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I'm trying to prove that $\mathbb{H_{\mathbb{F}_2}}$, the ring of quaternions over the finite field $\mathbb{F}_2$, is isomorphic to the group ring $\mathbb{F}_2[V_4]$, where $V_4$ is the Klein-four group.

As $\mathbb{H_{\mathbb{F}_2}} \cong M_2(\mathbb{F}_2)$, it's enough to establish that $M_2(\mathbb{F}_2) \cong \mathbb{F}_2[V_4]$. It's clear to me that these rings have the same size -- $2^4$ elements each. Furthermore, it's apparent that the $\mathbb{F_2}$-coefficients of the partial sums in $\mathbb{F}_2[V_4]$ correspond to $2$ x $2$ matrix entries: We can map the upper-left entry of any matrix in the ring to the coefficient of $a$, the lower-left one to that of $ab$, etc.

However, I'm not totally sure how to more formally define this mapping and show that it respects the operations of the group ring. Some guidance would be appreciated.

cloudchamber
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1 Answers1

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Remember that $V_4$ is abelian. Then it follows that $\Bbb F_2[V_4]$ is a commutative ring. Consider the following: $$ \begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}1&0\\1&1\end{pmatrix} = \begin{pmatrix}0&1\\1&1\end{pmatrix}\neq \begin{pmatrix}1&1\\1&0\end{pmatrix} = \begin{pmatrix}1&0\\1&1\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix}. $$ Thus, $M_2(\Bbb F_2)$ is a noncommutative ring. Hence, $\Bbb F_2[V_4]$ cannot be isomorphic to $M_2(\Bbb F_2)$.

Stahl
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  • Oh, alright: As 2 isn't an odd prime, I presumably can't assume that $\mathbb{H}{\mathbb{F}{2}}$ is isomorphic to $M_2(\mathbb{F}_2)$ in the first place. Then the strategy is to construct an isomorphism directly from the quaternions into the group ring, defined on the basis of the quaternions and mapping into the group elements of $\mathbb{F_2}(V_4)$ taken as a basis (treating the group ring as an algebra/vector space)? – cloudchamber Nov 17 '16 at 18:47
  • What's your definition of quaternions over $\Bbb F_2$? Normally a quaternion algebra over $K$ is a central simple algebra over $K$ of dimension $4$. While $\Bbb F_2[V_4]$ is an algebra over $\Bbb F_2$ of dimension $4$, it is not a central simple algebra, because the center of $\Bbb F_2[V_4]$ is $\Bbb F_2[V_4]$ itself (again because $\Bbb F_2[V_4]$ is commutative). – Stahl Nov 17 '16 at 18:53