Is M_n(K) isomorphic to K \otimes_F M_n(F) as F-algebras?

[math_grl]

So this is supposed be an introductory problem for tensor products that I was trying to do to verify I am understanding tensor products...turns out I'm not so much

Show that $$M_n(K)$$ is isomorphic as an $$F$$-algebra to $$K \otimes_F M_n(F)$$ where $$F$$ is a field and $$K$$ is an extension field of $$F$$ and $$M_n(K)$$ means all the nxn matrices that have entries in K.

So I figure as F-algebras we need to show that we have a ring homomorphism that is linear (preserving the scalar multiplication) or showing they are isomorphic as F-modules (vec. sp's) then showing preservation of the multipication. Either way, my attempts are fruitless.

 

[math_grl]

i think i solved it...nevermind

 

[math_grl]

actually i might have gotten ahead of myself...I don't think i have it.

I was trying $$\phi: K \otimes_F M_n(F) \rightarrow M_n(K)$$ where $$\phi (k \otimes A) = kA$$ as my mapping, then checking to see if it satisfied all the homomorphism conditions. This must be wrong tho.

 

[TMM]

K tensor over F of the n dimensional F-vector space is isomorphic to the n dimensional K-vector space by easily proven properties of the tensor product over the direct sum. Perhaps this fact is useful?

I'm new to this, as well.

 

[math_grl]

K tensor over F of the n dimensional F-vector space is isomorphic to the n dimensional K-vector space by easily proven properties of the tensor product...

Sounds like a path to try but I would refrain from saying easily proven since we are not experts in this stuff yet...

 

[TMM]

Sounds like a path to try but I would refrain from saying easily proven since we are not experts in this stuff yet...

Fair enough. I think this would lead to a nicer proof, but I think your method works anyway.

You need to consider it acting on all the tensors in your product, not just the simple ones. Every tensor in the matrix group over K can be expressed as a sum of its values in each index. These can be decomposed into their magnitude and the matrix which has a 1 in the specific index and zero elsewhere. This matrix is in the matrix group over F so the map is surjective. If the images of two simple tensors are the same, you can cancel the factor in K, showing that the tensors are the same, so it is also injective.

 

[math_grl]

Thanks for the help. I noticed you mentioned that you said all tensor elements (finite sums of basis elements of the tensor product)...how come it doesn't suffice to just show it for the pure (simple) tensor elements since these span the tensor product as a vector space?

 

[TMM]

Well it does, it's just easier to show surjectivity using a sum of simple tensors. In fact you need to since arbitrary elements of the codomain do not have preimages that are simple tensors.