if E is a splitting field for a division algebra D, is it always true that E can be embedded in D? Jacobson (BA II, Th. 4.8) only states that E can be embedded into Mat(r,D) for some r.


1 Answer 1


It depends on what you mean by "splitting field". What is true is that for a division algebra $D$ (over $F$) of dimension $n^2$, if $E$ is a field extension of degree $n$ over $F$ that splits $D$, then $E$ is isomorphic to a subfield of $D$. See the Theorem on page 16 of Paul Garrett's notes.

Obviously, an algebraic closure of $F$ will split $D$, but won't be isomorphic to a subfield for dimension-counting reasons.

  • 2
    $\begingroup$ Or flip ahead several pages to read Theorem 4.12 in BA II... $\endgroup$
    – grp
    Oct 5, 2012 at 3:41
  • $\begingroup$ Are all such fields equivalent? All I can tell from the references, they have the same [E:F] = deg(D). $\endgroup$
    – George
    Oct 5, 2012 at 10:55
  • $\begingroup$ I asked this as a separate question: mathoverflow.net/questions/108907/… $\endgroup$
    – George
    Oct 5, 2012 at 12:12
  • $\begingroup$ George, just so you know, it is ok to ask for clarification to an answer in the comments; you don't necessarily need to ask it as a separate question. A quaternion algebra over $\mathbb Q$ is split by any quadratic extension of $\mathbb Q$ that does not split at the places the quaternion algebra is ramified. E.g., if a quaternion algebra is ramified at $\infty$ and $2$, then any imaginary quadratic extension of $\mathbb Q$ such that $2$ does not split in the extension will split the quaternion algebra. $\endgroup$
    – B R
    Oct 5, 2012 at 12:37

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