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A basis for the tensor product

April 18, 2012

Suppose M_1, \ldots M_k are all free R-modules of respective rank n_1, \ldots n_k. Then, from the distributivity of tensor products over direct sums, we can deduce that M_1 \otimes \cdots \otimes M_k is the free R-module of rank \prod_{i=1}^k n_i.. Indeed, first recall that given a set of free generators (a basis) \{ m_1, \ldots m_n \} of the free R-module of rank M, we can write M = Rm_1 \oplus \cdots \oplus Rm_n \simeq R^n just as in the case of vector spaces. In this case, we get, with B_i = \{m_{i_j}\} a basis for M_j (i_j = 1, \ldots, n_j),

M_1 \otimes \cdots \otimes M_k \simeq \displaystyle\bigoplus_{j=1}^{n_1}Rm_{1_j} \otimes \cdots \otimes \bigoplus_{j=1}^{n_k} Rm_{k_j}

\simeq \displaystyle\bigoplus_{j1=1}^{n_1} \left( Rm_{1_{j1}} \otimes \bigoplus_{j=1}^{n_2}Rm_{2_j}\otimes \cdots \otimes \bigoplus_{j=1}^{n_k}Rm_{k_j} \right)

\simeq \displaystyle\bigoplus_{j1=1}^{n_1} \bigoplus_{j2=1}^{n_2} \left( Rm_{1_{j1}} \otimes Rm_{2_{j2}} \otimes \cdots \otimes \bigoplus_{j=1}^{n_k} Rm_{k_j} \right)

\simeq \cdots

\simeq \displaystyle\bigoplus_{(j_1, \ldots, j_k)} \left( Rm_{1_{j_1}} \otimes \cdots \otimes Rm_{k_{j_k}} \right)

where the last sum is taken on all multi-indices (j_1, \ldots, j_k) \in \textbf{N}^k such that m_{i_{j_r}} is an element of the basis B_i of M_i for all 1 \leq r \leq k and all 1 \leq i_{j_r} \leq n_i (fixed j_r). This is the free R-module on the set

\{ m_{1_j} \otimes \cdots \otimes m_{k_j} \mid m_{i_j} \in B_i \}

which indeed has cardinality \prod_{i=1}^k n_i.

I recommend writing down the particular case where k = 2 to clarify this as I’m sure the indices are confusing. Another way of showing that this set is a basis for M_1 \otimes \cdots \otimes M_k is to show that the free R-module over that set satisfies the universal property of M_1 \otimes \cdots \otimes M_k, namely that every k-multilinear maps from M_1 \times \cdots \times M_k to some R-module U factors uniquely through this set. Then since the universal property defines the tensor product of the M_i‘s uniquely up to isomorphism, we’re done.

As a corollary of this discussion, we get

Proposition: If M is a free R-module of rank n (for R a commutative ring), then, if B = \{ m_1, \ldots, m_n \} is a set of free generators for M, the set of k-tensors \{ m_{i_1} \otimes \cdots \otimes m_{i_k} \mid m_{i_j} \in B\} is a basis for T^k(M). In particular, T^k(M) is free of rank n^k.

Note that since a vector space of dimension n over a field K can be identified with the free K-module of rank n, this proposition (and the whole discussion) is applicable to vector spaces.

Reference: D. S. Dummit, R. M. Foote, Abstract Algebra, 3rd ed.

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