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Let $k$ be a commutative ring and $V$ be a $k$-module (possibly graded). Let $TV$ denote the free (graded) algebra on $V$. The following question could just as well be applied to the free (graded) commutative algebra on $V$, but since it is a quotient of the former it suffices to consider the non-commutative case.

For a free $V$, what $TV$ is is easy to describe: by choosing a basis, we get an isomorphism $V\cong k^{(I)}$, and then $TV\cong K\langle x_i:i\in I\rangle$ where this denotes polynomials in non-commuting variables.

But what about non-free $V$? Then not every module is free, but at least it is a quotient of a free module. So we have an exact sequence $0\to E\to F\to V\to 0$ where $F$ is free; does this help in describing $TV$ as a quotient of $TF$ (which is easy since $F$ is free)? I'm unsure because the tensor product being a right exact functor, I don't expect there to be an associated short exact sequence of tensor algebras.

I would ideally like to see some concrete examples as well... I get the impression that $T(\mathbb{Z}/2)$ (over $\mathbb{Z}$) is $\mathbb{Z}\oplus \mathbb{Z}/2[X]^+$ where $+$ denotes the augmentation ideal, and similarly for the other cyclic rings, for example.

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    If $V = F / J$ (for some submodule $J$ of a $k$-module $F$), then $TV = TF / \left$, where $\left$ is the ideal of $TF$ generated by $J$. How useful this description is might depend on how explicit $J$ is. – 2017-02-15
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    @darijgrinberg why is this the case? Don't you run into problems with the fact that tensor is not left exact? – 2017-02-16
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    I'm going to take the easy way out and argue that this follows from Theorem 32 in my *A few classical results on tensor, symmetric and exterior powers* ( http://www.cip.ifi.lmu.de/~grinberg/algebra/tensorext.pdf ) and the surjectivity of $TF \to TV$. But there are definitely better sources. – 2017-02-16
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    @darijgrinberg thanks. There are definitely more *well-established* sources, but I doubt they go into the amount of detail that you do. – 2017-02-17

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