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We can say that $\mathbb{C}^{2}$ is not isomorphic to $\mathbb{C}$ when both are considered as Vector spaces over the field of Complex numbers or Real numbers. But is $\mathbb{C}^{2}$ isomorphic to $\mathbb{C}$ when both are considered as Vector Spaces over Field of Rational numbers?

Two Finite dimensional Vector spaces over same field are isomorphic iff they are of same dimension. But here $\mathbb{C}^{2}$ and $\mathbb{C}$ are not Finite dimensional over rational numbers. So how can we tackle this problem? Is there any general methodology to establish such isomorphisms in Non finite dimensional case?

Through the comments I got to know Two vector spaces over the same field are isomorphic if and only if their bases have the same cardinality and this is true even in non finite dimensional case. But this requires finding a basis for the above two vector spaces and comparing their cardinalities, this further uses Axiom of choice. Is there a way to avoid this? (because isn't it difficult to find basis for Non finite dimensional vector spaces?) Is there a way to just establish isomorphism between the vector spaces and show they are isomorphic?

Shash
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1 Answers1

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It is both simple and difficult to answer your new question. If you believe that in any vector space over a field there exists a basis (finite or infinite), then for the case of a finite or countable field the answer is.

Two infinite-dimensional vector spaces are isomorphic if and only if these spaces have the same cardinality. (Note not the bases, but the spaces.)

kabenyuk
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  • How this can be the case? For instance, infinite-dimensional split-complex numbers ($i_n^2=1$) are not isomorphic to infinite-dimensional tessarines ($i_{2n}^2=-1, i_{2n+1}^2=1$) and not isomorphic to infinite-dimensional Grassmann numbers. – Anixx Nov 01 '23 at 16:44
  • However, as vector spaces over $\mathbb{Q}$ they are all pairwise isomorphic. – kabenyuk Nov 01 '23 at 17:13
  • How they can be isomorphic if split-complex numbers and tessarines are commutative and Grassmann numbers are not? – Anixx Nov 01 '23 at 17:18
  • @Anixx Because you're considering an algebraic structure (multiplication) which is not part of a vector space. As vector spaces (when the only available multiplication is multiplication by rational numbers) they are isomorphic. – Arthur Nov 01 '23 at 19:11
  • @Arthur the properties of moduluses are also different – Anixx Nov 01 '23 at 20:00
  • @Anixx That's still not relevant to vector spaces. Those don't appear until you start to talk about normed vector spaces. – Arthur Nov 01 '23 at 20:01