[Math] A ring without the Invariant Basis Number property

abstract-algebralinear algebramodules

I was reviewing my homework and it seems I overlooked something crucial while proving some ring has no Invariant Basis Number property. This is exercise VI.1.12 in Aluffi's Algebra: Chapter 0

The setup: $V$ is a $k$-vector space and let $R = \mathrm{End}_{k}(V)$.

  1. Prove that $\mathrm{End}_{k}(V\oplus V) \cong R^4$ as an $R$-module
  2. Prove that $R$ doesn't satisfy the IBN property if $V = k^{\oplus \mathbb N}$.

For the first, I used to the fact that $V \oplus V$ is both the product and coproduct (in $k$-Vect) of $V$ with itself to get the isomorphism. What I just realized is I only showed that the two are isomorphic as groups not $R$-modules. So what would be the $R$-module structure on $\mathrm{End}_{k}(V \oplus V)$?

For the second, I used the fact that $V = k^{\oplus \mathbb N}$ implies $V \cong V \oplus V$ which in turn implies $R = \mathrm{End}_{k}(V) \cong \mathrm{End}_{k}(V \oplus V)$. Again, I just realized that I only showed the latter two are isomorphic as groups.

It may be obvious (and maybe why my professor let it pass?) but I can't come up with a good $R$-module structure that makes the two group isomorphisms $R$-linear.

Edit:

Explicitly, these are the isomorphisms I'm dealing with. Let $\pi_j, i_j$ be the natural projection/inclusion maps of the $j$-th factor resp. and $\psi: k^{\oplus \mathbb N} \oplus k^{\oplus \mathbb N} \to k^{\oplus \mathbb N}$ the isomorphism given by $\psi(e_i, 0)=e_{2i-1}$ and $\psi(0, e_i)=e_{2i}$.

Then the first isomorphism $\mathrm{End}_k(V \oplus V)\to R^4$ is given by $\varphi \mapsto (\pi_1\varphi i_1,\pi_2\varphi i_1,\pi_1\varphi i_2,\pi_2\varphi i_2)$

The second isomorphism $R \to \mathrm{End}_k(V \oplus V)$ is given by $\alpha \mapsto \psi^{-1} \alpha \psi$

The composition doesn't seem to be $R$-linear if I use the obvious $R$-module structure on $R$ and $R^4$.

Best Answer

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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