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This is an example from Karen Smith's notes.

Let $V_1, V_2 \subset k^n$ be linear subspaces (defined by some collection of linear polynomials). Then $V_1 \cong V_2$ as algebraic sets if and only if $\dim(V_1)=\dim(V_2)$.

I am not sure how to start. I think they are defined by linear polynomials does not mean there exist a linear map between them. And they might not be able to be defined by finitely many linear polynomials, right?

Thanks for your help!

KittyL
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  • I don't know what definition you are using for equivalence as algebraic sets. But any subspace of a finite-dimensional vector space is defined by finitely many linear equations. – Gerry Myerson Feb 12 '15 at 12:14
  • They are isomorphic if there exists a regular map (morphism) from $V_1$ to $V_2$ which has an inverse regular map from $V_2$ to $V_1$. So the dimensions here mean dimension of vector spaces instead of dimension of algebraic set? I had a hard time understanding the connection between them. Or are they the same? – KittyL Feb 12 '15 at 12:25
  • I don't have those notes in front of me, so I don't know for sure what Karen Smith intended by "dimension". I was assuming that as $V_1$ and $V_2$ were introduced as linear subspaces (which I take to mean, as vector spaces) "dimension" meant dimension as a vector space, but I could be wrong. What definition do you have for dimension of an algebraic set? – Gerry Myerson Feb 12 '15 at 22:24
  • She mentioned once in her book that the dimension is of a variety is the length $d$ of the longest possible chain of distinct nonempty irreducible subvarieties of $V$. But she never mentioned it in her notes. And by the context of this example, I guess your first assumption is right. It makes much more sense if the dimension here means the dimension of a vector space. Thank you very much for your help! – KittyL Feb 12 '15 at 22:56

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