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Jul 31 at 3:53 comment added Skip Thanks! You are right about the compactness. You choose a metric on $\mathbb{R}^n$ and consider the unit sphere $S$ in that metric. For each division algebra $d \in \mathcal{D}$ and $s \in S$ you can talk about $\det L_s$ where $L_s$ is left multiplication by $s$ in $d$. The image of ${d} \times S$ is compact in $\mathbb{R} \setminus \{ 0 \}$, so it is closed and bounded away from 0. Because $\det L$ is continuous in $d$, there is an open nbhd of $d$ consisting of other algebras $d'$ such that ${d'} \times S$ has image also missing 0. Each of those algebras $d'$ is a division algebra.
Jul 29 at 18:19 vote accept Akiva Weinberger
Jul 29 at 18:19 comment added Akiva Weinberger What a beautiful argument! I'll admit that I fully forgot I asked this question, but thinking of an entire algebra as being just a single tensor is genius. It seems to me that the fact that $\cal D$ is open in $\cal A$ is the key idea in this argument. My intuition tells me that it should follow from some compactness shenanigans.
Jul 29 at 11:36 history edited Skip CC BY-SA 4.0
typo correction: added inequality for dimension of D/P
Jul 29 at 0:49 history answered Skip CC BY-SA 4.0