Let $R$ be finite commutative local ring. Let $U(R)$ be the group of units and $M$ the maximal ideal and denote by $k$ the degree of nilpotence of $M$. I'm trying to find an example of a finite commutative local ring $R$ where the order of $U(R)$ is less than $k$. I do not know how to find it, can someone help me out?

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    $\begingroup$ What makes you think such an example is possible? $\endgroup$ – quasi May 24 '18 at 15:57

Look no further.

Let $M^n=\{0\}$ but $M^{n-1}\neq\{0\}$. Then in the series $\{0\}\subset M^{n-1}\subset\ldots\subset M\subset R$, we can see that $2|M^i|\leq |M^{i+1}|$, because $M^i$ has to have at least two cosets in $M^{i+1}$.

Working up the chain this way, you can work out that $|M|\geq 2^{n-2}|M^{n-1}|$, and $M^{n-1}$ has at least two elements, so $|M|\geq 2^{n-1}$.

There is also an injective map $M\to U(R)$ given by $m\mapsto 1+m$. This means that $|U(R)|\geq |M|\geq 2^{n-1}$.

But if you check, $2^{n-1}\geq n$ for every positive integer $n\geq 1$.


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