The rational simple quaternion block

The quaternion multiplication gives $z$ central, $u^2=v^2=z$, and $vu=zuv$. Thus $e^2=(1-2z+z^2)\mathrm{/}4=e$ and $ze=-e$. The four elements $e,ue,ve,uve$ span $D$, since $(zh)e=-he$ for $h=1,u,v,uv$. They are rationally independent: $he=(h-zh)\mathrm{/}2$ has its two nonzero coefficients on one of four disjoint pairs of the eight group-basis elements. Hence they form a basis. Write $x=ue$, $y=ve$, $w=uve$; then $x^2=y^2=w^2=-e$, $xy=w=-yx$, $yw=x=-wy$, and $wx=y=-xw$.

For $q=ae+bx+cy+dw$, put $\bar{q}=ae-bx-cy-dw$. Using the multiplication table, all mixed terms cancel and $q\bar{q}=\bar{q}q=(a^2+b^2+c^2+d^2)e$. For rational coefficients this sum is positive when $q\mathrm{\ne }0$. Thus $q^{-1}=\bar{q}\mathrm{/}(a^2+b^2+c^2+d^2)$ exists on both sides, and $D$ is a division algebra. Every nonzero left ideal contains an invertible element and hence $e$, so its left regular module is simple. F2 for a single $1\times 1$ factor says this is its unique simple class. The zero quaternion needs no inverse; the unit is $e$.

Set $U=\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)$ and $V=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. We have $U^2=V^2=-I$ and $VU=-UV$, so the multiplication table defines an $E$-algebra map $E{\otimes }_{\mathbb{Q}}D\to M_2(E)$ sending $e,x,y,w$ to $I,U,V,UV$. A linear combination of these images is $\left(\begin{array}{cc}a+bi& c+di\\ -c+di& a-bi\end{array}\right)$. Every matrix occurs uniquely: its entries $r,s,t,h$ give $a=(r+h)\mathrm{/}2$, $b=(r-h)\mathrm{/}(2i)$, $c=(s-t)\mathrm{/}2$, $d=(s+t)\mathrm{/}(2i)$. Thus the map is an algebra isomorphism.

Let $f=1-e=(1+z)\mathrm{/}2$. The complementary block has basis $f,uf,vf,uvf$ by the same disjoint-pair argument. Here $zf=f$, so $uf,vf$ commute and square to $f$. Evaluating them independently at $ϵ,\eta \in \{1,-1\}$ gives a map $\mathbb{Q}[Q_8]f\to {\mathbb{Q}}^4$. Its sign matrix has rows $(1,ϵ,\eta ,ϵ\eta )$; the inner product of a row with itself is $4$, and with a different row is $(1+ϵ{ϵ}^{\mathrm{\prime }})(1+\eta {\eta }^{\mathrm{\prime }})=0$. Its inverse is one quarter its transpose. Hence $\mathbb{Q}[Q_8]\cong D\times {\mathbb{Q}}^4$ and, extending the displayed basis maps, $E[Q_8]\cong M_2(E)\times E^4$. The rational regular module is the direct sum of $D$ and four copies of $\mathbb{Q}$, each simple over its factor; the extended regular module is the sum of the two simple columns of $M_2(E)$ and four one-dimensional factors. Thus both regular modules are semisimple, and the displayed extended product is split, without an omitted block. The column simples over this product have only scalar endomorphisms (commute with matrix units), so it also satisfies the group splitting-field convention.

The subspaces of matrices supported in the first column and in the second column are left ideals, each isomorphic to $W=E^2$ by reading that column. They have zero intersection and their sum is all of $M_2(E)$. F2 makes $W$ simple. Consequently $E{\otimes }_{\mathbb{Q}}D\cong W\oplus W$ as group modules, with multiplicity exactly two: its $E$-dimension is four and ${\dim }_{E}W=2$.

The polynomial $X^2+1$ has no rational root, and its distinct roots $i,-i$ lie in $E$. Thus $E\mathrm{/}\mathbb{Q}$ is finite normal separable, with nontrivial automorphism $\sigma (i)=-i$. Coefficient conjugation sends $U$ to $-U$ and fixes $V$. Since $VU{V}^{-1}=-U$ and $VV{V}^{-1}=V$, conjugation by $V$ intertwines the representation with its coefficient-conjugate; these equations on the generators suffice on every group element. Therefore $[W]$ is Galois-stable. The hypotheses for F1 are all met by step 3.1 and this finite Galois extension. F1 identifies $D$ with this singleton orbit, and step 3.2 computes its common multiplicity as two.

The matrices for $1,z,u,v,uv$ are $I,-I,U,V,UV$. Their traces are $2,-2,0,0,0$ respectively; multiplying the last three by $-I$ keeps their traces zero. The class list follows directly from the relations: conjugation preserves each pair $\{h,zh\}$ for $h=u,v,uv$, and conjugation by a different generator exchanges its two members, whereas $1,z$ are central. Thus this list exhausts all eight elements and gives exactly the stated class values. All values lie in $\mathbb{Q}$. The trace of the rational regular module $D$ is twice this character after extension, since its extension is the displayed two column copies. [F3, F4, F5, step 2.2, step 3.2, algebra] QED