I have just begun learning about tensor products. Given are the following relations of the tensor product between vector spaces $A$ and $B$:

  1. $\lambda(a\otimes b)=(\lambda a)\otimes b=a\otimes(\lambda b) $ for $\lambda \in \mathbb{R}$.
  2. $a\otimes b_1 + a\otimes b_2 = a\otimes(b_1+b_2)$.
  3. $a_1\otimes b + a_2\otimes b=(a_1+a_2)\otimes b$.

For $a\in A$ and $b\in B$.

Now I want to show that $\mathbb{R}\otimes\mathbb{R}\cong \mathbb{R}$, but I am not sure where to start. Is it necessary to explicitly construct an isomorphism between the two spaces? Is there something else I should know about tensor products to get this started?

  • 2
    $\begingroup$ I might not be strictly necessary to construct an explicit isomorphism, but I would still advice you to try doing so. It is fairly straight-forward to define, and checking that it really is an isomorphism is a good exercise in how tensor products work. $\endgroup$ – Tobias Kildetoft Aug 25 '17 at 11:02

Just consider the map$$\begin{array}{ccc}\mathbb{R}\bigotimes\mathbb{R}&\longrightarrow&\mathbb R\\\displaystyle\sum_{k=1}^na_k\otimes b_k&\mapsto&\displaystyle\sum_{k=1}^na_kb_k.\end{array}$$It is an isomorphism.

  • $\begingroup$ And, actually, it suffices to just tell what the map is on tensor monomials, by at least one characterization of tensor product... $\endgroup$ – paul garrett Aug 25 '17 at 15:29
  • $\begingroup$ @paulgarrett Of course, but I was aware that I was writing to a beginner. $\endgroup$ – José Carlos Santos Aug 25 '17 at 15:50
  • $\begingroup$ Indeed, but/and I think too many people misunderstand what it takes to specify such a map, which sometimes sends them down a slippery slope where they think they need generators and relations (for modules over commutative rings, rather than vector spaces, etc.), which is a non-trivial... but irrelevant... issue. So in addressing my own students I quickly try to steer them away from picking bases for vector spaces to talk about tensor products... although the very first, transitional, step is invariably that... $\endgroup$ – paul garrett Aug 25 '17 at 16:02
  • $\begingroup$ I see, thank you very much :) $\endgroup$ – F. Zegers Aug 25 '17 at 19:24
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    $\begingroup$ It might be useful to note, for intuition sake, that whenever $R$ a commutative ring with $1$, that $R\otimes_R R \cong R,$ because the elementary tensors of the form $a\otimes_R b = ab\otimes_R 1.$ Hopefully this helps with where the isomorphism defined above originates from. $\endgroup$ – Chickenmancer Aug 26 '17 at 19:48

$\dim A\otimes B=\dim A\dim B$ so when $\dim A=1$ and $\dim B$ is finite then $A\otimes B\cong B$, for there are no two distinct linear spaces of finite dimension over the same field, up to isomorphisms.

If you do not want to use dimensions, the next is another proof, equally obvious.

If $A$ is a field and $B$ a linear space of finite dimension over $A$ the multiplication of a vector by a scalar defined in $B$ satisfies the axioms $1$-$3$. That means by definition of tensor product that there exists and is unique a homomorphism of linear spaces $\eta:A\otimes B\rightarrow B$ such that $\eta(a\otimes b)=ab.$ You now see that in this case it is also an isomorphism. Indeed it is surjective because $\forall x\in B,~x=1x=\eta(1\otimes x)$, and it is injective because if $\sum_i a_i\otimes b_i\ne \sum_i c_i\otimes d_i$, that means that there are no relations among axioms $1$-$3$ through which equality can be proved, then, being the axioms of the multiplication of a vector by a scalar formally the same as axioms $1$-$3$, there are no relations that can prove the equality of $ab$ and $cd$, that means that $\eta(\sum_i a_i\otimes b_i)\ne \eta(\sum_i c_i\otimes d_i)$.


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