# Why is $F[x]/(x^n)$ a local ring?

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.

• The ideals of a quotient ring are in bijection with ideals containing the defining ideal May 22, 2019 at 15:51
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## 3 Answers

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, $$M=(x)$$.

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.

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 $$A$$.