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43m

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(Non trivial) coidempotents(Co$K$theory)
Moreover, as soon as a group G is nontrivial is should be obvious to you that $\mathbb CG$ has nontrivial coidempotents! 
47m

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(Non trivial) coidempotents(Co$K$theory)
The isomorphism type of that coalgebra depends only on the cardinal of the group and not on the group structure. 
1h

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(Non trivial) coidempotents(Co$K$theory)
How do you consider $\mathbb CG$ as a coalgebra? Its standard coalgebra structure is the one in which all elements of G are grouplikes and depends only on the cardinal of G, so it should be more or less expected that the answer to your last question is no... 
7h

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Weyl algebras $A_n(k)$ as tensor product of the first Weyl algebra
deleted 2 characters in body 
8h

answered  Weyl algebras $A_n(k)$ as tensor product of the first Weyl algebra 
14h

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What are finite groups $H$ such that $H^n(H,\mathbb{Q/Z}) \cong H_n(G,\mathbb{Z})$?
The question wants an isomorphism of a cohomology group and a homology group in the same degree (I would guess the OP did not write what he intended, in any case) 
14h

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What are finite groups $H$ such that $H^n(H,\mathbb{Q/Z}) \cong H_n(G,\mathbb{Z})$?
There is a shift in degree, though. 
23h

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(Non trivial) coidempotents(Co$K$theory)
over itself, so there are no nontrivial coidempotents. 
23h

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(Non trivial) coidempotents(Co$K$theory)
Your definition of idempotent in an algebra is simply that of an idempotent in the endomorphism álgebra of its right regular module; your definition of a coidempotent is, similarly, that of an idempotent in the álgebra of endomorphisms of the regular right comodule. In particular, coidempotents correspond to direct summands of the regular right comodule. In the case of comatrix coalgebras, the right regular comodule is a direct sum of n simple comodules —just as in the álgebra case); $\mathbb C[x]$ with the coalgebra structure which makes x primitive, is indecomposable as a comodule 
23h

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(Non trivial) coidempotents(Co$K$theory)
The álgebra of matrices with entries in an algebra is just the tensor product of that álgebra with $M_n(\mathbb C)$; it is natural to define the comatrix coalgebra with entries in a coalgebra C as the tensor product of C and the comatrix coalgebra (which is just the coalgebra dual to the álgebra $M_n(\mathbb C)$) 
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Weyl algebras $A_n(k)$ as tensor product of the first Weyl algebra
@Csa, your comment is rather unhelpful, as it leaves the OP to browse the whole book (which has a rather unsatisfying index...) to find what you believe is there. 
1d

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Weyl algebras $A_n(k)$ as tensor product of the first Weyl algebra
What have you tried? 
1d

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Bipartite Graph
This site is for questions related to math research. I will move yours to math.stackexchange.com, where it fits immensely better. 
1d

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is the tensor product of projective modules again projective?
This is, in fact, the argument given by Cartan and Eilenberg. 
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is the tensor product of projective modules again projective?
See Proposition 2.3 in Chapter IX of the book by CartanEilenberg. 
2d

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Homological criteria for finite generation and finite presentation of modules?
If $A$ is a nonnegatively graded and connected over a field $k$, and your modules are graded with bounded below graduations, then you can use $H_i (M)=Tor^A(k,M)$: a module is $FL_n$ if $H_i(M)$ is finite dimensional for $i \leq n$. If you want zero cohomology groups, you can notice that $H_i$ takes values in graded vector spaces, and compose it with the functor which localizes at the Serre subcategory of finite dimensional graded vector spaces —call $H'$ the composition; now a module is in $FL_n$ if $H_i'(M)=0$ for $i\leq n$. 
Apr 13 
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A simple proof of the Weyl algebra's rigidity.
It is not true that HH^1(A)=0 implies A is a projective bimodule. It is very easyy to construct counterexamples for that claim. (It is true that if H^1(A,M)=0 for all Abimodules M, then A is projective, but that is rather irrelevant here) In any case, the Weyl álgebras are definitely not projective as bimodules over themselves. 
Apr 13 
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Rising Sun Inequality (DunfordSchwartz maximal inequality)
It is rather rare that posting problems from textbooks here is acceptable. I don't think this is an instance of that. 
Apr 13 
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For which maps $S^1\to S^1$ is the winding number defined?
Any measurable function with $\infty$norm strictly less than $\text{radius}/2$ is in BMO and has BMOnorm strictly less than the radius. It'd be surprising that such a thing has a winding number! 
Apr 13 
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Geometric meaning of the Euler sequence on \mathbb{P}^n (Example 8.20.1 in Ch II of Hartshorne)
Notice that the identity matrix sometimes has zero trace —you need to be careful with the characteristic. 