Prove that $n \in \mathbb{Z}^\star \Leftrightarrow n \mid 1$ and $n-1 \mid 1$ or $n+1 \mid 1$, and $(x-1)/(t-1) \equiv n \pmod {t-1}$

Question

I want to prove the following lemma:

Let $F[t,t^{-1}]$ be the ring of the polynomials in $t$ and $t^{-1}$ with coefficients in the field $F$ and assume that the characteristic of $F$ is zero.

Then for any $n$ in $F[t, t^{-1}]$, $n$ is a nonzero integer if and only if

• $n$ divides $1$
• either $n-1$ divides $1$ or $n+1$ divides $1$, and
• there is a power $x$ of $t$, so that $\dfrac{x-1}{t-1} \equiv n \pmod {t-1}$.

Could you give me some hints how we could show that?

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EDIT:

I thought that we could use the following, but I am not sure:

an integer is nonnegative iff it is the sum of four integer squares and $x$ is nonzero iff there is a nonnegative integer $y$ so that $x=y+1$ or $x=−y−1$.

Does this help?

Answer 1

I had wanted to give some hints and then put the solution into the "hidden" mode. However, the command >! didn't seem to be working with my text (this might have something to do with LaTeX commands), so I will just put my solution then. I would greatly appreciate if anybody can suggest a way to hide my text below.

First, suppose $n\in\mathbb{Z}\setminus\{0\}$. Clearly, $n\mid 1$, and either $n+1\mid 1$ or $n-1\mid 1$. Furthermore, if $n>0$, then $$\frac{t^n-1}{t-1}=\sum_{j=0}^{n-1}\,t^j=n+(t-1)\left(\sum_{j=0}^{n-1}\,\frac{t^j-1}{t-1}\right)=n+(t-1)\left(\sum_{j=1}^{n-1}\,\sum_{k=0}^j\,t^k\right)\equiv n\pmod{t-1}\,.$$ If $n<0$, then $$\frac{t^n-1}{t-1}=-t^{n}\left(\frac{t^{-n}-1}{t-1}\right)\equiv -t^n(-n)\equiv nt^n\equiv n\pmod{t-1}\,.$$

Now assume that $f(t)\in F\left[t,t^{-1}\right]$ is such that
(a) $f(t)\mid 1$,
(b) $f(t)+1\mid 1$ or $f(t)-1\mid 1$, and
(c) there exists $n\in\mathbb{Z}$ such that $\frac{t^n-1}{t-1}\equiv f(t)\pmod{t-1}$.
We shall prove that $f(t)=n \neq 0$. Condition (c) implies that $f(t)-n$ is divisible by $t-1$. Hence, $$f(t)=n+(t-1)\,g(t)$$ for some $g(t)\in F\left[t,t^{-1}\right]$. Since $f(t)\mid 1$, $f(t)$ is a unit in $F\left[t,t^{-1}\right]$, whence $f(t)=at^k$ for some $a\in F\setminus\{0\}$ and $k\in\mathbb{Z}$. If $f(t)\pm 1 \mid 1$, then $f(t)\pm 1$ is also a unit. That is, $at^k\pm 1$ is of the form $bt^l$, where $b\in F\setminus\{0\}$ and $l\in\mathbb{Z}$. If $k\neq 0$, we note that $at^k\pm1$ has a nonzero root in the algebraic closure of $F$, but $bt^l$ does not. Therefore, $k=0$. Hence, $f(t)=a$ is a nonzero, constant polynomial. Now, from $f(t)=n+(t-1)\,g(t)$, we have $0\neq a=f(1)=n$. Ergo, $f(t)=n \neq 0$ as desired.