Freshman's dream in the quotient Let $k$ be a field of characteristic $p$ and let $A$ be a $k$-algebra. Let $S$ be the subspace of $A$ generated by the commutators, that is, the $k$-span of elements of the form $[a,b] = ab-ba$ (I think $S$ is called the derived algebra of the Lie algebra $A$ with Lie bracket given by the commutator).
I saw in a few places that since $ab \equiv ba \pmod{S}$, we can deduce
$$(a+b)^p \equiv a^p + pa^{p-1}b + \cdots + b^p \equiv a^p + b^p \pmod{S} \, .$$ 
This would be a direct application of freshman's dream in the quotient $A/S$, if $S$ were an ideal of $A$. But it is in general just a subspace so I don't know how the claim can be justified. I think we need rearrangements like $aba \equiv aab$ for the above binomial expansion to work and the multiplicative structure need not be preserved like that in the quotient.
Is the claim $(a+b)^p \equiv a^p + b^p$ true at all?
 A: This is actually true. To see this, notice that when you expand $(a+b)^p$, we only need to analyze terms with a fixed number of $a$s and $b$s. Clearly, we have $a^p$ and $b^p$ and each occurs only once. The problem lies with the other terms.
However, notice that if we have two words in alphabet $a,b$, any two words which are obtained from one another by cyclic permutation (repeatedly taking the last letter to the front) are $S$-equivalent. There are exactly $p$ cyclic permutations, so for them to cancel out, it is enough for all $p$ to be distinct for a given word. This is not the case for $a^p$ and $b^p$, as those are invariant under cyclic permutations (in fact, under any permutations). On the other hand, any other word is not invariant under cyclic permutations, as there is always a cyclic permutation which has $a$ in front and one which has $b$. From this it follows that they are all distinct.
More formally, consider the action of the cyclic group $C_p$ on $p$-letter words in alphabet $a,b$ by cyclic permutation. Any element other than identity of $C_p$ generates it, so the stabilizer of a given word $W$ is either just identity or the entire $C_p$. It is clearly not the entire $C_p$ if $W$ is not $a^p$ or $b^p$, so it must be trivial, so we have $p$ cyclic permutations of each word and they are $S$-equivalent and cancel each other out.
In other words, the set of all $p$-letter words (whose sum is exactly $(a+b)^p$) is divided into orbits of $C_p$ which are $\{a^p\}$, $\{b^p\}$ and $(2^p-2)/p$ orbits with $p$ elements each, and each $p$-element orbit sums to zero modulo $S$.
A: You do in fact need $aab - aba$ to be in $S$. And it is! Let $a' = a$ and $b' = ab$. Then $[a', b']$ is in $S$. But $[a', b'] = [a, ab] = aab - aba$.
Edit: The comment below makes a good point: as far as I can tell $aabb-abab$ will not be in the vector space $S$ (though I haven't proven this). So let me backtrack here and ask: in the equation mod $S$ above, what does $(a+b)^p$ (for example) mean?
