Timeline for Can transcedence degree be defined for arbitrary ring homomorphism?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 15, 2019 at 12:51 | history | edited | darij grinberg | CC BY-SA 4.0 |
clearer start of proof
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| Jan 15, 2019 at 12:43 | history | edited | darij grinberg | CC BY-SA 4.0 |
add corollary 3 for citing
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| Apr 3, 2015 at 4:00 | history | edited | darij grinberg | CC BY-SA 3.0 |
I had my commutative triangles wrong
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| Apr 2, 2015 at 17:23 | history | edited | darij grinberg | CC BY-SA 3.0 |
now with even more generality
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| Apr 2, 2015 at 16:49 | history | edited | darij grinberg | CC BY-SA 3.0 |
added 2848 characters in body
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| Oct 12, 2013 at 16:05 | comment | added | Sasha Anan'in | Sure, when checking, I had "constructive" in mind, "the axiom of choice" is just a bad wording. | |
| Oct 12, 2013 at 15:58 | comment | added | darij grinberg | ... combination of the given generators of $I$.) | |
| Oct 12, 2013 at 15:58 | comment | added | darij grinberg | Yes, I started writing it as a sketch but then I ended up making so many mistakes that I could only convince myself of the proof being correct by extending it to this level of detail. Thanks for checking it! By "constructive" I mean not only avoiding the axiom of choice but also avoiding the tertium non datur, hence being entirely algorithmic. (Of course, the algorithmic value of Theorem 1 only becomes noticeable when, for instance, $A$ is constructed as a quotient ring of a ring $A'$ by some ideal $I$, in which case the triviality of $A$ gives an algorithm to write $1_{A'}$ as a linear ... | |
| Oct 12, 2013 at 6:12 | comment | added | Sasha Anan'in | Nice! I checked all the details (including those in the reference) and found that you successfully avoided the use of the axiom of choice. (What you wrote is not really a sketch. You might be much more sketchy and save time for preparing your talk.) Yes, most of the mathematics is constructive. (A sort of a counter-example is given by the answer $(\sqrt2^{\sqrt2})^{\sqrt2}=2$ to the question whether is it possible that $a^b$ is rational with $a,b$ irrational.) I have no option to accept your answer, so can only upvote. Have a nice talk! | |
| Oct 12, 2013 at 4:04 | history | edited | darij grinberg | CC BY-SA 3.0 |
deleted 1 characters in body
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| Oct 12, 2013 at 1:59 | history | edited | darij grinberg | CC BY-SA 3.0 |
added 1027 characters in body
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| Oct 12, 2013 at 1:46 | history | answered | darij grinberg | CC BY-SA 3.0 |