What is this ring $\mathbb{Z}[x,1/x]$?

Question

I can see that it is the ring of polynomials over $\mathbb{Z}$ with variables $x,1/x$, i.e. things like $x+1/x$. Is there a better way to visualize the elements of this ring? I am wondering if it is Noetherian and I think I can figure that piece out if I can see the elements of this ring better.

I want to say that this is the ring of rational functions over $\mathbb{Z}$ i.e. $p(x)/q(x)$ where $p,q$ are taken over $\mathbb{Z}$ but I am not sure because $\mathbb{Z}$ does not have division.

Answer 1

In fact, $\mathbb{Z}[x,1/x]$ is not equal to the ring of rational functions over $\mathbb{Z}$.

Proof: If it were, then $1+x$ would have a multiplicative inverse in $\mathbb{Z}[x,1/x]$, namely $\frac{1}{1+x}$. So, we would have $$(1+x) (a_n x^n + \dots + a_1 x + a_0 + a_{-1} x^{-1} + \dots + a_{-n} x^{-n}) = 1$$ for some $a_{-n}, \dots, a_n \in \mathbb{Z}$. Now evaluating at $x = -1$ yields $0 = 1$, contradiction!

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It is true that $\mathbb{Z}[x,1/x]$ is a subring of the ring of rational functions over $\mathbb{Z}$. It is not hard to show that $\mathbb{Z}[x,1/x]$ consists of the rational functions of the form $p(x)/x^n$ with $p(x) \in \mathbb{Z}[x]$ and $n \in \mathbb{N}$.

Answer 2

You want to know whether or not the ring $\Bbb Z[x,\frac{1}{x}]$ is Noetherian. Since for a Noetherian ring $R$ also the ring $S^{-1}R$ is Noetherian, the claim follows from $$ S^{-1}\Bbb Z[x]=\Bbb Z[x,\frac{1}{x}], $$ with the multiplicatively closed set $S=\{x^n\mid n\in \Bbb N\}$.