Let $K$ be a field with characteristic $p>0$ and set $q=p^{n}$ for $n \in \mathbb{N}^*$. Prove $(a+b)^{q}=a^{q}+b^{q}\ \forall a,b \in K$.

Question

Let $K$ be a field with characteristic $p>0$ and set $q=p^{n}$ for $n \in \mathbb{N}^*$. Prove $(a+b)^{q}=a^{q}+b^{q}\ \forall a,b \in K$.

I can prove using the coefficient theorem that $(a+b)^{p}=a^{p}+b^{q}$, by saying that $p\mid {p\choose i}$ for $0\leq i \leq p-1$. The problem is we have to prove it for $q$ that is not a prime number, which doesn't allow the same reasoning.

score 0 · Accepted Answer · edited Nov 08 '21 at 17:19

0

Let $q=np$ for $n\in \mathbb{Z}^+$ and consider the expression $$(a+b)^{np}=\sum_{k=0}^{np}{np \choose k} a^{np-k}b^k$$ and again notice that if $p\mid {np \choose k}$ for $k\in \lbrace 1,2,3,\dots, np-1\rbrace$ then the claim follows, but it is false in general.

Let $n=2$ and $p=3$. Then $3$ doesn't divide ${6 \choose 3}=20$.

Probably you want write $q=p^n$ for $\mathbb{Z}^+$ in which case the assertion is true, and the proof consist of give a easy generalization for $$p\mid {p^n \choose k}$$ for $k\in \lbrace 1,2,\dots, p^n-1\rbrace$ which can be done by Lucas Theorem.

edited Nov 08 '21 at 17:19

user26857

53,190

answered Nov 08 '21 at 13:52

Brien Navarro

787

You are right, I meant $q=p^{n}$ – Nov 08 '21 at 14:13
Right, now only notice that $p|{p^n \choose k}$ is a generalization of the first proof that you made, it is you prove the case $n=1$, but the general case is a easy consequence of apply Lucas´s Theorem as well. – Brien Navarro Nov 08 '21 at 14:16
I'll take a look into that theorem. I don't think I'm familiar with it, but thanks in advance – Nov 08 '21 at 14:18
Here is the proof https://math.stackexchange.com/questions/751908/proof-that-p-pn-choose-k-for-any-prime-p-and-k-pn – Brien Navarro Nov 08 '21 at 14:19
This answer is unnecessarily complicated. What the OP wants can be proved by a straightforward mathematical induction on $n$. – user26857 Nov 08 '21 at 17:22

score 0 · Answer 2 · edited Nov 08 '21 at 17:17

0

Let $\varphi(x)=x^p$. You said you proved that $\varphi$ is a homomorphsim of $K$. The of course also $\varphi^n$ is one.

edited Nov 08 '21 at 17:17

user26857

53,190

answered Nov 08 '21 at 14:17

principal-ideal-domain

6,324

score -1 · Answer 3 · answered Nov 08 '21 at 13:57

-1

Hint:

Try to find a counterexample with $p=3$ or $p=5$ and $n=2$.

answered Nov 08 '21 at 13:57

supinf

13,593

Let $K$ be a field with characteristic $p>0$ and set $q=p^{n}$ for $n \in \mathbb{N}^*$. Prove $(a+b)^{q}=a^{q}+b^{q}\ \forall a,b \in K$.

3 Answers3