MHB Prove Nonzero Integer Lemma in $F[t, t^{-1}]$

[mathmari]

Hey!

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 $\dfrac{x-1}{t-1}\equiv n \pmod {t-1}$

I need some help to prove the direct direction...

We suppose that $n \in F[t, t^{-1}] \land n \in \mathbb{Z}\setminus \{0\}$.

That means that $n$ is a nonzero constant polynomial of $F[t, t^{-1}]$. Since $F$ is a field, $n$ is invertible, i.e., $n \mid 1$.

Is this correct so far?

How can we show the other two points?

 

[mathmari]

Is the proof as follows?

From an other lemma we have that $(t^n-1)/(t-1)\equiv n \pmod {t-1} \tag {1}$.

Direct:

Let $n \in \mathbb{Z} \setminus \{0\}$, then $n \mid 1$, since $F$ is a field.

It also holds that $n-1 \mid 1$ or $n+1 \mid 1$, otherwise it would be $n-1=0 \Rightarrow n=1$ and $n+1=0 \Rightarrow n=-1$.

From the relation $(1)$ we have that $(t^n-1)/(t-1)\equiv n \pmod {t-1}$, so there is a power $x$ of $t$ such that $(x-1)/(t-1)\equiv n \pmod {t-1}$.

Converse:

We suppose that $n\in F[t,t^{-1}]$ and $n \mid 1$ and $( n-1 \mid 1 \text{ or } n+1 \mid 1 )$ and $(t^k-1)/(t-1)\equiv n \mod (t-1)$.

Since $n\in F[t,t^{-1}]$ is invertible ($n \mid 1$) we have that $n=at^i$, where $a \in F \setminus \{0\}, i \in \mathbb{Z}$.

Since also $n-1$ or $n+1$ is invertible ($ n-1 \mid 1 \text{ or } n+1 \mid 1 $) we have that
$$n-1=b_1t^{j_1}, b_1\in F \setminus \{0\}, j_1 \in \mathbb{Z} \Rightarrow at^i-1=b_1t^{j_1} \\ \text{ or } \\ n-1=b_2t^{j_2}, b_2\in F \setminus \{0\}, j_2 \in \mathbb{Z} \Rightarrow at^i-1=b_2t^{j_2}$$

If $i \neq 0$, then $at^i \pm 1$ has a nonzero root , which is not a root of $bt^j, b \in \{b_1, b_2 \}, j \in \{j_1, j_2 \}$.

So, it must be $i=0$.

So, $n=a \in F \setminus \{0\}$.

Since $(t^k-1)/(t-1)\equiv n \mod (t-1), k \in \mathbb{Z}$ we have that $$k \equiv n \mod (t-1) \Rightarrow n-k \equiv 0 \mod (t-1) \Rightarrow \exists y \in F[t, t^{-1}]: n-k=y(t-1) \Rightarrow n=k+y(t-1)$$
Since $n \in F\setminus \{0\}$, we have that $n$ is nonzero constant polynomial, so it must be $y=0$.

So, we have that $n=k \in \mathbb{Z} \setminus \{0\}$. Is this correct? Could I improve something?

 

[mathmari]

If $i \neq 0$, then $at^i \pm 1$ has a nonzero root , which is not a root of $bt^j, b \in \{b_1, b_2 \}, j \in \{j_1, j_2 \}$.

Is this root in $F$ or in an extension of $F$ ?