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I would like to know if there are some central-simple algebras $D_1$, $D_2$ and $D_3$ over a field $k$ satisfying the following properties :

  • $ind(D_1)=exp(D_1)=4$ ($ind$ is the Schur index and $exp$ the exponent);
  • $D_2$ and $D_3$ are two non-isomorphic quaternion algebras;
  • $ind(D_1^{\otimes2} \otimes D_2)=ind(D_1^{\otimes 2} \otimes D_3)=4$ (recall that $ind(D_1^{\otimes 2})$ is always $2$ in this setting).

To give a little motivation, i'm interested in the existence of "square roots" (in the Brauer group of a field) of quaternion algebras.

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  • $\begingroup$ There aren't any conditions on $D_2$, and $D_4$ seems to appear out of nowhere. $\endgroup$
    – S. Carnahan ♦
    Commented Aug 11, 2011 at 8:41
  • $\begingroup$ oh my god sorry, i meant D_2 but wrote D_4... I corrected the typo, thank you. $\endgroup$
    – Louis
    Commented Aug 11, 2011 at 13:50
  • $\begingroup$ Could you explain what you mean by $D_1^2D_2$? Are you taking the tensor product of $D_1$, $D_1$, and $D_2$? $\endgroup$
    – André Henriques
    Commented Aug 11, 2011 at 13:59
  • $\begingroup$ exactly, i also correct this. Thank you. $\endgroup$
    – Louis
    Commented Aug 11, 2011 at 14:01
  • $\begingroup$ If $D$ is a quaternion algebra, then its local Hasse invariant is $1/2$ at a finite even number of places and zero elsewhere, so you can get a square root of the Brauer class by changing the nonzero invts to $1/4$ at half the places and $3/4$ at the other half. $\endgroup$
    – David Hansen
    Commented Aug 11, 2011 at 22:45

1 Answer 1

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You can find such division algebras over $\mathbb{Q}(x_1,x_2,x_3)$ or $k(x_1,x_2,x_3,x_4)$ where $k$ is any field using the results of "Nakayama, T. Über die direkte Zerlegung einer Divisionsalgebra. Japanese J. of Mathematics 12 (1935), 65–70". A simplified proof is given in the book "Associative algebras" by Pierce, Corollary c, p. 381.

In the notation of Pierce, let $r=3$, $n_1 = 4$, $n_2 = 2$, $n_3=2$ and set $D_i$ to be the cyclic algebras denoted as $B_i$ by Pierce. The only thing one needs to observe is that $B_1^{\otimes 2}$ is equivalent to the cyclic algebra associated to the quadratic subfield of $E_1$ (and the element $x_1$) so we may apply the corollary to $D_1^{\otimes 2} \otimes D_2$ and $D_1^{\otimes 2} \otimes D_3$.

(Sorry for not giving more details but the theorem is long to write out in full. If anything is not clear please leave a comment.)

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  • $\begingroup$ Thanks a lots for the hint and the great reference. The only missing thing for me is the observation that B_1^2 is equivalent to the cyclic algebra (K,<s>,x_1), where K/F is a quadratic extension inside E_1. $\endgroup$
    – Louis
    Commented Aug 12, 2011 at 14:26
  • $\begingroup$ This follows, for example, from Corollary a) and b) on p. 277 of Pierce. $\endgroup$
    – naf
    Commented Aug 12, 2011 at 15:59
  • $\begingroup$ Well, it's perfect. Thank you so much. $\endgroup$
    – Louis
    Commented Aug 12, 2011 at 16:34

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