# Field of definition of a finite dimensional division algebra and how to reduce it

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.

• 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 $K[x]$. – 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.)