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Given Two Fields $F,K$, and two vector spaces $V,W$ over $F$, what does tensor product $$V\otimes_{K} W$$ mean? I am not certain whether this is defined in general. I came across it in cases wheh $F=\mathbb{R},K=\mathbb{C}$.

I found some material in Wikipedia. However, I am approaching tensor product not with "free vector space" approach but with "universal property" approach. So, I could not follow the material.

I would appreciate the meaning of $$V\otimes_{K} W$$ and difference with $$V\otimes W$$

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  • $\begingroup$ Is not this Wikipedia article what you're looking for then? $\endgroup$
    – fuglede
    Apr 16, 2014 at 14:29
  • $\begingroup$ @fuglede, perhaps not. There both vector spaces are over R and it says $\otimes$ and $\otimes_{R}$ are same and gives a statement somewhere that $V\otimes_Z W$ and $V\otimes_R W$ are not same. $\endgroup$
    – user143587
    Apr 16, 2014 at 14:42

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What you describe: $V \otimes_\mathbb{C} W$ when $V, W$ are merely real vector spaces is not generally possible. It's only possible if the vector spaces you consider are already complex vector spaces.

What is possible though is the reverse: $V \otimes_\mathbb{R} W$ when $V,W$ are complex vector spaces. Indeed, a complex vector space is canonically a real vector space; this is because there is a canonical embedding of fields $\mathbb{R} \subset \mathbb{C}$. More generally given a field extension $F \subset K$, a $K$-vs is always an $F$-vs, and then you can do $V \otimes_F W$ when $V, W$ are $K$-vs.

You should note that $V \otimes_\mathbb{R} W$ is different from $V \otimes_\mathbb{C} W$ when both are defined. In other words the base field is important. See this for example.

The difference between $\otimes$ and $\otimes_F$ is that in $\otimes$, the base field (or ring) isn't specified. So it implies that at the beginning, you fixed some field $F$ and implicitly when you write $V \otimes W$ the tensor product is actually over $F$. But this depends on context. Sometimes the tensor product is taken by default over $\mathbb{Z}$, because every module over anything is always a $\mathbb{Z}$-module (aka an abelian group).

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Regardless of your definition of a tensor product what you have written is true only if you have a ring homomorphism $~f:K\rightarrow F$. Then you can turn $V$ and $W$ into $K$-vector spaces by the rule $kv=f(k)v$, $k\in K$, $v\in V$ and $kw=f(k)w$, $k\in K$, $w\in W$. Then the tensor product of $V$ and $W$ over $K$ is well-defined. This in fact is true for general modules over rings, of which vector spaces are merely a special case.

For instance, as the above user has explained, every complex vector space is a real vector space (by the natural inclusion $\mathbb{R}\rightarrow\mathbb{C}$) and thus one can define real tensor products between complex linear spaces. The other way around however, as Najib Idrissi explains, does not work.

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