A nonsplit simple has Hom-pairing diagonal two

By [F1], $A=\mathbb{C}$ is a field containing the embedded copy of $\mathbb{R}$, and every element of $A$ is uniquely $a+bi$ with $a,b\in \mathbb{R}$. By [F3] every nonzero element of the field $A$ is a unit with two-sided inverse, so $A$ is a division ring; by [F4], with the embedding of $\mathbb{R}$ in the commutative field $A$, it is a unital associative $\mathbb{R}$-algebra whose scalar action is restriction of complex multiplication. By [F2], $1,i$ is an $\mathbb{R}$-basis, so ${\dim }_{\mathbb{R}}A=2$.

Let $N\le S=A$ be a submodule of the regular module [F6] and suppose $0\mathrm{\ne }x\in N$. For $y\in A$ the element $y{x}^{-1}$ lies in $A$, so closure under scalars [F6] gives $y=(y{x}^{-1})x\in N$. Hence $N=A$, and the only submodules of $S$ are $0$ and $S$. Since $S\mathrm{\ne }0$, [F7] makes $S$ simple.

Let $T$ be a simple left $A$-module [F7] and choose $0\mathrm{\ne }t\in T$. The map $\phi :A\to T$, $\phi (x)=xt$, is $A$-linear: $\phi (x+y)=(x+y)t=xt+yt$ and $\phi (ax)=(ax)t=a(xt)$ for $a,x,y\in A$, by the module axioms [F5], which is exactly the two clauses of [F8]. If $\phi (x)=0$ with $x\mathrm{\ne }0$, then $t=1\cdot t=(x^{-1}x)t=x^{-1}(xt)=0$ by [F5], a contradiction; hence $\ker \phi =0$ and $\phi$ is injective. Its image is a submodule of $T$ containing $t\mathrm{\ne }0$, so simplicity [F7] forces $\mathrm{im}\phi =T$. Therefore $\phi$ is an isomorphism of $A$-modules and $T\cong A=S$: up to isomorphism, $S$ is the only simple left $A$-module.

The regular module $P=A$ is projective in the sense of [F12]: if $q:E\to M$ is a surjective $A$-module homomorphism and $f:A\to M$ is $A$-linear, choose $e\in E$ with $q(e)=f(1)$; then $\tilde{f}(x)=xe$ defines an $A$-linear map $\tilde{f}:A\to E$ by [F5], and $q(\tilde{f}(x))=xq(e)=xf(1)=f(x)$ for all $x\in A$, using the $A$-linearity of $f$ [F8]. The identity map $A\to A$ is surjective with kernel $0$, which is superfluous: if $N\le A$ and $N+0=A$, then $N=A$. Hence the identity $P\to S$ is a projective cover of $S$ in the sense of [F13], with projective finite-dimensional source $P=A=S$ of $\mathbb{R}$-dimension $2$ by [F2].

Evaluation at $1$ is an $\mathbb{R}$-linear bijection $\mathrm{ev}:{\mathrm{Hom}}_A(P,S)\to A$, $\mathrm{ev}(f)=f(1)$; here ${\mathrm{Hom}}_A(P,S)$ is an $\mathbb{R}$-vector subspace of ${\mathcal{L}}_{\mathbb{R}}(A,A)$ by [F15], and $\mathrm{ev}$ is $\mathbb{R}$-linear because addition and scalar multiplication of homomorphisms are pointwise. It is injective: if $f(1)=0$, then $f(x)=xf(1)=0$ for every $x\in A$ by the second clause of [F8]. It is surjective: for $\lambda \in A$ the map $x\mapsto x\lambda$ is $A$-linear by [F5] and has value $\lambda$ at $1$. Hence ${\mathrm{Hom}}_A(P,S)\cong A=\mathbb{C}$ as $\mathbb{R}$-vector spaces, and [F16] with ${\dim }_{\mathbb{R}}\mathbb{C}=2$ from [F2] gives ${\dim }_{\mathbb{R}}{\mathrm{Hom}}_A(P,S)=2$.

The algebra $A$ is finite-dimensional over $k=\mathbb{R}$ by step 1.1, and by step 1.3 the single module $S=A$ represents all simple left $A$-modules; step 1.4 supplies a finite-dimensional projective cover $\mathrm{i}\mathrm{d}:P\to S$ of that representative. Applying [F14] to this data: the split Grothendieck group $K_0(A)$ of finite-dimensional projective left $A$-modules [F18] is the free abelian group with basis $[P]$, the cover $P$ is indecomposable, and every finite-dimensional projective left $A$-module is a finite direct sum of copies of $P$ [F11]. If a nonzero finite-dimensional projective $X$ is isomorphic to ${⨁}_{i=1}^{m}P$ with $m\ge 0$, then $m\ge 1$ because $X\mathrm{\ne }0$; if $m\ge 2$, then $X=P\oplus {⨁}_{i=2}^{m}P$ exhibits $X$ as a direct sum of two nonzero submodules, contradicting indecomposability. Hence $m=1$ and $X\cong P$: up to isomorphism, $P$ is the only indecomposable finite-dimensional projective left $A$-module.

We verify the hypotheses of [F19] for the full subcategory $\mathcal{M}$ of finite-dimensional left $A$-modules [F18]. First, $\mathcal{M}$ is abelian: the category of all left $A$-modules is abelian [F10]; inside $\mathcal{M}$ the zero module and finite biproducts exist, a finite biproduct of finite-dimensional modules having finite-dimensional underlying space by [F26]; and kernels, images and cokernels of $A$-linear maps of finite-dimensional modules are again finite-dimensional — kernels and images are $\mathbb{R}$-linear subspaces of finite-dimensional spaces, hence finite-dimensional and of no larger dimension by [F25], while a cokernel $N\mathrm{/}\mathrm{im}f$ is the image of the quotient map, so [F9] and [F17] give ${\dim }_{\mathbb{R}}N={\dim }_{\mathbb{R}}(\mathrm{im}f)+{\dim }_{\mathbb{R}}(N\mathrm{/}\mathrm{im}f)$ — so the abelian-category clauses of [F10] hold in the full subcategory. Second, $\mathcal{M}$ is essentially small: for each $n\ge 0$ the module structures on the $\mathbb{R}$-vector space ${\mathbb{R}}^n$ are given by the $\mathbb{R}$-bilinear maps $A\times {\mathbb{R}}^n\to {\mathbb{R}}^n$ satisfying the axioms [F5], and these maps form a set; every finite-dimensional module is isomorphic to one of these models after choosing an $\mathbb{R}$-basis. Third, every object of $\mathcal{M}$ has finite length in the sense of [F23], by induction on ${\dim }_{\mathbb{R}}M$: for $M=0$ the empty chain is a composition series [F24]; for $M\mathrm{\ne }0$ choose a proper submodule $N\le M$ of maximal $\mathbb{R}$-dimension [F6] among the finite set of dimensions of proper submodules (the zero submodule is proper because $M\mathrm{\ne }0$). If $N<N^{\mathrm{\prime }}<M$, then $N^{\mathrm{\prime }}\mathrm{/}N\mathrm{\ne }0$ and the quotient map $N^{\mathrm{\prime }}\to N^{\mathrm{\prime }}\mathrm{/}N$ is $\mathbb{R}$-linear with kernel $N$, so [F17] gives ${\dim }_{\mathbb{R}}{N}^{\mathrm{\prime }}={\dim }_{\mathbb{R}}N+{\dim }_{\mathbb{R}}(N^{\mathrm{\prime }}\mathrm{/}N)>{\dim }_{\mathbb{R}}N$, contradicting maximality among proper submodules. If $M\mathrm{/}N$ had a proper nonzero submodule $U$, its inverse image $N^{\mathrm{\prime }}$ under the quotient map $q:M\to M\mathrm{/}N$ would be a submodule by [F6], [F8] and [F9]; surjectivity of $q$ and $\ker q=N$ would give $N<N^{\mathrm{\prime }}<M$, which was just excluded. Since $N<M$, the quotient $M\mathrm{/}N$ is nonzero and therefore simple [F7]. Also ${\dim }_{\mathbb{R}}N<{\dim }_{\mathbb{R}}M$ by [F25], so by induction $N$ has a finite composition series, and appending the top object $M$, whose quotient $M\mathrm{/}N$ is simple, gives a composition series of $M$ in the sense of [F24]. Since by step 1.3 the single module $S$ represents all simple classes, [F18] and [F19] give $G_0(A)\cong \mathbb{Z}[S]$, with the class $[S]$ of the regular module as the only basis element.

By steps 2.1 and 2.2, $[P]$ is the single basis class of $K_0(A)$ and $[S]$ the single basis class of $G_0(A)$, so the well-defined pairing of [F21] takes the value $⟨[P],[S]{⟩}_A=h(P,S)={\dim }_{\mathbb{R}}{\mathrm{Hom}}_A(P,S)=2$ of [F20] and step 1.5. By [F22] the pairing matrix in these bases has the single entry ${\dim }_k{\mathrm{End}}_A(S)={\dim }_{\mathbb{R}}{\mathrm{End}}_{\mathbb{C}}(\mathbb{C})$, and the evaluation argument of step 1.5 with $P=S=A$ identifies ${\mathrm{End}}_A(S)\cong A=\mathbb{C}$ as $\mathbb{R}$-vector spaces, of dimension $2$ by [F2]; so the entry is $2$. The bases $[P]$ and $[S]$ would be dual exactly if this single matrix entry were $1$, which it is not. In particular the hypothesis of [F22] that ${\mathrm{End}}_A(S_i)=k$ for every $i$ is false here: ${\mathrm{End}}_A(S)\cong \mathbb{C}$ has $\mathbb{R}$-dimension $2$, so it is not the scalar field $k=\mathbb{R}$, and the dual-basis conclusion fails for this input. Hence that hypothesis cannot be dropped from the theorem. [F1, F2, F16, F20, F21, F22, step 2.1, step 2.2, step 1.5, algebra] $\mathrm{\square }$