Take the 2-minute tour ×

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.

share|improve this question
$Z$ 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. –  Arturo Magidin Mar 1 '12 at 3:29
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). –  Georges Elencwajg Mar 1 '12 at 8:44

1 Answer 1

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.

share|improve this answer
This is a more hands on approach. You should really look at Arturo's proof. It is more slick. –  Rankeya Mar 1 '12 at 4:02

Your Answer


