Timeline for Can you compute the quotient set below?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| May 5, 2013 at 13:10 | answer | added | Gigel Militaru | timeline score: 0 | |
| May 4, 2013 at 18:37 | comment | added | Mariano Suárez-Álvarez | I did not say that you would arrive at a different problem: but the version with polynomials is expressed in terms of familiar things —polynomials and linear changes of variables— instead of weird ordered pairs under an unmotivated equivalence relation! | |
| May 4, 2013 at 18:34 | answer | added | Mariano Suárez-Álvarez | timeline score: 1 | |
| May 4, 2013 at 18:33 | comment | added | Gigel Militaru | No! I can not do that :) You will arrive at the same problem. :) Btw the bijection between $K \times K/ \equiv$ and the set of isomorphism classes of all $2$-dimensional algebras over a field of characteristic $2$ is clear for anybody: $\overline{(a, b)} \mapsto k_{(a, b)}$, where $k_{(a, b)}$ is the algebra having $\{1, x\}$ as a basis and the multiplication is given by $x^2 = a + b x$. Now, I understand why everyone avoided the characteristic $2$ case in their classification results. | |
| May 4, 2013 at 18:08 | comment | added | Mariano Suárez-Álvarez | An associative two-dimensional unitary algebra is commutative, and it is therefore the quotient of $K[X]$ by the ideal generated by a monic polynomial of degree $2$. So you want to classify polynomials of this type under the automorphism group of $K[X]$. Maybe stating it this way would make your problem more enticing :-) | |
| May 4, 2013 at 16:27 | answer | added | Dietrich Burde | timeline score: 1 | |
| May 4, 2013 at 8:33 | comment | added | Gigel Militaru | I should say a little bit more related to the question. The quotient set $K\times K/≡$ parametrize the isomorphism class of all $2$-dimensional (unitary, associative) algebras over $K$. What is strage here is the fact that I am not able to give an explicit description of a set of representatives; since I am stupid :). If we have such a system of representative (a discution about K is mandatory -- there are two main cases: $K^2 = K$ or not) the we can list the types of isomorhism of all $2$-dimensional algebras over $K$. | |
| May 3, 2013 at 21:16 | history | edited | François G. Dorais | CC BY-SA 3.0 |
typo & tags
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| May 3, 2013 at 18:21 | history | asked | Gigel Militaru | CC BY-SA 3.0 |