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Let $k$ and $\ell$ be division rings such that $\ell$ contains $k$, and $[\ell : k] = 2$. When do I know that there is an element $a \in k$ such that $x^2 = a$ has solutions in $\ell$, but not in $k$?

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  • $\begingroup$ Since the dimension of a division ring over its centre is a square, we have that $2[k : Z(k)] = [\ell : k][k : Z(k)]$ equals $[\ell : Z(\ell)][Z(\ell) : Z(k)]$, and hence that $[Z(\ell) : Z(k)]$ is $2$ modulo squares. In particular, if $Z(k)$ is not of characteristic $2$, then there should be an element of $Z(k)$ that is a square in $Z(\ell)$ but not in $Z(k)$. But of course that doesn't mean that it isn't a square in $k$. $\endgroup$
    – LSpice
    Nov 16 at 1:10
  • $\begingroup$ Do you have a condition you like in the commutative case? $\endgroup$
    – Kimball
    Nov 16 at 17:51

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