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.
 A: 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.
A: 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 $A$. 
A: 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, $M=(x)$.
A: Hint: The non-units form an ideal.
$f(x)+I$ is invertible when there exists $g(x)+I$ such that $f(x)g(x)-1\in I$. This implies $f(x)g(x)+h(x)x^n=1$ and hence $(f,x^n)=1$.
Hence an element $f(x)+I$ is non-unit whenever $(f,x^n)\neq 1$ and hence it must be the case that $x|f(x)$ and now it should be clear that the set of non-units forms an ideal.
