Let $R$ be a ring of prime characteristic. Say $a\in R$ and $a$ is nilpotent. Prove that $1+a$ is unipotent.

Question

Let $R$ be a ring of prime characteristic. Say $a\in R$ and $a$ is nilpotent. Prove that $1+a$ is unipotent.

What I have done so far:

Since $a$ is nilpotent, $a^n=0$ for some $n\geq5.$ Let $p=char(R)$.

Then $(1+a)^{p^n} =$ ${{p^n}\choose0}a^0$ $+$ ${{p^n}\choose1}a^1$ $+$ ... $+$${{p^n}\choose n}a^n$

Now, if I can prove that each of these binomial coefficients (except $p^n\choose0$) is divisible by p, I should be done. How do I do this?

Answer 1

Hint Show first that if $a,b$ commute then $$(a+b)^p=a^p+b^p$$

Then use $$(1+a)^{p^{n+1}}=((1+a)^{p^n})^{p}$$ and induction