A regular module with bases of sizes one and two

Define ${\alpha }_0(e_k)=e_{2k}$ and ${\alpha }_1(e_k)=e_{2k+1}$. Define ${\beta }_0(e_{2k})=e_k$, ${\beta }_0(e_{2k+1})=0$, ${\beta }_1(e_{2k})=0$, and ${\beta }_1(e_{2k+1})=e_k$. Extending the formulas across the unique finite basis expressions from [F5] gives four endomorphisms.

On every basis vector, ${\beta }_i{\alpha }_j$ is the identity when $i=j$ and zero otherwise, while ${\alpha }_0{\beta }_0+{\alpha }_1{\beta }_1={\mathrm{id}}_V$. Unique finite basis expressions from [F5] make these identities hold on all of $V$.

The operations from [F2] and [F3] satisfy the ring axioms [F4], so $R={\mathrm{End}}_F(V)$ is a unital ring. Define left $R$-linear maps $T:R\to R^2$ and $S:R^2\to R$ by $T(f)=(f{\alpha }_0,f{\alpha }_1),S(g,h)=g{\beta }_0+h{\beta }_1\mathrm{.}$ Associativity and distributivity give left linearity.

The identities in step 2.1 give $T(S(g,h))=(g,h)$ and $S(T(f))=f$, so $T$ and $S$ are inverse isomorphisms.

Therefore ${}_{R}R\cong {}_R{R}^2$, refuting finite rank invariance in this ring as [F1] records.