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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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    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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Just take all polynomials in your nilpotent element $x$ (the generator of $Z$ as a $K$ algebra), with constant term zero. This is clearly an ideal of $Z$, say $m$. Note that $m$ is a proper ideal because $1$ is not in $m$.

Now an element not in $m$ is a polynomial $a_nx^n + ... + a_1x + a_0$, where $a_i \in K$, and $a_0 \neq 0$, so a unit in $Z$. But, $a_nx^n + ... + a_1x$ is nilpotent being a sum of nilpotent elements. If $a_nx^n + ... + a_1x + a_0$ is not a unit, then there is some max ideal m' such that a_nx^n + ... + a_1x + a_0 \in m'. But, nilpotents are in every max ideal. So, a_nx^n + ... + a_1x \in m', which shows that a_0 \in m', which is impossible because $a_0$ is a unit. Thus, $a_nx^n + ... + a_1x + a_0$ must be a unit. Hence, any element of $Z/m$ is a unit, and $m$ is the only maximal ideal.

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    This is a more hands on approach. You should really look at Arturo's proof. It is more slick.2012-03-01