A ring without the Invariant Basis Number property

Solution 1:

I know how to prove that $R$ does not satisfy IBN (still thinking about the first question). Take a basis $\{e_i\mid i\in\mathbb{N}\}$ for $V$ as a $k$-vector space. Define $f_1,f_2\in R$ by $f_1(e_i)=e_{2i-1}$ and $f_2(e_i)=e_{2i}$. Then $\{f_1,f_2\}$ generates $R$ as a right $R$-module, and this set is $R$-linearly independent. So $R^{2}$ and $R$ are isomorphic as $R$-modules, because $\{1\}$ is also a basis for $R$ as $R$-module.

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