Timeline for Skew fields inside quaternion division algebras
Current License: CC BY-SA 3.0
17 events
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| Mar 5, 2013 at 10:11 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Mar 5, 2013 at 9:44 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Feb 5, 2013 at 17:43 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Feb 5, 2013 at 16:16 | comment | added | Filippo Alberto Edoardo | Thanks! I am sure with some more paper I’ll be 100% convinced... | |
| Feb 5, 2013 at 15:51 | comment | added | Venkataramana | Sorry: a small correction; does not alter anything. I should have taken the $K$ span of the elts $1,a,b,ab,ba,aba, bab$. The product of any two of these is a linear combination of these elts, with $K$ coefficients. | |
| Feb 5, 2013 at 15:43 | comment | added | Venkataramana | how do we test that it is an algebra? by seeing that the product of two generating elts is a linear combination. for example, take $a$ and $ab$. Their product is $a^2b=(tr (a)ab+det (a)b$ is a linear combination. The coefficients lie in $K$ as was proved. If $D$ were not quaternionic, then over the algebraic closure of its centre, it becomes an $n\times n$ matrix algebra with $n\geq 3$. But then it is not true that the square of $xy-yx$ is scalar; you can easily get two $n\times n$ matrices $x,y$ with the square of $xy-yx$ non-scalar. | |
| Feb 5, 2013 at 15:33 | comment | added | Filippo Alberto Edoardo | @Aakumadula: I am sorry, but I do not see the point (I must be missing something) and got interested in the business... In the first part you say that the condition $(xy-yx)^2$ lies in the center forces $D$ to be quaternionic, right? And at the end you claim that the $K$-vector space spanned by $1,a,b,ab,ba$ is an algebra. I do not understand neither of these claims, and apologize for being slow... ;-) | |
| Feb 5, 2013 at 13:05 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Feb 5, 2013 at 12:42 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Feb 5, 2013 at 11:59 | comment | added | Filippo Alberto Edoardo | I am a bit lost. The OP seemed to ask for very general $D\hookrightarrow Q$ and you assume that $D$ be finite dimensional over its center $K$ (which, as the OP said in his answer to a comment, is his definition of division algebra...) | |
| Feb 5, 2013 at 11:31 | vote | accept | Tom De Medts | ||
| Feb 5, 2013 at 10:26 | comment | added | Venkataramana | @Tom De Medts: you are right. I should have said $Deg (D)\geq 3$. As to the other observation, I think (I have not thought it through), that you can find $x,y \in D$ of trace zero such that they do not commute, and since $(xy-yx)^2$ is central, the algebra generated by the centre $K$ of $D$ and $x,y$ is finite dimensional over $K$ because if you see $x^2$ in a (multiplicative) word in $x,y$ you may replace it a scalar; if you see yxyx in the word, it can be replaced by $xyxy$ and a sum of lower degree terms... . You can then argue that this algebra is 4 dimensional (i.e. quaternionic). | |
| Feb 5, 2013 at 9:15 | comment | added | Tom De Medts | @Aakumadula: Thanks for your answer! The fact that $(xy - yx)^2$ is a central element is a nice observation. Two questions/comments though: (1) I assume you mean $\deg(D) \geq 3$ instead of $\dim(D) \geq 3$ (where the degree is the square root of the dimension)? (2) I am not assuming $D$ to be finite-dimensional over its center, and the Zariski density argument seems to fail in the infinite-dimensional case. | |
| Feb 5, 2013 at 1:22 | comment | added | Venkataramana | the space $V=D\otimes {\overline K}$ is a vector space and hence an affine space. If $K$ is infinite, then $D$ is Zariski dense in the affine space $V$. Let $x,y,z\in V$ be arbitrary and let $w=(xy-yx)^2$. Then the condition $wz=zw$ for all $x,y,z \in V$ is a collection of polynomial equations on $V$; if this does not hold on $V$, then by Zariski density, it does not hold on $D$ either. | |
| Feb 5, 2013 at 0:30 | comment | added | Filippo Alberto Edoardo | What do you mean by "Zariski dense"? Are you speaking of the Zariski density of $D$ or of $\mathrm{Spec}(D)$, which is a bit weird as $D$ is not commutative? | |
| Feb 4, 2013 at 18:10 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Feb 4, 2013 at 17:48 | history | answered | Venkataramana | CC BY-SA 3.0 |