How are R^2 and R^4 isomorphic as VS's over Q?

[joeblow]

I came across a problem asserting that the C-tensor product of C and C and the R-tensor product of C and C are isomorphic as Q-modules. How does this begin to make sense as they do not have the same dimension over R? The first is isomorphic to R² and the second is isomorphic to R⁴ over R.

Also, I'm struggling with tensors. Does anyone have a good source I should check out? (I've tried many)

 

[lavinia]

I came across a problem asserting that the C-tensor product of C and C and the R-tensor product of C and C are isomorphic as Q-modules. How does this begin to make sense as they do not have the same dimension over R? The first is isomorphic to R² and the second is isomorphic to R⁴ over R.

Also, I'm struggling with tensors. Does anyone have a good source I should check out? (I've tried many)

Over Q these vector spaces are infinite dimensional. They will be isomorphic if the cardinality of their bases is the same - which I think is the cardinality of the Continuum.

The tensor product is a way to extend the field of scalars of a vector space. A real vector space of dimension n (n can be infinite) becomes a complex vector space of dimension n when tensored with C. It becomes a real vector space of twice the dimension over R.

If one views R as a 1 dimensional vector space over itself with single basis vector v, the all elements are of the form rV for real numbers,r. Over C the basis is also v and the resulting vector space is 1 dimensional over C. But over R the basis is V and iV and so is 2 dimensional over R.

 

[Landau]

Over Q these vector spaces are infinite dimensional. They will be isomorphic if the cardinality of their bases is the same - which I think is the cardinality of the Continuum.

Indeed; we discussed this here.