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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 $R^4$.

user26857
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genepeer
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  • For the conclusion I want, i.e., $R$ doesn't have the IBN property since $R \cong R^4$, I need isomorphic as $R$-modules. – genepeer May 03 '15 at 21:38
  • Now that I think some more, it could be that the two isomorphism are just group isomorphisms but the composition $R \cong R^4$ becomes $R$-linear trivially. – genepeer May 03 '15 at 21:47
  • The composition doesn't seem to be R-linear: Let $\psi:V \oplus V \to V$ be the isomorphism $\psi(e_i, 0) = e_{2i-1}$ and $\psi(0, e_i) = e_{2i}$. Then the composition $R \cong R^4$ is given by $$\alpha \mapsto (\pi_1 (\psi^{-1} \alpha \psi) i_1, \pi_2 (\psi^{-1} \alpha \psi) i_1, \pi_1 (\psi^{-1} \alpha \psi) i_2, \pi_2 (\psi^{-1} \alpha \psi) i_2)$$ – genepeer May 03 '15 at 23:26

2 Answers2

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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.

user26857
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Studzinski
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  • I'll let it stay. I'm a bit confused by why the set is linearly independent. Let $\alpha_1, \alpha_2, f_1, f_2 , z \in R$ where $z$ is the zero function. You're saying $$\alpha_1 \circ f_1 + \alpha_2 \circ f_2 = z$$ implies $\alpha_1 = \alpha_2 = z$? I'm assuming multiplication in $R$ is composition and addition point-wise. – genepeer May 03 '15 at 22:00
  • yes, that is right. Do you want me to complete the proof? – Studzinski May 03 '15 at 22:08
  • Wait. What if $\alpha_1(e_1) = e_1$ and vanishes on other bases, and $\alpha_2(e_2) = -e_1$ and vanishes on other bases. Then $$(\alpha_1 \circ f_1 + \alpha \circ f_2 )(r_1, r_2, \ldots) = \alpha_1(r_1, 0, r_2, 0, \ldots) + \alpha_2(0, r_1, 0, r_2, 0, \ldots) = (r_1, 0, 0, \ldots) + (-r_1, 0, 0, \ldots) = (0, 0 , 0 \ldots)$$ – genepeer May 03 '15 at 22:40
  • Aren't $f_1, f_2$ zero-divisors? – genepeer May 03 '15 at 22:47
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    Sorry, you are correct. But if you consider the right $R$-module structure then ${f_1,f_2}$ will be $R$ linearly independent. – Studzinski May 04 '15 at 00:36
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    Thanks. Also, using $f_1(e_{2i-1})=e_i$ and $f_2(e_{2i})=e_i$ and they're zero elsewhere, then they are a basis for $R$ as a left $R$ module. – genepeer May 04 '15 at 20:40
  • Yes, nicely observed. – Studzinski May 04 '15 at 20:42
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The way the question is posed, it seems implied that we can use the first part to prove the second. But that is impossible as far I can tell: the induced composition $R \cong R^4$ is not $R$-linear.

However, the first isomorphism can be made $R$-linear by using the following structure:

$\alpha \in R, \varphi \in \mathrm{End}_k(V \oplus V) $ then $$ \alpha \cdot \varphi = (\alpha \oplus \alpha) \circ \varphi $$

This structure appears to be lost through any isomorphism $R \cong \mathrm{End}_k(V \oplus V)$.

user26857
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genepeer
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