tensor product of polynomial algebra

Question

Is $R[x] \otimes R[x]$ a free $R \otimes R$-module? Here $R$ is a $k$-algebra and $\otimes = \otimes_k$.

Answer 1

This result is also valid in a more case general, as follows.

Let $R$, $S$ be $k$-algebras which are free modules over $k$ (where $k$ is a field or a commutative ring) with basis $\mathcal{V}$ and $\mathcal{W}$, respectively. Let $A$ be a free left $R$-module with basis $\{a_i \}_{i \in I}$ and $B$ a free left $S$-module with basis $\{b_j \}_{j \in J}$. So, $A$ is a free $k$-module with basis $\{v_sa_i : v_s \in \mathcal{V}; i \in I \}$ and $B$ is a free $k$-module with basis $\{w_tb_j : w_t \in \mathcal{W}; j \in J \}$. Therefore, $A \otimes_k B$ is a free $k$-module with basis $\{v_sa_i \otimes w_tb_j : v_s \in \mathcal{V}; w_t \in \mathcal{W}; i \in I; j \in J \}$.

1º) $\{a_i \otimes b_j : i \in I; j \in J\}$ spans $A \otimes_k B$ as left $(R \otimes_k S)$-module.

2º) $\{a_i \otimes b_j : i \in I; j \in J\}$ is linearly independent over $R \otimes_k S$.

Let $0 = \sum_{i,j} \lambda_{ij} (a_i \otimes b_j)$, with $\lambda_{ij} \in R \otimes_k S$. Then $\lambda_{ij} = \sum_l x_{ij}^l \otimes y_{ij}^l $, $x_{ij}^l = \sum_p \lambda_{ij}^{lp} v_p $ and $y_{ij}^l = \sum_q \mu_{ij}^{lq} w_q $. So $$0 = \sum_{i,j} \lambda_{ij} (a_i \otimes b_j) = \sum_{i,j,l,p,q} (\lambda_{ij}^{lp} v_p \otimes \mu_{ij}^{lq} w_q) (a_i \otimes b_j) = \sum_{i,j,l,p,q} (\lambda_{ij}^{lp} v_pa_i \otimes \mu_{ij}^{lq} w_qb_j) = \sum_{i,j,l,p,q} \lambda_{ij}^{lp}\mu_{ij}^{lq} (v_pa_i \otimes w_qb_j)$$ But $A \otimes_k B$ is a free $k$-module, hence $\sum_l \lambda_{ij}^{lp}\mu_{ij}^{lq} = 0, \forall i,j,p,q$. Therefore,

$\lambda_{ij} = \sum_l x_{ij}^l \otimes y_{ij}^l = \sum_l (\sum_p \lambda_{ij}^{lp} v_p) \otimes (\sum_q \mu_{ij}^{lq} w_q) = \sum_{l,p,q} \lambda_{ij}^{lp}\mu_{ij}^{lq} (v_p \otimes w_q) = \sum_{p,q} (\sum_l \lambda_{ij}^{lp}\mu_{ij}^{lq}) (v_p \otimes w_q) = 0, \qquad \forall i,j.$

Answer 2

This is true, and follows from the fact that tensor products commute with arbitrary direct sums. This is a specific case of the more general category-theoretic result that left adjoint functors commute with colimits. If you have not come across the relevant category theory, this will not be particularly enlightening, but if you have, the argument is as follows:

From the $\hom-\otimes$ adjunction $\hom_k(A\otimes B,C)\cong \hom(A,\hom_k(B,C))$, we have that $B\otimes-$ is a left adjoint functor, and thus commutes with all colimits. In particular, it commutes with arbitrary direct sums. Now, in the generality of your answer, let $R$ and $S$ be $k$-algebras. Then, applying this twice, we have

$$\left(\bigoplus_{i\in I} R\right)\otimes \left(\bigoplus_{j\in J} S\right)\cong \left(\bigoplus_{i\in I} R\otimes \left(\bigoplus_{j\in J} S\right)\right) \cong \bigoplus_{i\in I} \left(\bigoplus_{j\in J} R\otimes S\right)\cong \bigoplus_{(i,j)\in I\times J}R\otimes S. $$

Therefore, the tensor product of a free $R$-module and a free $S$-module is a free $R\otimes S$-module.