# Charaterisation of quaternion algebras

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.)

• 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 $\le 2$.
– YCor
Mar 1, 2021 at 7:59
• @YCor: That is not a division algebra. Mar 1, 2021 at 8:02
• Ah sorry (I had reread carefully all your post to double check... except, apparently, at the right place!).
– YCor
Mar 1, 2021 at 8:13
• 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.
– YCor
Mar 1, 2021 at 8:15
• 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. Mar 1, 2021 at 8:27