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Let $k$ be a field, and $A$ an associative $k$-algebra with an identity element. Say that $A$ is quadratic if any subalgebra of $A$ generated by a single element has dimension at most two.

I am looking for a reference for the following (or any similar) result:

Assume that ${\rm char}(k)\ne2$, and let $A$ be a quadratic division algebra over $k$ with $\dim_k(A)\ge3$. Then $A$ is a quaternion algebra.

(I expect this to be well known, but am not aware of any reference.)

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  • $\begingroup$ A counterexample is the (commutative) algebra $A=k[x_i:1\le i\le n]/(x_ix_j:1\le i,j\le n)$ for $n\ge 2$. Every element has the form $t1_A+w$ with $t\in K$ and $w^2=0$, so generates a (unital) subalgebra of dimension $\le 2$. $\endgroup$
    – YCor
    Mar 1, 2021 at 7:59
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    $\begingroup$ @YCor: That is not a division algebra. $\endgroup$
    – Erik D
    Mar 1, 2021 at 8:02
  • $\begingroup$ Ah sorry (I had reread carefully all your post to double check... except, apparently, at the right place!). $\endgroup$
    – YCor
    Mar 1, 2021 at 8:13
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    $\begingroup$ At least in finite dimension it follows from the fact that a division algebra has square dimension $n^2$ over its center, over which it has an $n$-dimensional commutative subalgebra. $\endgroup$
    – YCor
    Mar 1, 2021 at 8:15
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    $\begingroup$ The point being that a commutative subalgebra can have dimension at most two in this situation? Indeed, that is a good argument. However, I would like to know if this or any similar result is available in the literature, for easy reference. $\endgroup$
    – Erik D
    Mar 1, 2021 at 8:27

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Yes, this is a great and well-studied question with a super nice answer! See Theorem 3.5.1 of my book (http://quatalg.org). You can also say something in characteristic 2 (see Theorem 6.2.8) if you refine "quadratic" = "degree 2" to "nonidentity standard involution".

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  • $\begingroup$ Thanks, this is what I was looking for. It seems that your proof follows the same lines as the one by Palais for the real case (Palais: The classification of real division algebras, Amer. Math. Monthly 75 (1968), 366–368.) $\endgroup$
    – Erik D
    Mar 2, 2021 at 4:35
  • $\begingroup$ Yes, the argument is not due to me, and I found echoes of it in many places! $\endgroup$ Mar 3, 2021 at 3:03

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