# The group of invertible elements of $\mathbb F_{p}[x]/(x^m)$ is not a cyclic group.

I am stuck in a question about finite fields and would like to ask you for some help.

Given an integer $m\geq 2$ and $p$ a prime number, show that $(\mathbb F_{p}[x]/(x^m))^{\times}$ (the group of all invertible elements of $\mathbb F_{p}[x]/(x^m)$) is not a cyclic group.

I don't really know how to start. I know a lemma which tells me that for every finite field $\mathbb F_{q}$ with $q$ elements, the multiplicative group of the nonzero elements $\mathbb F_{q}^{\times}$ is cyclic. But I don't think that I should use it here...

Does anyone have an idea? Thank you in advance!

• Related to this and this. – user26857 Oct 5 '13 at 18:47
• Sorry, but i am still new on MSE. I will correct my mistake. Thank for you editing my question! – Lullaby Oct 6 '13 at 15:08

Similar arguments to those I have already used here give the following group isomorphism: $$(\mathbb F_{p}[x]/(x^m))^{\times}\simeq\mathbb F_p^{\times}\times\mathbb F_p^{m-1}.$$ (This shows that you should assume $m\ge 3$ in order to get a non-cyclic group.) For $m\ge 3$ if you assume that the group of units is cyclic, then $\mathbb F_p^{m-1}$ is also cyclic (or $\mathbb F_p\times\mathbb F_p$ if you like), a contradiction.

Let $R = \mathbb{F}_p[X]/(X^m)$, then for any $f(X) = a_0 + a_1X + \ldots + a_nX^n \in \mathbb{F}_p[X]$, let $\overline{f}$ denote its image in $R$. Then $$\overline{f} \in R^{\ast} \Leftrightarrow (f,X^m) = 1 \Leftrightarrow X\nmid f(X) \Leftrightarrow a_0 \neq 0$$ Hence, if $\overline{f} \in R^{\ast}$, we may write $$g(x) = 1 + a_0^{-1}a_1X + a_0^{-1}a_2X^2 + \ldots + a_0^{-1}a_nX^n$$ Define $$G = \{1 + b_1\overline{X} + b_2\overline{X}^2 + \ldots + b_{m-1}\overline{X}^{m-1}\}$$ Then, $G$ is a multiplicative subgroup of $R^{\ast}$, and $$\overline{f} \mapsto (a_0, \overline{g})$$ defines an isomorphism $$R^{\ast} \cong \mathbb{F}_p^{\ast} \times G$$ In particular, $|R^{\ast}| = (p-1)p^{m-1}$

1. Note that for any $a \in \mathbb{F}_p^{\ast}$, $a^{p-1} = 1$
2. For any $\overline{g} \in G$, the Frobenius map gives $$\overline{g}^p = 1 + a_1\overline{X}^p + a_2\overline{X}^{2p} + \ldots + a_{m-1}\overline{X}^{p(m-1)}$$ But since $p > m$, $X^m \mid X^p$, and hence $$\overline{g}^p = 1$$ Hence,

The order of any element of $R^{\ast}$ is atmost $(p-1)p$.

Since $|R^{\ast}| = (p-1)p^{m-1}$, $R^{\ast}$ cannot be cyclic, provided $m>2$.

Added : If $m=2$, then $|G| = p$, and $G$ is cyclic. We also know that $\mathbb{F}_p^{\ast}$ is cyclic. Since $(p,p-1) =1$, it follows that

$R^{\ast}$ is cyclic if $m=2$

• Thank you very much, dear Prahlad! Now i am going to try to understand every step of your proof :) Thanks again! – Lullaby Oct 5 '13 at 15:28
• Yes, of course. I was so keen on finding the orders of elements that I didn't think of that! – Prahlad Vaidyanathan Oct 5 '13 at 15:58
• I don't think that $1+a_1 X + \dotsc + a_{m-1} X^{m-1} \mapsto (a_1,\dotsc,a_{m-1})$ (you certainly didn't mean $X^p$, which vanishes) is a group homomorphism (unless $m=2$). This already fails for products $(1+a_1 X)(1+ b_1 X)$. – Martin Brandenburg Oct 5 '13 at 16:05
• Hmm. Yes. My brain is refusing to cooperate now. Will just delete that bit, and think about it later. – Prahlad Vaidyanathan Oct 5 '13 at 16:08

A cyclic group has at most one subgroup of each order, to it is enough to show that there exists two invertible elements in $F_p[x]/(x^m)$ which have the same order and generate different subgroups. In particular, in a cyclic group there is at most one element of order two.

If $p$ is odd, there is a nonzero element $\alpha$ in the field which is different to $1$, and the elements $a=1+x^{n-1}$ and $b=1+\alpha x^{n-1}$ are two different units of order two.

If $p$ is even and $m\geq4$, then $a=1+x^{m-2}+x^{m-1}$ and $b=1+x^{m-1}$ are two different elements of order two.

We are left with $p=2$ and $m=3$, which you can do by hand :-)

• (Notice that in the case I did not do we do have a cyclic group) – Mariano Suárez-Álvarez Oct 5 '13 at 20:38