I'm searching for a proof of Witt's result that a biquadratic extension $K(\sqrt{a},\sqrt{b})/K$ extends to a Galois extension $L/K$ with quaternion group $Q_8$ iff the quadratic forms $<a,b,\frac{1}{ab}>, <1,1,1>$ are equivalent iff $(a,b)(a,a)(b,b) = 0 \in Br(K)$.

I know there is a proof in his original paper "Konstruktion von galoisschen Körpern der Charakteristik p zu vorgegebener Gruppe der Ordnung pf". However, I am searching for a proof in English. I would also like to find proofs from a more modern perspective as well as generalizations of the result.

Thank you.


A complete proof can be found in the first few pages of


Jensen, Christian U.(DK-CPNH); Yui, Noriko(3-TRNT) Quaternion extensions. Algebraic geometry and commutative algebra, Vol. I, 155–182, Kinokuniya, Tokyo, 1988.

I will scan the relevant pages and post them here as soon as electricity is restored in my office.

Addendum: Here is the paper :



The problem you raise is tied to the "embedding problem." There is chapter devoted to this in Malle and Matzat's "Inverse Galois Theory" text (here's a link to the publisher's page about this book).

There's a summary of the embedding problem when the kernel is $\mathbb{Z}/2\mathbb{Z}$ in the paper by Klueners and Malle here (see subsection 3.4). Their Proposition 7 is a "Hasse principle" for embedding problems with kernel $\mathbb{Z}/2\mathbb{Z}$, and they give the example that

"If $L/\mathbb{Q}$ is a Galois extension with Galois group $(\mathbb{Z}/2\mathbb{Z})^{2}$, then $L$ is embeddable into a $Q_{8}$ extension if and only if $L$ is totally real and all odd primes $p$ that are ramified in $L$ have the property that $p \equiv 1 \pmod{4} \iff p$ has odd inertia degree in $L$."

It's not obvious to me that this is equivalent to Witt's statement in the case that $K = \mathbb{Q}$, but I can imagine it should be possible to sort out a clear connection in the context of cohomology (since the Brauer group of $K$ is $H^{2}(K,G_{m})$, and group extensions are also classified by elements of $H^{2}$).

  • 2
    $\begingroup$ You can also see in the book "An introduction to Galois cohomology and its applications", G.Berhuy, LMS Lecture notes 377, $\S$ VII.19. $\endgroup$ – GreginGre Nov 23 '17 at 8:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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