Skip to main content
10 events
when toggle format what by license comment
Jun 25, 2020 at 11:01 comment added YCor Thinking twice, I tend to guess that "the two canonical maps $\mathbf{Z}_p\to \mathbf{Z}_p\otimes_\mathbf{Z} \mathbf{Z}_p$ are distinct", holds in ZF. Namely choose $x$ transcendental: the claim is that $x\otimes 1$ and $1\otimes x$ are distinct. Otherwise we have some "relation" saying that they are equal, and this relation holds in some finitely generated subring of $\mathbf{Z}_p$ containing $x$, and in the latter I guess that we distinguish by making an explicit pair of homomorphisms.
Jun 25, 2020 at 10:40 comment added Maxime Ramzi In fact I think YCor's example is in some sense "universal" (for lack of a better word), in that for a map of commutative rings $A\to B$, if the two maps $B\to B\otimes_A B$ agree, then the restriction $\mathrm{Mod}_B\to \mathrm{Mod}_A$ is fully faithful; so an $A$-module either is or isn't a $B$-module, but there is only one way to make it so.
Jun 25, 2020 at 8:36 comment added user160178 @YCor Thanks again!
Jun 25, 2020 at 8:35 comment added YCor An alternative is to consider the two obvious homomorphisms $\mathbf{Z}_p\to\mathbf{Z}_p\otimes_{\mathbf{Z}}\mathbf{Z}_p=:B$. While these homomorphisms are explicit, the fact that they are distinct (i.e., that $B$ is not reduced to $\mathbf{Z}_p$) makes use of some choice as far as I know.
Jun 25, 2020 at 8:34 comment added user160178 @YCor Thanks! (I knew I should have asked the logicians downstairs...)
Jun 25, 2020 at 8:32 comment added YCor No you won't explicitly exhibit any such homomorphism at all. But for $x$ transcendental in $\mathbf{Z}_p$, for every transcendental $t$ in $\mathbf{C}$ there exists a ring homomorphism $\mathbf{Z}_p\to\mathbf{C}$ mapping $x$ to $t$ (it makes use of AC).
Jun 25, 2020 at 8:30 comment added user160178 @Wojowu Thanks! I am almost ashamed to ask this but is it possible to explicitly describe two distinct morphisms $\mathbb{Z}_p \to \mathbb{C}$? All I can find is about set theory (e.g. axiom of choice).
Jun 25, 2020 at 8:24 comment added Wojowu Let $R$ be a commutative ring which admits two distinct ring homomorphisms $\mathbb Z_p\to R$, e.g. $R=\mathbb C$. Now take an abelian group $M$ such that $R$ embeds into the endomorphisms of $M$, e.g. $M=R$ viewed as a group.
Jun 25, 2020 at 8:16 review First posts
Jun 25, 2020 at 8:49
Jun 25, 2020 at 8:15 history asked user160178 CC BY-SA 4.0