Let F be a field, and E/F an infinite algebraic extension. Let D be a finite dimensional division algebra over E (meaning its center is also E).

Is it possible to somehow gow down to a finite dimensional division algebra over a finite algebraic extension of F? More precisely, does there exist a finite dimensional division algebra D' over K, where K/F is finite algebraic and $K \subseteq E$, such that $D' \otimes_K E= D$? What about the relationship between the degrees of $D$ and $D'$?

The naive approach I have tried: Let $m_1, \dots, m_n$ be a basis of D over E. Let K be a finite field extension of F containing all the elements of E needed to write the products and inverses of $m_1, m_2, \dots, m_n.$ Take the finite dimensional K-algebra generated by the $m_i.$ This is where I get stuck, because I don't see why this would be a division algebra.

Any ideas or references would be very helpful.

  • 2
    $\begingroup$ Let $D'$ be your finite-dimensional $K$-algebra, and suppose that $x \in D'$. Then $x$ satisfies its characteristic polynomial (i.e., the characteristic polynomial of the multiplication-by-$x$ map), which lies in $K[T]$. If the characteristic polynomial has $0$ constant coefficient, then $x$ is a $0$-divisor in $D$, hence $x = 0$. Otherwise, you can use the characteristic equation to find the inverse in $K[x]$. $\endgroup$ – LSpice May 23 '19 at 22:59

If $L/K$ is a field extension and $D/K$ is an algebra such that $D \otimes_{K} L = D_L$ is a division algebra, then $D$ is a division algebra. Note that the map

$$d \otimes 1: D_L \rightarrow D_L$$

is an isomorphism and that $L/K$ is faithfully flat. So $d: D \rightarrow D$ is also an isomorphism and $D$ is a division algebra. So your naive construction works perfectly. (In fact, you only even need to take $K$ so that the products of the $m_i$ are defined and then inverses will come for free.)

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    $\begingroup$ Great, thanks a lot! I assume d is an element of D and your maps are just multiplication by d? $\endgroup$ – Martin May 24 '19 at 8:24

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