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Consider two vector spaces $E$ and $F$ over the same field $K$. Now to form the tensor product $E\otimes F$ we typically take a particular vector space $V_1$ and quotient it. The vector space $V_1$ is from page 265 here, is defined as $V_1=\oplus_{(e,f)\in E\times F}K(e,f)$. My questions are:

(1) Is $K(e,f)$ the set of all formal expressions of the form $\alpha\cdot(e,f)$? If so, how is this a vector space? Alternatively is $K(e,f)$ the span of $(e,f)$ in the vector space $E\times F$?

(2) How do we visualize the direct sum? The way I see it, $K(e,f)$ the span of $(e,f)$ and the direct sum consists of all those elements of $\prod K(e,f)$ for which all but finitely many coordinates are zero. But then how are the basis elements of $V_1$ all the elements of $E\times F$?

Thanks

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$V_{1}$ is the so-called ''free-vector space''. Using the defining property of the direct sum, this can be represented as:

$$F(V\times W):=\bigoplus_{(v,w)\in V\times W}\mathbb{K}\cdot (v,w)\cong$$

$$\bigg\{\sum_{(v,w)\in I}\lambda_{(v,w)}(v,w)\mid I\subset V\times W\text{ with } \vert I\vert<\infty\land \lambda_{(v,w)}\in\mathbb{K}\bigg\}$$

More generally, the free vector space of a given set $X$ is the vector space $F(X)$, which is ''generated by the set $X$'', which means that $F(X)$ is defined to be a vector space in such a way that $X$ is a basis of $F(X)$. In other words, $F(X)$ is the set of finite linear combinations of elements of $X$. So you see that $V_{1}$ is the collection of all finite linear combinations of $(e,f)\in E\times F$.

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  • $\begingroup$ The free vector space is isomorphic to the direct sum. Is that correct? $\endgroup$ – Shahab May 17 '20 at 8:14
  • $\begingroup$ Exactly...In other words, for the construction of the tensor product you need the free vector space, which is the set of finite linear combination....The direct sum is just a more fancy way to write this.... $\endgroup$ – Udalricus.S. May 17 '20 at 8:16
  • $\begingroup$ Yes, I understand. Why the author did not simply use the free vector space is not clear though. Thanks. $\endgroup$ – Shahab May 17 '20 at 8:22
  • $\begingroup$ No problem....But as I said, the direct sum is just a more compact and fancier form of the free vector space.....Depending on your exact definition on $\oplus$, you may also write a $=$ instead of $\cong$, because $\oplus$ is often defined as sum of all finite linear combinations. $\endgroup$ – Udalricus.S. May 17 '20 at 8:28

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