Explicitly lifting the two simple labels in a formal rank-two algebra

Put $p(X):=X^2-S^2$. If a polynomial has degree $m\ge 2$ and leading coefficient $a_m$, subtracting $a_m{X}^{m-2}p(X)$ cancels its degree-$m$ term and lowers its degree; finitely many repetitions give a representative $a+bX$. If $g\mathrm{\ne }0$ has degree $n$ and leading coefficient $b_n\mathrm{\ne }0$, then the leading term of $p(X)g(X)$ is $b_n{X}^{n+2}$ because $p$ is monic, so no nonzero multiple of $p$ has degree at most one. Thus every quotient class has a unique representative $a+bX$, and $(1,x)$ is an $R$-basis. The principal ideal is $R[X]p$ by [F5], and the quotient ring operations are well defined by [F6]. Reducing coefficients modulo $h$ using [F3] gives $\bar{A}=\mathbb{Q}[X]\mathrm{/}(X^2-1)$. The elements ${\bar{e}}_{\pm }:=(1\pm \bar{x})\mathrm{/}2$ satisfy ${\bar{e}}_{\pm }^2={\bar{e}}_{\pm }$, ${\bar{e}}_{+}{\bar{e}}_{-}=0$, and ${\bar{e}}_{+}+{\bar{e}}_{-}=1$ by ${\bar{x}}^2=1$; their coefficient matrix in $(1,\bar{x})$ has determinant $-1\mathrm{/}2\mathrm{\ne }0$ by [F11]. Hence the map $\mathbb{Q}\times \mathbb{Q}\to \bar{A}$, $(a,b)\mapsto a{\bar{e}}_{+}+b{\bar{e}}_{-}$, is a unital algebra isomorphism. The quotient map $R[X]\to A$ is unital, and $A$ is commutative, so the coefficient map $R\to A$ is central; thus $A$ is an $R$-algebra as in [F10].

Define $S^{-1}$ by coefficients $[h^n]S^{-1}=(-1{)}^n$. By [F7], the constant coefficient of $(1+h)S^{-1}$ is $1$, and its coefficient at every $n\ge 1$ is $(-1{)}^n+(-1{)}^{n-1}=0$; hence $(1+h)S^{-1}=1$. Put $u:=S^{-1}x$; the relation $x^2=S^2$ gives $u^2=1$. Direct expansion now gives $e_{\pm }:=(1\pm u)\mathrm{/}2$ with $e_{\pm }^2=e_{\pm }$, $e_{+}{e}_{-}=e_{-}{e}_{+}=0$, and $e_{+}+e_{-}=1$. Their reductions are ${\bar{e}}_{\pm }$.

The identities $e_{+}+e_{-}=1$ and $e_{+}-e_{-}=S^{-1}x$ give $1=e_{+}+e_{-}$ and $x=S(e_{+}-e_{-})$. The definitions express each $e_{\pm }$ as an $R$-linear combination of $1,x$, while these identities express $1,x$ as combinations of $e_{+},e_{-}$. Since $2$ is a unit by [F11] and $S$ is a unit by [F7], the changes of coordinates are inverse, so $(e_{+},e_{-})$ is an $R$-basis by step 1.1 and [F9]. The map $R\times R\to A$, $(a,b)\mapsto a{e}_{+}+b{e}_{-}$, is multiplicative because the idempotents are orthogonal and idempotent, is unital because they sum to $1$, and is bijective because they form a basis.

Reducing the isomorphism from step 3.1 gives $\bar{A}\cong \mathbb{Q}\times \mathbb{Q}$; extending scalars to $K$ gives $K{\otimes }_{R}A\cong K\times K$ by [F3]. For a product of two fields, the central coordinate idempotents split any module into their two images; a simple nonzero module has exactly one nonzero image, on which that coordinate field acts. For any field $L$, if $M$ is a simple nonzero left $L$-module and $0\mathrm{\ne }m\in M$, then $L\to M$, $a\mapsto am$, is injective because nonzero $a$ is invertible by [F11], and its nonzero image is all of $M$ by [F8]. Thus each coordinate field has exactly one simple module up to isomorphism, namely itself. The labels $+$ and $-$ are the reductions of $e_{+}$ and $e_{-}$, in agreement with the split-label theorem [F1] and the matrix decomposition [F2].