Timeline for Can there be a non-trivial epimorphism (of rings) from a field? [closed]
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 12, 2015 at 22:52 | history | closed |
Alex Degtyarev David Roberts♦ Stefan Kohl♦ Dima Pasechnik Andreas Blass |
Not suitable for this site | |
| Apr 12, 2015 at 20:50 | comment | added | YCor | I guess that you mean "epimorphism" in the categorical sense (some people use it for "surjective homomorphism", which is stronger since the ring homomorphism $\mathbf{Z}\to\mathbf{Q}$ is a non-surjective epimorphism). An ambiguity is on what you call "ring": associative? commutative? there are people on mathOverflow using various conventions. | |
| Apr 12, 2015 at 19:16 | answer | added | tj_ | timeline score: 2 | |
| Apr 12, 2015 at 19:13 | answer | added | Todd Trimble | timeline score: 5 | |
| Apr 12, 2015 at 18:25 | comment | added | Paul Taylor | If this is so trivial it ought to have a rigorous proof. Why shouldn't the dimension of $A$ over $K$ be infinite? | |
| Apr 12, 2015 at 17:40 | review | Close votes | |||
| Apr 12, 2015 at 22:56 | |||||
| Apr 12, 2015 at 17:31 | comment | added | Todd Trimble | The typical way to approach this is to consider the pair of maps $i_1, i_2: A \to A \otimes_K A$ where $i_1(a) = a \otimes 1$ and $i_2(a) = 1 \otimes a$, show their restrictions along $K \to A$ agree, and show $i_1, i_2$ disagree if the dimension of $A$ as a vector space over $K$ is greater than 1. | |
| Apr 12, 2015 at 17:28 | comment | added | Alex Degtyarev | Yes, it is necessary, unless $A=0$. Not appropriate for this site. | |
| Apr 12, 2015 at 17:25 | review | First posts | |||
| Apr 12, 2015 at 17:28 | |||||
| Apr 12, 2015 at 17:23 | history | asked | stupidq75 | CC BY-SA 3.0 |