Proving that an element is nilpotent

Question

Let $m,n \in \mathbb N, n \geq 2$. $A$ is a ring with $|A|=n$ and $a \in A$ s.t. $1-a^k$ is invertible $\forall k \in \{m+1,m+2,...,m+n-1\}$. Prove that $a$ is nilpotent.

Can somebody help me, please? I have no idea how to solve it.

Answer 1

There are two possibilities. Note that $|A \setminus \{0\}| = n-1$.

If each element of $\{1-a^k~|~ k = m+1, m+ 2, \ldots, m+n-1 \}$ is distinct, it contains $n-1$ distinct elements of $A \setminus \{0\}$. That exhausts all of the possibilities, so one of those elements must be $1$, in which case $1-a^k=1$ for some $k$, $a^k=0$ and we are done.

Otherwise, there exist $k \lt l$ with $1-a^{m+k}=1-a^{m+l}$, so $a^{m+k} = a^{m+l}$. But then $a^{m+k}(1-a^{l-k})=0.$ Let $j=l-k$ and note that $j \leq n-2$. If $1-a^j$ is invertible, then the right hand term of the product is invertible and we are done.

But for all positive integers $i, 1-a^j$ is a factor of $1-a^{ij}$. For some $i, m+1 \leq ij \leq m+n-1$ so $1-a^{ij}$ is invertible by hypothesis. For this $i$, let's say $(1-a^j)y = 1-a^{ij}$ and assume $z=(1-a^{ij})^{-1}$. Then $(1-a^j)yz =1$ so $1-a^j$ is invertible, completing the proof.