Is it true that any finite dimensional division algebra over a pseudo-algebraically closed field is trivial? We know that this is true for algebraically closed field.
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3$\begingroup$ Presumably the answer below has to do with the following issue. Some mathematicians use "division algebra over $k$" to mean a finite dimensional $k$-algebra that is a division algebra and whose center equals $k$. That is presumably also what you intend. However, other mathematicians make no such hypothesis about the center of the division algebra. In that case, also field extensions of $k$ are "division algebras over $k$". The term "central simple algebras" clarifies this, because "central" is part of the name. $\endgroup$– Jason StarrCommented Aug 24, 2017 at 13:03
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$\begingroup$ @JasonStarr Yeah, if I remember correctly, EGA or SGA makes this distinction between R-rings, R-algebras, and R-extensions, the first being noncentral ring maps, the second being central ring maps, and the third being central maps into commutative rings in the section about something involving derived functors, possibly the degree 1 Quillen-Andre ones. $\endgroup$– Harry GindiCommented Aug 24, 2017 at 16:39
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$\begingroup$ Oops ran out of time to edit it, but yeah it's in the section on ExAlComm $\endgroup$– Harry GindiCommented Aug 24, 2017 at 16:45
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$\begingroup$ $\operatorname{EGA}_0 \S 18$ got it for ya $\endgroup$– Harry GindiCommented Aug 24, 2017 at 17:11
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2 Answers
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The fact that every finite dimensional division algebra is trivial is not only true over an algebraically closed field but it is in fact equivalent to the field being algebraically closed. Remember that (extension-)fields are just a special case of division algebras. Thus, if it would be true over pseudo-algebraically closed fields, we would have that a field is algebraically closed if and only if it is pseudo-algebraically closed.
However, this is not the case, as for example finite fields are pseudo-algebraically closed.
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$\begingroup$ It seems from the proof of Proposition $6.2.3$, page $171$ of "Central simple algebras and Galois cohomology" by Gille and Szamuely, that this is true for $C_1$-fields. But I am not an expert. $\endgroup$ Commented Aug 24, 2017 at 12:53
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3$\begingroup$ @user43198: yes, if you consider division algebras whose center is the given field, then these things are classified by the Brauer group, and the Brauer group is trivial for PAC fields. $\endgroup$ Commented Aug 24, 2017 at 12:55
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$\begingroup$ No. As mentioned in the answer, finite fields are $C_1$, and every extension field of them is such an algebra with itself (so not the base field) as center. $\endgroup$– DirkCommented Aug 24, 2017 at 13:10
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The fact that the Brauer group of a pseudo-algebraically closed field is trivial is noted in: J.Ax, The elementary theory of finite fields, Annals Math. 88 (1968), p. 269. A more detailed proof is given in: M. Fried and M. Jarden, Field Arithmetic (3rd edition), Springer, p. 209.
