Timeline for Can there be a non-trivial epimorphism (of rings) from a field?
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9 events
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| Apr 12, 2015 at 20:31 | comment | added | Todd Trimble | @tj_ I expect what you mean is that $Tor$ commutes with filtered or directed colimits, not to be confused with direct limits = colimits. But your argument still goes through. And perhaps more to the point, I take it that you meant to offer a choice-free argument (I already knew modules over fields are flat assuming choice), and the argument seems valid in that respect as well. Thanks! | |
| Apr 12, 2015 at 20:27 | comment | added | tj_ | @Todd Trimble: Over a field each module is flat. For, a module $A$ is flat, iff $Tor_1(A,-)=0$. But $A$ is the direct limit of finitely generated modules $A_i$ and $Tor$ commutes with direct limits. But f.g. modules over a field a free and thus flat. Hence $Tor_1(A_i,-)=0$ showing $Tor_1(A,-)=0$. | |
| Apr 12, 2015 at 20:13 | comment | added | Todd Trimble | @EmilJeřábek Yep! | |
| Apr 12, 2015 at 20:09 | comment | added | Emil Jeřábek | I convinced myself that it should work without choice. One can construct $A\otimes_KA$ as the quotient of the space with basis $\{u\otimes v:u,v\in A\}$ over a subspace generated by $uk\otimes v-u\otimes kv$ and friends. If this subspace contained $i_1(a)-i_2(a)$, this would be witnessed by a finite linear combination involving only finitely many elements of $A$, hence it would already show up for a finite-dimensional subspace of $A$, where we can do the algebra as usual. | |
| Apr 12, 2015 at 19:33 | comment | added | Todd Trimble | @Emil Yeah, I was just wondering that myself. Need that all modules over a field are flat for that part. I'm not sure. | |
| Apr 12, 2015 at 19:28 | comment | added | Emil Jeřábek | Is it clear that without a basis of $A$ that the map $V\otimes_KV\to A\otimes_KA$ is injective, which you are implicitly using in the argument? | |
| Apr 12, 2015 at 19:23 | comment | added | Todd Trimble | @FanZheng Presumably not; I'm using just basic notions of independence etc. and cutting down to a finite-dimensional subspace $V$ to finish the argument. I didn't need a basis for all of $A$. Can you say more explicitly what you have in mind? | |
| Apr 12, 2015 at 19:20 | comment | added | Fan Zheng | So it still requires axiom of choice? | |
| Apr 12, 2015 at 19:13 | history | answered | Todd Trimble | CC BY-SA 3.0 |