An identity in Ring of characteristic $p$ prime

Question

Is it true that in a ring of prime characteristic $p$ results that

$(x-1)^{p-1}=1+x+x^2+...+x^{p-1}$ ?

If this is not true in general, the assumption that $x$ is a nilpotent element (let's say $x^{p^n}=0$) make it works?

Answer 1

The identity does indeed hold in general. One way to see this is from the binomial coefficient identity

$$\binom{p-1}{n}\equiv (-1)^n\pmod p$$

To see that this identity holds, notice that

$$\binom{p-1}{n}=\frac{(p-1)(p-2)\ldots(p-n)}{1\cdot2\cdot\ldots\cdot n}\equiv\frac{-1\cdot-2\cdot\ldots\cdot-n}{1\cdot2\cdot\ldots\cdot n}\pmod{p}$$

Answer 2

Yes this is true. As the polynomial ring $\Bbb F_p[x]$ is a domain we may cancel $(x-1)$ from

$$\color{Green}{(x-1)}(x-1)^{p-1}=(x-1)^p=x^p-1=\color{Green}{(x-1)}(x^{p-1}+\cdots+x+1).$$

Since the identity holds for polynomials in an indeterminate, it must also hold for any value of $x$ in an $\Bbb F_p$-algebra (even if $x-1$ were not cancellable in it like it is in $\Bbb F_p[x]$).

Answer 3

Yes this is true, the binomial theorem gives $$(x - 1)^{p-1} = \sum_{k = 0}^{p-1} \binom{p - 1}{k}(-1)^{p - 1 - k}x^k$$ and in characteristic $p$ one can show that $$\binom{p - 1}{k} = (-1)^k$$ and $$(-1)^{p - 1 - k} = (-1)^k$$ so the above simplifies to $$(x - 1)^{p-1} = \sum_{k = 0}^{p-1}x^k.$$