Timeline for Can you compute the quotient set below?
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| May 5, 2013 at 3:42 | comment | added | Gigel Militaru | The same is valid with the elements that are nonsquare in $K$. If $a$ is nonsquare, then $(a,0)$ is not equivalent to $(0,0)$ or $(0,1)$. The number of such elements is the index of the group $(K^2)^*$ in $(K^*, \cdot)$. That is exactly what you say; and the same thing follows trivialy from the relation of equivalence. | |
| May 5, 2013 at 3:32 | comment | added | Gigel Militaru | My english... What I what to say is the following: dealing with polinomials or with the above relation is exactely the same thing. Is just a matter of taste. The fact that the polionomials $X^2$ and $X (X+1)$ are not in the same orbit is nothing else that a reformulation of the fact that $(0,0)$ and $(0, 1)$ are not equivalent in $K \times K$. In both cases, we arrive at the same algebras: $k<x | x^2 = 0>$ or $k<x | x^2 = 1>$. | |
| May 4, 2013 at 20:18 | comment | added | Mariano Suárez-Álvarez | Your comment makes me think that you misunderstand what I wrote. I did noy say that the two polynomials $X^2$ and $X(X+1)$ are in the same orbit ---I said exactly the opposite. The rest of the comment, I don't understand it. | |
| May 4, 2013 at 19:09 | comment | added | Gigel Militaru | Than you! That is probably the best thing that you can say on the problem :( On the other hand, as a said, is a little bit ... a reformulation of the equivalent relation. The first two polynomial coresspound in fact to the non-equivalent pairs $(0, 0)$ and $(0, 1)$. The third one to $(a, 0)$, where $a$ is not a square in $K$, while the last case (the one that make trouble) is the first part of the equivalent relation up to a normalization. | |
| May 4, 2013 at 19:00 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| May 4, 2013 at 18:40 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| May 4, 2013 at 18:34 | history | answered | Mariano Suárez-Álvarez | CC BY-SA 3.0 |