Suppose $A$ is a $F$-central simple algebra with maximal subfield $E$ such that $[A:F] = 4$. if $N_{E/F}(E^*) \ne F^*$, then $A$ is a division algebra.

Is this even true? If it is true how i can prove it? I know we have $A \simeq M_2(F)$ or $A$ is a division ring so we must show that $A$ isn't isomorphic with $M_2(F)$.

  • $\begingroup$ No, it's not true. All quadratic extensions $E/F$ are contained in $M_2(F)$. $\endgroup$ – Kimball Jul 23 '18 at 0:42
  • $\begingroup$ For example consider $\mathbb{C}/\mathbb{R}$ and set: $$A = \left\lbrace \begin{pmatrix} x & y\\\overline{y} & \overline{x} \end{pmatrix} :x,y\in \mathbb{C}\right\rbrace \subseteq M_2(\mathbb{C}) $$. so A is a $\mathbb{R}$-central simple algebra and isn't division ring. is this counter-example correct? $\endgroup$ – s.Bahari Jul 23 '18 at 16:18
  • 1
    $\begingroup$ Yes, your $A$ is isomorphic to $M_2(\mathbb R)$. $\endgroup$ – Kimball Jul 24 '18 at 1:33
  • $\begingroup$ Thanks for answer. $\endgroup$ – s.Bahari Jul 24 '18 at 15:09

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