and also that $p$ is prime, since then there's no factor of $p$ on the bottom when $i < p$ and $p-i < p$. You can give elementary proofs of all three of those, but hopefully don't need to every time you use them ;-) – Steve Jessop Apr 17 '15 at 09:30
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    @Asvin Let me convince you it's not inelegant, doesn't require induction to understand (only counting with symmetry, which is beautiful), and is germane to the problem at hand. Expanding $(x+y)^p$ yields all possible words in the letters $x$ and $y$; every one of these words can get its letters cyclically shifted to the right and wrapped back around yielding $p$ different words all of which are like terms and cancel - except for $xx\cdots x$ and $yy\cdots y$. The theorem may not be about finite fields, but it is about binomials, which is what we're talking about so it's perfectly germane. – anon Apr 17 '15 at 12:58
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    Also, Asvin, you have to use induction to say that any ring homomorphism is finitely additive, that is not at all specific to finite fields. But it's perfectly intuitive and obvious so who cares? | BTW, @SteveJessop, the full factorial formula for binomial coefficients is unnecessary to prove $p\mid\binom{p}{i}$ for $0<i<p$. There is a combinatorial argument using symmetry that also works; see my comment above to Asvin. (The argument can also be cast in terms of groups actions and the orbit-stabilizer theorem, but I decided to keep it simple and focused on binomial expansions.) – anon Apr 17 '15 at 13:00
  • @whacka I am sorry for replying so late. I really like your proof but it somehow does not seem fundamental to the theory. I really like the proof by derivatives, maybe because derivatives appear so many more times in finite field theory. Your proof is very similar to the proof for the existence of a p-subgroup in any group though. – Asvin Apr 21 '15 at 22:55
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Let $F$ be a field of characteristic $p$. Let $f = (1 + x)^p \in F[x]$. We want to show that $f = 1 + x^p$.

Take the formal derivative: $f' = p(x+1)^{p-1} = 0$

Now we know that $f$ has degree $p$, and its derivative is $0$, so $f$ must be in the form $A + Bx^p$ with $A$, $B \in F$.

$f(0) = 1$ so $A = 1$.

A product of monic polynomials is always monic so $B = 1$.

Q.E.D.

The "freshman's dream" is a corollary of this fact.

The fact that the binomial coefficient $\binom p i$ is divisible by $p$ for $1 \leq i \leq p-1$ is also a corollary.

The binomial theorem itself can be proved by taking derivatives of $(1 + x)^n$.

Fermat's little theorem follows easily: $\left( \sum_{i=1}^n 1 \right)^p = \sum_{r=1}^n (1^p) = \sum_{r=1}^n 1$

wlad
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    This requires proving the power-chain rule $(u(x)^e)'=eu(x)^{e-1}u'(x)$ for the formal derivative, which initially might seem to involve a lot of algebra. There is a clever definition of the formal derivative (not just the artificial $(x^n)':=nx^{n-1}$ one) that allows us to only use the geometric series formula to prove it, and it's elegant in my opinion. The definition is that $f'(x)$ equals the two-variable polynomial $\frac{f(x+h)-f(x)}{x-h}$ simplified to an element of $K[x,h]$ first and then evaluated at $h=0$ second. – anon Apr 17 '15 at 13:13
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    @whacka: that should be divided by $h$, not $x - h$ – wlad Apr 17 '15 at 13:22
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    Oops, yes, that's what I meant. – anon Apr 17 '15 at 13:23
  • A really nice way to do it is to map $F[X]$ into $F[X][Y]/Y^2$. then $f(X + aY) = f(X) + af'(X)Y$ where $a \in F[X]$. the chain rule falls out with no effort. – wlad Apr 17 '15 at 13:49
  • I suppose computing $(f(X)+f'(X)Y)^n=f(X)^n+nf(X)^{n-1}f'(X)Y$ mod $Y^2$ is "easier" than computing $(x+y)^p$ mod $p$, but perhaps the OP could disagree with that. – anon Apr 17 '15 at 13:53
  • I decided to accept this comment because derivatives seem much more fundamental to the theory due to how often they appear... – Asvin Apr 21 '15 at 22:56