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.

  • $\begingroup$ There aren't any conditions on $D_2$, and $D_4$ seems to appear out of nowhere. $\endgroup$
    – S. Carnahan
    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
    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 $D_2$? $\endgroup$ Aug 11, 2011 at 13:59
  • $\begingroup$ exactly, i also correct this. Thank you. $\endgroup$
    – Louis
    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. $\endgroup$ Aug 11, 2011 at 22:45

1 Answer 1


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.)

  • $\begingroup$ 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
    Aug 12, 2011 at 14:26
  • $\begingroup$ This follows, for example, from Corollary a) and b) on p. 277 of Pierce. $\endgroup$
    – naf
    Aug 12, 2011 at 15:59
  • $\begingroup$ Well, it's perfect. Thank you so much. $\endgroup$
    – Louis
    Aug 12, 2011 at 16:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.