# free modules being isomorphic with different dimension

This question is from the book A course in Homological algebra by Hilton and Stammbach

Let $V$ be a vector space of countable dimension over the field $K$. Let $\Lambda=\rm{Hom}_K(V,V)$. show that, as $K$ vector spaces $V$, is isomorphic to $V\oplus V$. We therefore obtain $$\Lambda=\rm{Hom}_K(V,V)\cong \rm{Hom}_K(V\oplus V,V)\cong\rm{Hom}_K(V,V)\oplus \rm{Hom}_K(V,V)$$

Conclude that, in general, the free module on a set of $n$ elements may be isomorphic to the free module on a set of $m$ elements, with $n\neq m$.

Let $V$ be a vector space of countably infinite dimension. We then have a basis for $V$ as $\{e_n: n\in \mathbb{N}\}$. This gives a basis for $V\oplus V$ as $\{(e_n,e_m):n,m\in \mathbb{N}\}$. As we have a bijection between $\mathbb{N}$ and $\mathbb{N}\times\mathbb{N}$, we have an isomorphism between $V$ and $V\oplus V$.

By properties of homomorphism ring, we have $$\Lambda=\rm{Hom}_K(V,V)\cong\rm{Hom}_K(V\oplus V,V)\cong\rm{Hom}_K(V,V)\oplus \rm{Hom}_K(V,V)$$

Now, $\Lambda$ is free as $\Lambda$ module of dimension $1$. We have $\Lambda\oplus \Lambda$ free as $\Lambda$ module of dimension $2$.

Nevertheless we have isomorphism between $\Lambda$ and $\Lambda\oplus \Lambda$.

Thus, the free module on a set of $n$ elements may be isomorphic to the free module on a set of $m$ elements, with $n\neq m$.

Please let me know if this justification is correct.

Provide some more examples of this kind.

• In this case $n=1, m=2$.. @arctictern – user311526 Jul 27 '16 at 1:18
• Ah, right, modules over $\Lambda$, not $K$. Looks good. – arctic tern Jul 27 '16 at 1:20
• @arctictern : Yes. Modules over $\Lambda$. – user311526 Jul 27 '16 at 1:22
• wasn't $Hom_K(V,V) \cong Hom_K(V,V)\oplus Hom_K(V,V)$ an isomorphism of $K$-module? This sort of make sense because the dimensions are both countably infinite. It is hard to believe that $\Lambda = \Lambda^2$ as $\Lambda$-module, but I might misunderstood something. – user113988 Jul 27 '16 at 2:17
• Good point, you have to show these are isomorphisms of $\Lambda$ modules not just $K$ modules. – Gregory Grant Jul 27 '16 at 2:21