Description of ideals of ring $F[x]/(x^n)$?

Question

What is a description of the ideals of the ring $F[x]/(x^n)$, where $F$ is a field?

Answer 1

Hint: You can pull back ideals from the quotient to $F[x]$, and so you can convert the problem to classifying ideals of $F[x]$ that contain $(x^n)$.

Answer 2

By the Lattice Isomorphism Theorem, there is a 1-1 correspondence between ideals in $F[x]$ containing $(x^n)$ and ideals in $F[x] / (x^n)$. So you are reduced to considering ideals $I \subseteq F[x]$ that contain $(x^n)$, i.e. $(x^n) \subseteq I$.

Now, $F$ is a field so $F[x]$ is a Euclidean domain. In particular, it is a PID. Thus $I = (p(x))$ for some $p(x) \in F[x]$. Then $(x^n) \subseteq (p(x))$ if and only if $p(x) \mid x^n$ if and only if $p(x) = cx^d$ for some integer d satisfying $0 \le d \le n$ and $c \in F^*$.

Answer 3

It is possible to proceed directly by determining the structure of the ring $R = F[x]/(x^{n})$. Note that when $n =1,$ the quotient ring is just $F$ itself. In general, by the Euclidean property of polynomial rings (over fields), every element of $R$ has a unique expression in the form $a_{0} + a_{1}y + \ldots + a_{n-1}y^{n-1}$, where $y$ is the element $x + (x^{n})$ in $R$. The product of such elements is easily described, determined by the rule $y^{i}.y^{j} = 0$ if $i+j \geq n$.

An element $a_{0} + a_{1}y + \ldots + a_{n-1}y^{n-1}$ is a unit of $R$ if and only if $a_{0} \neq 0$ ( I omit details, but it is consequence of the Euclidean algorithm for $F[x]$).

Thus every proper non-zero ideal of $R$ is contained in $(y)$. Let $I$ be an ideal of $R$ (proper and non-zero), and choose $i$ maximal such that $I \subseteq (y^{i})$ but $I \not \subseteq (y^{i+1})$. Then $I$ contains an element of the from $y^{i}(a_{0} + \ldots a_{n-1-i}y^{n-1-i})$ with $a_{0} \neq 0$. Since $(a_{0} + \ldots a_{n-1-i}y^{n-1-i})$ is a unit of $R$, we see that $(y^{i}) \subseteq I$, so that $I = (y^{i})$.

Hence the ideals of $R$ are $0 \subset (y^{n-1}) \subset (y^{n-2}) \subset\ldots \subset (y^{2}) \subset (y) \subset R$, (and all inclusions are strict).

Answer 4

Let A be the quotient ring and t be the image of x in A. Then A is isomorphic to F[t]. Since A/(t) is isomorphic to F, which is a field, (t) is of course a maximal ideal of A. But it is actually the only one, because any element outside (t) is invertible: up to a non zero multiplicative constant, any such element can be written 1 + tz, which has an inverse because t is nilpotent (think of the formal inverse of 1 + tz). So the strict ideals of A are the principal ideals (t^k), k between 1 and n.

PS. After writing this, I realize that it's exactly Geoff Robinson's answer, under a slightly different guise !