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Let be $R$ a local conmutative ring of dimension 0, I want to prove that ring has only units and nilpotent elements. I managed to get that in the case of non-zero divisors are units. Nevertheless, I couldn't get that there are also nilpotent elements.

user26857
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Hakim Ziyech
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1 Answers1

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Krull dimension $0$ means that there are no prime ideals besides the maximal ideal $M$.

An element of $R$ outside $M$ is clearly a unit, as it is not contained in any proper ideal.

Now use the fact that the intersection of all prime ideals coincides with the nilradical, i.e., the set of nilpotent elements.

Andreas Caranti
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