# Axiom of choice and algebraic tensor product

The first part of the question was asked on Math-stackexchange.

Let $$V$$, and $$W$$ be vector spaces. By the universal property of the tensor product, there is a canonical map from $$V^*\otimes W^*$$ into $$(V\otimes W)^*$$ (since the map $$(\omega_1,\omega_2)\mapsto \omega_1\otimes\omega_2$$ is bilinear from $$V^*\times W^*$$ into $$(V\otimes W)^*$$.

I have read that this map is actually injective by using some basis on $$V$$ and $$W$$.

Since the existence of basis for any arbitrary vector space relies on the axiom of choice, my questions are

1) is the axiom of choice necessary to prove the injectivity of the canonical map $$V^*\otimes W^*\to(V\otimes W)^*$$ ?

2) A (possibly) connected question is the following : is it necessary to use the axiom of choice to prove that if $$(v_i)$$ is a linearly independent family in $$V$$, and $$(w_j)$$ is a linearly independent family in $$W$$, then $$(v_i\otimes w_j)_{i,j}$$ is linearly independent in $$V\otimes W$$.

• "Is AC necessary to prove X" technically means "does X implies AC" (probably, in ZF). But this is probably not what you mean. Probably you're asking whether the given statement is a theorem of ZF. – YCor Mar 9 '19 at 23:03
• My question can be broadly intepreted as : what is the status of those statements in ZF ? Are they theorems ? are they equivalent to AC ? are they equivalent to a known weaker axiom ? – Phil-W Mar 9 '19 at 23:11
• It is probably worth linking to the analogous question for modules over a (commutative) ring: it turns out, as explained in several answers there, that the map $M^\vee\otimes N^\vee\to(M\otimes N)^\vee$ can fail to be injective (even in the presence of AC, that is). – Gro-Tsen Mar 9 '19 at 23:44

I think both can be proved without choice, essentially because, in both cases, whenever you're tempted to choose a basis, you can manage with a little care to get by with a basis of a finite dimensional subspace.

For (2), if there's a linear dependence between the $$v_i\otimes w_j$$ then it involves only finitely many $$v_i$$ and $$w_j$$. Also, the linear dependence must be a (finite) linear combination of the usual relations such as $$(u+u')\otimes v-u\otimes v-u'\otimes v$$ for the tensor product, so there are finite dimensional subspaces $$V'\leq V$$ and $$W'\leq W$$ so that you have the same linear dependence in $$V'\otimes W'$$. And now you can use bases without invoking choice.

For (1), an element of the kernel is a finite sum of simple tensors $$\varphi\otimes\psi$$. By choosing a basis of the finite-dimensional subspaces of $$V^*$$ and $$W^*$$ spanned by the $$\varphi$$ and $$\psi$$ that occur, we can write the element of the kernel as a linear combination of $$\{\varphi_i\otimes\psi_j\}_{i,j}$$, where $$\{\varphi_i\}_i$$ and $$\{\psi_j\}_j$$ are finite linearly independent subsets of $$V^*$$ and $$W^*$$.

Now, again without choice, we can find finite dimensional subspaces $$V'\leq V$$ and $$W'\leq W$$ together with (finite) bases $$\{v_i\}_i$$ and $$\{w_j\}_j$$ that are dual bases to the restrictions of $$\{\varphi_i\}_i$$ and $$\{\psi_j\}_j$$ to $$V'$$ and $$W'$$, and prove that the kernel element is zero using these bases.

[To add a bit more detail to the last step, suppose that $$U$$ is a vector space over $$k$$, and $$\alpha_1,\dots,\alpha_d$$ a finite linearly independent list of elements of $$U^*$$. Then the subspace $$S=\left\{\left(\alpha_1(u),\dots,\alpha_d(u)\right)\mid u\in U\right\}$$ of $$k^d$$ must be the whole of $$k^d$$, or else there would be a nonzero linear functional $$k^d\to k$$ vanishing on $$S$$, and hence a linear dependence between the $$\alpha_i$$.

Hence there are elements $$u_1,\dots,u_d\in U$$ with $$\alpha_i(u_j)=\delta_{ij}$$ and so $$U$$ has a finite dimensional subspace $$U'=\langle u_1,\dots,u_d\rangle$$ with the $$u_i$$ forming a basis dual to the basis of $$(U')^*$$ consisting of the restrictions of the $$\alpha_i$$ to $$U'$$.]

In fact, answering a question asked in comments, there is a common generalization of (1) and (2). For any vector spaces $$V$$, $$V'$$, $$W$$, $$W'$$, the fact that the natural map $$\text{Hom}(V,V')\otimes\text{Hom}(W,W')\to\text{Hom}(V\otimes W,V'\otimes W')$$ is injective can be proved without the axiom of choice.

To see this, note that an element of the kernel can be written in terms of $$\alpha_i\otimes\beta_j$$, where $$\alpha_1,\dots,\alpha_s$$ are finitely many linearly independent elements of $$\text{Hom}(V,V')$$ and $$\beta_1,\dots,\beta_t$$ finitely many linearly independent elements of $$\text{Hom}(W,W')$$.

I claim that there is a finite dimensional subspace $$V''\leq V$$ such that the restrictions of $$\alpha_1,\dots,\alpha_s$$ to $$V''$$ are still linearly independent (and similarly for $$W$$ and the $$\beta_j$$). This follows by induction on $$s$$. Suppose there is a finitely generated subspace $$U$$ so that the restrictions of $$\alpha_1,\dots,\alpha_k$$ to $$U$$ are linearly independent. Then either the restrictions of $$\alpha_1,\dots,\alpha_{k+1}$$ to $$U$$ are linearly independent, or there is a linear dependence $$\sum_{i=1}^{k+1}\lambda_i(\alpha_i|_U)=0$$, which is unique up to a scalar multiple. But since $$\alpha_1,\dots,\alpha_{k+1}$$ are linearly independent, there is some $$v\in V$$ such that $$\sum_{i=1}^{k+1}\lambda_i\alpha_i(v)\neq0$$, and then the restrictions of $$\alpha_1,\dots,\alpha_{k+1}$$ to $$U+\langle v\rangle$$ are linearly independent.

Replacing $$V$$ and $$W$$ by $$V''$$ and $$W''$$, and $$V'$$ and $$W'$$ by $$\sum_i\alpha_i(V'')$$ and $$\sum_j\beta_j(W'')$$ reduces the original question to one about finite dimensional vector spaces, which can easily be solved without invoking choice.

• In the last paragraph's first sentence, you write "dimensional", I think you meant "finite dimensional"? – Asaf Karagila Mar 10 '19 at 11:25
• @AsafKaragila Thanks, fixed. (I think “find” looked too much like “finite” to my tired old eyes.) – Jeremy Rickard Mar 10 '19 at 12:07
• If you can also get rid of excluded middle (which I think is needed to show that finite-dimensional spaces have bases), I'll upvote this lovely answer twice. – Andrej Bauer Mar 10 '19 at 12:21
• @AndrejBauer Since you can’t upvote my answer twice, that presumably implies that I can’t get rid of excluded middle. Or am I assuming excluded middle ...? – Jeremy Rickard Mar 10 '19 at 13:07
• @Andrej Buaer: "...to show that finite-dimenasional vector spaces have bases" - What do you mean by a finite dimensional vector space here? Doesn't a f.d. v.s. have a finite basis by definition? – Qfwfq Mar 10 '19 at 13:26