If $A$ is of characteristic $p$ :Then in $A[x]$ we have, $[a(x)+b(x)]^p=a(x)^p+b(x)^p$

Question

Why, if $A$ has characteristic $p$, is the following part of a proof incorrect: Proving if $\text{char}(A)=p$ then in $A[x]$, $(x+c)^p=x^p+c^p$: $(x+c)^p=x^p+c^p$ in $A[x]$ because $(a+c)^p=a^p+c^p$ for all $a,c\in A$.

Then, use this to prove $[a(x)+b(x)]^p=a(x)^p+b(x)^p$ for any $a(x),b(x) \in A[x]$.

So I don't think it works because $A[x]$ is in respect to $x$, so $x^p \neq a^p$ because $x \in A[x]$ and $p\in A$. Is there a better way of stating this?

Then for the proof: $[a(x)+b(x)]^p=p=(a_0+a_1x+...+a_nx^n)^p+(b_0+b_1x+...+b_nx^n)^p$ thus each term except the first and last for both $a(x)$ and $b(x)$ is $0$ beause we have characteristic $p$, thus $[a(x)+b(x)]^p=a(x)^p+b(x)^p$. QED.

Answer 1

That is, Frobenius Morphism:

More generally, If $a(x),b(x)\in A[x].$ then, $$(a(x)+b(x))^p =\sum_{j=0}^{p}{p\choose j}x^jc^{p-j}=\color{red}{a^p(x)+b^p(x)+} \underbrace{\sum_{j=1}^{p-1}{p\choose j}a^j(x)b^{p-j}(x)}_{=0} $$

But for $1\le j\le p-1$ we have $$p~\text{divides}~{p\choose j}~~~\text{ that is}~~~ {p\choose j} =pm~~\text{for some $m\in\Bbb N$}$$ Since $p$ is the characteristic of $A$ we have $$ \forall~z \in A \implies {p\choose j}z=pm\cdot z= 0~~~\text{for $1\le j\le p-1$}\tag{I}$$

However, $ a(x),b(x)\in A[x]$ implies that $a^j(x)b^{p-j}(x)\in A[x]$ then it can be written as $$ a^j(x)b^{p-j}(x) = \sum_{k=0}^{d_{p,j}} c_{j,k} x^k\qquad\text{with $c_{j,k}\in A$}$$

It springs from (I) that $$ {p\choose j}a^j(x)b^{p-j}(x) = \sum_{k=0}^{d_{p,j}} {p\choose j}c_{j,k} x^k = 0 ~~\text{for $1\le j\le p-1$}$$ Since $$\forall, ~k~~{p\choose j}c_{j,k} = 0 ~~\text{for $1\le j\le p-1$}$$ Finally, $$\color{red}{(a(x)+b(x))^p =a^p(x)+b^p(x)} $$

Answer 2

The proof that, in characteristic $p$, we have $(x+y)^p =x^p + y^p$ follows from the fact the binomial numbers $\binom pj$ are divisible by $p$ for $j\neq 0,p$. Here $x,y\in R$ where $R$ is a ring of characteristic $p$.

Indeed, the binomial theorem gives that $$(x+y)^p = \sum \binom pj x^j y^{p-j} $$ and all terms in the right vanish except possibly $x^p+y^p$.

Now note that if $R$ has characteristic $p$, so does the polynomial ring $R[X]$, and hence the same proof applies.