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Give an example of a non-commutative algebra $D$ such that $\dim_{\mathbb{Q}} (D) = |\mathbb{R}|$.

I'm trying to figure out what this creature could be. Note that $\mathbb{R} \cong 2^{|\mathbb{Q}|}$. So I want to construct something such that:

the set of basis of the desired $D$ is the power set of $\mathbb{Q}$.

This amounts to taking the basis to be all the real numbers. Then this $D$ could in fact be a finite dimensional non-commutative algebra over $\mathbb{R}$, which by Frobenius theorem, is just $\mathbb{R}$, $\mathbb{C}$ and $\mathbb{H}$ the quaternions.

Is this the correct thought?

nekodesu
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2 Answers2

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Pick your favourite non-commutative algebra $A$ over $\mathbb{Q}$ (with $\operatorname{dim}_{\mathbb{Q}}(A)\leq \vert \mathbb{R}\vert$), then

$$ A \oplus \bigoplus_{r\in \mathbb{R}} \mathbb{Q} $$

is a non-commutative algebra over $\mathbb{Q}$ with the desired dimension.

Added: Of course you can also just take $\mathbb{H}$. It is a non-commutative $\mathbb{Q}$-algebra and it has the right dimension as

$$ \operatorname{dim}_{\mathbb{Q}} (\mathbb{H}) = \dim_{\mathbb{Q}} (\mathbb{R}) \cdot \operatorname{dim}_{\mathbb{R}} (\mathbb{H}) = \vert \mathbb{R} \vert. $$

Where we used that $\dim_{\mathbb{Q}} (\mathbb{R}) = \vert \mathbb{R} \vert$ (see https://mathoverflow.net/questions/23202/explicit-big-linearly-independent-sets/23206#23206).

Severin Schraven
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Let $G$ be any non-commutative group of order $|\Bbb R|$. Then you can let $D$ be the group ring $\Bbb Q [G]$.

Or just take the direct product of sufficiently many rational quaternion algebras, $\Bbb Q[i,j,k]^{\Bbb Q}$.

Hagen von Eitzen
  • 374,180