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How is $\frac{F[x]}{(x^n)}$ a local ring?

I was trying to show the elements which are not units are nilpotent. But not being able to prove it properly. Please give some hint.

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    $\begingroup$ The ideals of a quotient ring are in bijection with ideals containing the defining ideal $\endgroup$
    – leibnewtz
    May 22, 2019 at 15:51
  • $\begingroup$ Hi: the original title seemed to indicate you had confused the title field for the tags field. It looks like Arnaud D. has kindly rectified that for you. In the future, please try to use the question in the title, if possible. $\endgroup$
    – rschwieb
    May 22, 2019 at 16:38

3 Answers 3

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More generally, if you take any maximal ideal $M$ in a commutative ring $R$, $R/M^k$ is local for any positive integer $k$.

A maximal ideal of $R/M^k$ would have to be of the form $A/M^k$ where $A$ is a maximal ideal of $R$ containing $M^k$, by ideal correspondence.

But since $A$ is maximal it is prime, so $M^k\subseteq A$ would imply $M\subseteq A$. But then by maximality of $M$, $M=A$.

In your case, $M=(x)$.

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The maximal ideal is $(x)$. An element not in $(x)$ has the form $a-xg(x)$ where $a$ is a nonzero element of $F$. It has the inverse $\sum_{k=0}^{n-1}a^{-k-1}x^kg(x)^k$. As all elements outside $(x)$ are invertible, the ring is local.

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I guess you are assuming $F$ being a field.

So let $F$ be a field. Let $\mathfrak m$ denote the ideal generated by $x$ in $A:=F[x]/(x^n)$.

Let us prove that $\mathfrak m$ is the maximal ideal of $F[x]/(x^n)$. I propose a constructive trick. Notice that the elements of $A\setminus \mathfrak m$ are of the form $k + xf$ where $k\in F^\times$, $f\in A$. Thus $k + fx$ is invertible if and only if $1+xf/k$ is invertible.

For simplicity let us invert $1-x$. Consider the sequence defined by $$ a_1=1 - x $$ and $$ a_{m+1} = \left(1+x^{2^{m-1}}\right)a_m. $$

Hence $$ a_2 = (1-x)(1+x) = 1-x^2, $$ $$ a_3 = (1-x)(1+x)(1+x^2)= 1-x^4, $$ $$ a_{m} = (1-x) (1+x)\cdots (1+x^{2^{m-2}}) = 1-x^{2^{m-1}} $$ are multiples of $1-x$. Since $x^t=0$ in $K[x]/(x^n)$ for every $t\geq n$ one has that $a_m=1$ for $m$ large enough. Thus the inverse of $1-x$ in $K[x]/(x^n)$ is $$ (1+x) \cdots (1+x^{2^{m-1}}) $$ for $m$ large enough ($m$ large so that $2^{m-1} \geq n$).

The case $1-x f$ follows similarly.

This is proves a general property of nilpotent elements. If $\alpha$ is a nilpotent element in a ring $A$ then $1 + \alpha$ is a unit in $A$.

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