TUT ALG3 2

Problem 2

Question

Let $n\ge 1$ be an integer. Determine $U((\mathbb{Z}/n\mathbb{Z})[x])$.

Let $\psi (n)$ be the squarefree part of $n\in \mathbb{N}$. Call $a\in \mathbb{Z}/n\mathbb{Z}$ nilpotent if $\psi (n)|a$. Call $g(x)\in \mathbb{Z}/n\mathbb{Z}[x]$ nilpotent if $g(x{)}^k=0$ for some $k\in \mathbb{N}$, called the order of $g$.

Lemma 1(Lemma).

The sum of two nilpotent polynomials is nilpotent.

Proof.

Let $g(x)$ and $h(x)$ be nilpotent polynomials of orders $\alpha$ and $\beta$. Then, $(g(x)+h(x){)}^{\alpha +\beta }=0$, since every term in its expansion is of the form $cg(x{)}^{i}h(x{)}^j$, where $i\ge \alpha$ or $j\ge \beta$.□

Lemma 2(Lemma).

$g(x)={\sum }_{i\ge 0}{g}_i{x}^i$ is nilpotent iff every $g_i$ is nilpotent for $i\ge 0$.

Proof.

$(⟸)$ Let $s$ be the highest power of a prime factor in the prime factorization of $n$. It is clear that any product of $s$ or more nilpotent elements in $\mathbb{Z}/n\mathbb{Z}$ is $0$. Thus, $g(x{)}^s=0$, since every coefficient in $g(x{)}^s$ is a product of $s$ nilpotent elements.

$(⟹)$ If $h(x)$ is a nilpotent polynomial of degree $d$, it is clear that $h_0$ and $h_d$ must be nilpotent. Thus, any nilpotent polynomial of degree $1$ has nilpotent coefficients!

Now, we will proceed to prove the claim by induction. The claim is trivially true for $d=0$, is true for $d=1$ by the preceding observation. Next, assume that the claim is true for nilpotent polynomials of degree $d$. Let $h(x)=h_0+\dots h_{d+1}{x}^{d+1}$ be a nilpotent polynomial of degree $d+1$. Again, we know that $h_0$ and $h_{d+1}$ are nilpotent. Now, notice that

$$h_0+h_{1}x+\cdots +h_d{x}^d=h(x)+(-h_{d+1}{x}^{d+1}).$$

Since both $h(x)$ and $-h_{d+1}{x}^{d+1}$ are nilpotent, it follows from Lemma 1 that $h_0+h_{1}x+\cdots +h_d{x}^d$ is nilpotent. It follows from the induction hypothesis that $h_0,\dots ,h_d$ are nilpotent.□

Now, I claim the following: $f(x)=\sum f_i{x}^i$ is a unit in $\mathbb{Z}/n\mathbb{Z}[x]$ $⟺$ $f_0\in U_n$ and $f_i$ is nilpotent for $i>0$.

Proof.

$(⟸)$ If $f_0\in U_n$ and $f_i$ is nilpotent for $i>0$, $f$ can be expressed as

$$f_0(1+f_0^{-1}\stackrel{g(x)\equiv }{\overbrace{(f_{1}x+f_2{x}^2+\dots f_k{x}^k)}}).$$E1

Note that by Lemma 2, $g(x)$ is nilpotent.

If we attempt to construct an inverse $h(x)$ of $1+f_0^{-1}g(x)$, we will observe $h(x)$ is forced to be

$$1-f_0^{-1}g(x)+(f_0^{-1}{)}^{2}g(x{)}^2+\cdots =\sum (-f_0^{-1}{)}^{i}g(x{)}^i.$$E2

It is clear that $h(x)\in \mathbb{Z}/n\mathbb{Z}[x]$ only if $g(x{)}^i=0$ for all $i>l$ for some integer $l$. Since $g(x)$ is nilpotent, this condition is satisfied. Thus, $f(x)$ is a unit, with inverse

$$f_0^{-1}\sum _{i=0}^l(-f_0^{-1}{)}^{i}g(x{)}^i.$$

$(⟹)$ $f_0\in U_n$ is immediate. We again express $f(x)$ as in (E1). Since $f_0$ is invertible and $f(x)$ in invertible, so is $1+f_0^{-1}g(x)$, the inverse of must then be given by (E2), because any inverse would need to satisfy the recursive relation from multiplying out, which formally yields the series. Since the inverse is a (finite-degree) polynomial, the series must terminate, implying $g(x)$ must be nilpotent, which in turn implies $f_i$ is nilpotent for $i>0$ by Lemma 2.□