# Abstract Nonsense

## Tensor Product of Free Modules

Point of Post: In this post we discuss more particularly about the tensor product of free modules, and some of the consequences of this.

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Motivation

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We have proved theorems before to the effect that the tensor product of free modules are free. In this post we’d like to solidify this by discussing precisely how, given bases for two free modules, construct a basis on their tensor product. This will enable us to give an explicit example of elements of a tensor product which are not simple tensors.

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Tensor Product of Free Modules

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Let us first get out of way the obvious: tensor products of free modules are free. In particular, we have the following little gem:

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Theorem: Let $R$ be a unital ring. Then, $R^{\oplus\lambda}\otimes R^{\oplus\kappa}\cong R^{\oplus\lambda\kappa}$ as left $R$-modules.

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This follows immediately from our earlier proof that, in general, $M\otimes_R R^{\oplus\lambda}\cong M^{\oplus \lambda}$. But, what’d we really like is to be able to actually find a basis for such a tensor product given bases for the individual factors. In particular, we have the following:

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Theorem: Let $M$ and $N$ be free left $R$-modules with bases $\{m_\alpha\}_{\alpha\in\mathcal{A}}$ and $\{n_\beta\}_{\beta\in\mathcal{B}}$ respectively. Then, $\{m_\alpha\otimes n_\beta\}_{(\alpha,\beta)\in\mathcal{A}\times\mathcal{B}}$ is a basis for $M\otimes_R N$.

Proof: We have the following chain of isomorphisms

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$M\otimes_R N\to R^{\oplus\mathcal{A}}\otimes_R R^{\oplus\mathcal{B}}\to R^{\oplus(\mathcal{A}\times\mathcal{B})}$

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with

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$m_{\alpha_0}\otimes n_{\beta_0}\mapsto (\delta_{\alpha_0})\otimes(\delta_{\beta_0})\mapsto (\delta_{(\alpha_0,\beta_0)})$

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and so the inverse of the composition of these isomorphisms is an isomorphism $R^{\oplus(\mathcal{A}\times\mathcal{B})}\to M\otimes_R N$ taking $\delta_{(\alpha,\beta)}$ to $m_\alpha\otimes n_\beta$. But, $\{\delta_{(\alpha,\beta)}\}_{(\alpha,\beta)\in\mathcal{A}\times\mathcal{B}}$ is a basis for $R^{\oplus(\mathcal{A}\times\mathcal{B})}$ and since bases are preserved under isomorphism, the conclusion follows. $\blacksquare$

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With this theorem we can finally show that not all tensors are simple. Indeed, consider the very simple case of $\mathbb{R}^2\otimes_\mathbb{R}\mathbb{R}^2$. We have the canonical basis $\{e_1,e_2\}$ for $\mathbb{R}^2$, what we claim is that $e_1\otimes e_2+e_2\otimes e_1$ is not a simple tensor in $\mathbb{R}^2\otimes_\mathbb{R}\mathbb{R}^2$. Indeed, suppose for a second that it was, then we’d have that

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\begin{aligned}e_1\otimes e_2+e_2\otimes e_1 &=(a e_1+b e_2)\otimes (c e_1+d e_2)\\ &=ac (e_1\otimes e_1)+ad (e_1\otimes e_2)+bc (e_2\otimes e_1)+bd(e_2\otimes e_2)\end{aligned}

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But, we know that $\{e_i\otimes e_j\}_{i,j\in[2]}$ is a basis for $\mathbb{R}^2\otimes_\mathbb{R}\mathbb{R}^2$ so that this equality implies that $ac=bd=0$ and $ad=bc=1$. Clearly this cannot happen.

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print.

[4] Lang, Serge. Algebra. Reading, MA: Addison-Wesley Pub., 1965. Print.

[5] Grillet, Pierre A. Abstract Algebra. New York: Springer, 2007. Print.

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January 25, 2012 -

## 1 Comment »

1. […] it’s a ring map emanating from a field!) and since is an -map this implies that (since we know ) , but since the reverse inequality holds (by above discussion) we may conclude that as desired. […]

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