Let $R$ be a commutative ring with identity and let $a\in R$ such that $a^{(n−1)}$ is not zero, but $a^n$ is zero for some positive integer $n$. Show that the factor ring $R[x]/(1−ax)$ is a zero ring, that is,
$$
R[x]/(1−ax)=\{(1−ax)\}
$$
I tried:
let $b\in(1−ax)$, then $b=(1-ax)c=c-cax$ for some $c\in R$. Since $a^n$ is zero and $a^{(n-1)}$ is not zero, $a$ is a zero divisor. then
$$
(b+(1-ax))^n=b^n+(1-ax)
$$
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Daniele Tampieri
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JJJeong
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Saying that $A/I$ is a zero ring, when $A$ is a ring with identity and $I$ is an ideal thereof, means that $I=A$.
In order to prove $(1-ax)=R[x]$, you just need to show that $1\in(1-ax)$, so you need to find an inverse of $1-ax$.
Now, if you were dealing with series in calculus, you'd write $$ \frac{1}{1-ax}=\sum_{k=0}^\infty a^kx^k $$
Can you make a guess now?
egreg
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