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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    $\begingroup$ If $V = F / J$ (for some submodule $J$ of a $k$-module $F$), then $TV = TF / \left<J\right>$, where $\left<J\right>$ is the ideal of $TF$ generated by $J$. How useful this description is might depend on how explicit $J$ is. $\endgroup$ – darij grinberg Feb 15 '17 at 15:41
  • $\begingroup$ @darijgrinberg why is this the case? Don't you run into problems with the fact that tensor is not left exact? $\endgroup$ – user46225 Feb 16 '17 at 10:17
  • $\begingroup$ 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 ( cip.ifi.lmu.de/~grinberg/algebra/tensorext.pdf ) and the surjectivity of $TF \to TV$. But there are definitely better sources. $\endgroup$ – darij grinberg Feb 16 '17 at 17:23
  • $\begingroup$ @darijgrinberg thanks. There are definitely more well-established sources, but I doubt they go into the amount of detail that you do. $\endgroup$ – user46225 Feb 17 '17 at 8:58

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