Skip to main content
4 events
when toggle format what by license comment
Jul 24, 2020 at 10:37 comment added Sean Eberhard Yes you are right, my mistake, thanks.
Jul 24, 2020 at 10:33 comment added Emil Jeřábek Homomorphisms do not preserve the order of elements. A preimage of an element of order $n$ may have order $mn$, or even infinite.
Jul 24, 2020 at 10:12 comment added Sean Eberhard If every $p_n$ is a Fermat prime then $(\mathbb{Z}/p_n\mathbb{Z})^\times \cong \mathbb{Z}/2^{k_n}\mathbb{Z}$, so $K^\times$ contains a unique element of order 2, and it has a square root, so there is no epimorphism $K^\times \to \mathbb{R}^\times$.
Jul 24, 2020 at 9:58 history answered Sean Eberhard CC BY-SA 4.0