# Tensor of Hom spaces over an algebra

Given a field $k$ of null caracteristic, an associative/commutative/unital $k$-algebra $A$ and two free $k$-modules $M,N$, denote by $Hom_k(M,A)$ the space of $k$-linear maps from $M$ to $A$.

I want to show that the following map is injective:

$Hom_k(M,A)\otimes_A Hom_k(N,A) \longrightarrow Hom_k(M \otimes_k N,A)$

$f \otimes_A g \mapsto \Big( v \otimes_k w \mapsto f(v)g(w)\Big)$

I might understand this in the finite dimensional case using the following string of isomorphisms

$Hom_k(V,A) \otimes_A Hom_k(W,A) \simeq (V^* \otimes_kA)\otimes_A(W^*\otimes_kA)\simeq V^* \otimes_k W^* \otimes_k A \simeq (V \otimes_k W)^* \otimes_k A \simeq Hom_k(V\otimes_kW,A)$

But for some reason I can't figure it out in the general case.

Can someone give me some hint or a reference for it ?

• Put $M=k^{(I)}$ and $N=k^{(J)}$. What you are asking is whether the natural map $u:A^{I}\otimes _A A^{J}\rightarrow A^{I\times J}$ satisfying $u((x_i)\otimes (y_j))=(x_iy_j)$ is injective. I would guess this is true, but I don't see an argument right now. – abx May 15 '18 at 16:36
• As @abx comment says, if not injective, you can reduce to the case of finite dimensional case easily. For example, if $u$ as in the comment is not injective, you have $u(\sum \alpha_i\otimes \beta_i)=0$, a finite sum and each of the $\alpha_i,\beta_i$ involve only finitely many basis elements. – Mohan May 15 '18 at 18:41

For infinite-dimensional algebras this is not true.

As it is mentioned in the comments, the claim is equivalent to the injectivity of the coordinate-wise product map $u:A^I \otimes_A A^J \to A^{I\times J}$ (where now the exponent stands for products, not coproducts).

For a counterexample, consider the commutative, unital, associative algebra $A=\Bbbk [x_i \; (i \in \mathbb{N})]/(x_i x_j \; (i,j \in \mathbb{N}))$, let $I=J=\mathbb{N}$ and $\alpha = (x_i)_{i \in \mathbb{N}} \in A^I$. Then clearly $u(\alpha \otimes \alpha)=0$, but one can prove that $\alpha \otimes \alpha$ is nonzero in $A^I \otimes_A A^I$.

Indeed, $A^I \otimes_A A^I = (A^I \otimes_{\Bbbk} A^I)/U$, where $U$ is spanned by the elements of the form $$\lambda x_i \otimes \mu - \lambda \otimes x_i \mu,\quad \lambda x_i \otimes \beta,\quad \beta \otimes x_i \lambda$$ for all $i \in \mathbb{N}$, $\lambda, \mu\in \Bbbk^I$, $\beta \in N^I$ denoting $N=\mathrm{Span}(x_k \mid k \in \mathbb{N})$. These elements do not span $\alpha \otimes \alpha$ in $A^I \otimes_{\Bbbk} A^I$.

(Edit, response to the comment: If $A$ is finite-dimensional then $u$ is injective. Indeed, given a $\Bbbk$-basis $b_1, \dots, b_d$ in $A$, the elements of the form $\lambda \otimes \mu\, b_k$ span $A^I \otimes_A A^J$, where $k=1,\dots,d$, $\lambda \in \Bbbk^I$, $\mu \in \Bbbk^J$. Hence it is enough to prove the statement for $A=\Bbbk$.

Assume that $u$ vanishes on the nonzero element $\sum_{k=1}^m \lambda^{(k)} \otimes \mu^{(k)}$ where $\lambda^{(k)},\mu^{(k)}$ are as above. We may assume that $\lambda^{(1)},\dots,\lambda^{(m)}$ are independent and similarly for $\mu$. Consequently, $\{(\lambda_i^{(k)})_k\mid i\in I\}$ spans $\Bbbk^m$ by a rank-argument. On the other hand, as $u$ vanishes on the sum, we have $\sum_{k=1}^m \lambda_i^{(k)}\cdot \mu_j^{(k)}=0$ for all $i,j$, i.e. $(\mu_j^{(k)})_k$ is orthogonal to a spanning set with respect to the obvious bilinear form on $\Bbbk^m$. This is a contradiction.)

• Thank you ! This is really helpfull. Although I'm not sure to really understand your first statement "For infinite-dimensional algebras", for me it seems that the main issue come from the fact that you use infinite-dimensional $k$-modules through the definition $I = J = \mathbb{N}$. Anyway, the key of your counter example is the relation in your algebra, so would you think that the statement is true in the case of a free algebra $A$ and free $k$-modules $M,N$ ? – J.Dawn May 16 '18 at 12:43
• I would guess that it is injective in the case of the free algebra, but unfortunately I don't see a proof. For the first part of your comment, see the edit. – SzM May 24 '18 at 20:44