Automorphisms of the ring of Laurent polynomials Is the group of automorphisms of the ring $\mathbb{F}[t,t^{-1}]$ of Laurent polynomials known? Here, $\mathbb{F}$ is an algebraically closed field of characteristic $0$ and I am considering not necessarily unital automorphisms. Thanks in advance.
 A: First of all I think the unitality is automatic, since there is the unique element which multiplies trivially with any other (and the image of 1 satisfies this).
All invertible elements in $\mathbb F[t,t^{-1}]$ are of the form $\alpha\cdot t^n$ where $\alpha\in \mathbb F^\times $ and $k\in \mathbb Z$. Since the image of $\mathbb F$ should be closed under addition and multiplication and all elements in $\mathbb F\setminus 0$ are invertible it is easy to see that the image of $\mathbb F$ should be contained in $\mathbb F$ (otherwise it is easy to cook up some non-invertible Laurent polynomial in the image of $\mathbb F\setminus 0$). The same is true for the inverse map, thus any automorphism of $\mathbb F[t,t^{-1}]$ induces an automorphism of $\mathbb F$. Then I guess the question is whether you want these automorphisms to be $\mathbb F$-linear or not. For the linear ones you just need to understand the image of $t$ which can only be $\alpha\cdot t$ or $\alpha\cdot t^{-1}$. This gives an equivalence $\mathrm{Aut}_{\mathbb F}(\mathbb F[t,t^{-1}])\simeq \mathbb F^\times\rtimes \mathbb Z/2\mathbb Z$ where non-trivial $\sigma\in \mathbb Z/2\mathbb Z$ acts sending $\alpha \in \mathbb F^\times$ to $\alpha^{-1}$. In the non-linear case we need to add the group of automorphisms of $\mathbb F$ over $\mathbb Q$, then we get $\mathrm{Aut}_{\mathbb F}(\mathbb F[t,t^{-1}])\simeq \mathbb F^\times\rtimes \left(\mathbb Z/2\mathbb Z\times \mathrm{Aut}_{\mathbb Q}(\mathbb F)\right)$
