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Let $Z$ be an algebra over a field $K$ such that $Z$ is generated by a single nilpotent element. Why is $Z$ a local ring?

Would this follow from the Chinese Remainder theorem?

Let $m$ be a maximal ideal of $Z$ then since $Z$ is Artinian then $Z \cong Z/m^{j}$ (by counting dimensions over $K$) for some natural $j$. But $Z/m^{j}$ is a local ring so we are done.

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    $Z$ is a quotient of $K[x]/(x^n)$; but $K[x]/(x^n)$ is a local ring: every (image of a) polynomial with nonzero constant coefficient is a unit, so $(x)/(x^n)$ is an ideal and exactly the collection of all nonunits of the ring. Thus, $Z$ is a quotient of a local ring, and hence is local.2012-03-01
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    Dear @Arturo, you can even say that $Z$ is isomorphic to $K[x]/(x^n)$, without any need to take a further quotient (but what you wrote is perfectly correct, of course).2012-03-01

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