I am searching an example of (non-negative) chain complexes over a commutative ring $R$ which are not isomorphic but homotopy equivalent. Such an example could come from Algebraic Topology (for instance singular homology) but not necessarily.
Thank you!
Take any abelian group $\rm A$, then any two different free resolutions of $\rm A$ are homotopy equivalent.
Here is another example from algebraic topology. The singular chain complexes of the real line and the one point space are chain homotopic (our commutative ring will be taken to be the integers) , but they are definitely not isomorphic because the module of q-dimensional chains will be uncountable in case your space is the real line and countable in case your space is the one point space
From algebraic topology, the different ways of obtaining ordinary homology usually give homotopic, non-isomorphic chain complexes, e.g. the cellular chain complex and the singular chain complex of a space.
To get an example, I would try computing these complexes for a simple but non-trivial space.
Here is a specific example. As Damien L points out, any two free resolutions of some fixed module are homotopy equivalent. More generally, any two projective resolutions of a fixed module are homotopic as well.
Take $R = \mathbb{Z}\times\mathbb{Z}$ and $M = \mathbb{Z}\times\{0\}\subseteq \mathbb{Z}\times \mathbb{Z}$. Then $M$ is an ideal of $R$ and has an $R$-module structure, and it is projective, being the direct summand of a free $R$-module (it is not free).
Then $0\to M\to M\to 0$ is a projective resolution and $0\to \{0\}\times\mathbb{Z}\to \mathbb{Z}\times \mathbb{Z}\to M\to 0$ is another projective resolution, so they must be homotopic. It's a good idea to write down the homotopy.
Take $R=SX^{*}$ to be the symmetric algebra on a finite dimensional vector space $X^{*}$ on the ground field $\mathbb K$ (take $\mathbb R$ or $\mathbb C$ for simplicity). Let us consider for the time being just one grading, called the cohomological grading : w.r.t. the cohomological grading $R$ is concentrated in degree $0$.
The bar resolution $(\mathcal B(R;\mathbb K),d)$ of $\mathbb K$ as left $R$ module (the action is just the evaluation of polynomials at $0$) is given by $\mathcal B(R;\mathbb K)=\oplus_{n\geq 0} R \otimes R[1]^{\otimes n} \otimes\mathbb K$ with natural differential $d$ (you can find it on standard books on homological algebra or alg. topology). Note that $n$ is not a grading; it is called "weight". Such resolution is homotopy equivalent to the Koszul complex $K(R;\mathbb K)=R\otimes \wedge(X^{*})$, denoting by $\wedge(X^{*})$ the exterior algebra over the ground field (concentrated in negative cohomological grading) and endowed with differential induced by the Euler vector field.
You can write an explicit morphism $\Phi: K(R;\mathbb K) \rightarrow \mathcal B(R;\mathbb K)$ given by $\Phi(1\otimes x_1 \otimes \dots \otimes x_p)= \sum_{p\in \Sigma(p)} 1 \otimes x_{\Sigma(p)} \otimes \dots \otimes x_{\sigma(p)}$ (on monomials $x_i \in X^{*} $ and summing over $p$-permutations)
