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 at 16:36
  • 2
    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 at 18:41
up vote 3 down vote accepted

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 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 at 20:44

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.