My discovery about mathematics

Naoto Meguro. Amateur.
MSC 2010. Primary 13C10 ;Secondary 13C99 .
Key Words and Phrases. A direct sum, countable sums.
The abstract. A direct sum of the commutative ring becomes a zero module.

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I indicate defect of the homological algebra. You may have to correct many theorems.

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Let A be a commutative ring.
Put e(1)=(1,0,0,…(infinitely)),e(2)=(0,1,0,0,…),…, ae(1)=(a,0,0,…),ae(2)=(0,a,0,0,…),….
0e(n)=0e(m)=(0,0,…) (n,m ∈N). Define ae(m)+a"e(m)=(a+a")e(m) (a,a"∈A, m∈N).
Define ∑_n∈Na(n)=a(1)+a(2)+…(infinitely)=…((a(1)+a(2))+a(3))+…(infinitely),
(∑_n∈N a(n))+(∑_n∈N b(n))=∑_n∈N(a(n)+b(n)),
F=⊕_n∈NAe(n)={∑_n∈N a(n)e(n) | a(n)∈A, ∃p∀n((n∈N)∧(n≥p)→(a(n)=0))}, 0"=∑_n∈N0 e(n). 0" is the zero of F.
Theorem 1. F={0"}.
Proof. For a∈A and m∈N,
∑_n∈N ae(m)=…((ae(m)+ae(m))+ae(m))+ae(m))+ …(infinitely)=…(2ae(m)+ae(m))+ae(m))+…(infinitely)
=2ae(m)+ae(m)+ae(m)+…(infinitely). (∑_n∈N -ae(m))+(∑_n∈N ae(m))=(-ae(m)+ae(m))+(-ae(m)+ae(m))+…
=(-ae(m)+2ae(m))+(-ae(m)+ae(m))+(-ae(m)+ae(m))+…=0e(m)+0e(m)+…=ae(m)+0e(m)+0e(m)+…
=0e(1)+0e(2)+…=0"=0e(1)+…+0e(m-1)+ae(m)+0e(m+1)+….
The elements of F are finite sums of ones like the last formula and are 0". ♦
Let M={m₁,m₂,…} be a countable set. Define e(m_i)=e(i).By theorem 1,F"=⊕_m∈M Ae(m)={0"}.
When M is an A-module, F"∋(∑_1≤n≤pa(n)e(m_n))+∑_n≥p+1 0e(m_n)→(∑_1≤n≤p a(n)m_n)+0∈M doesn't become a mapping.
Let L and L" be A-modules. If L×L" is a countable set, L⊗_AL"=⊕_{(x,y)∈L×L"} Ze((x,y))/0⊗_A0 is a zero module.
For 0→2Z→Z→Z/2Z→0 (exact),
2Z⊗_Z(Z/2Z)(=0)→Z⊗_Z(Z/2Z)(=0)→(Z/2Z)⊗_Z(Z/2Z)(=Z/2Z)→0 (exact) isn't formed.
You cannot get the tensor product except the cases like (A/I)⊗_AM=M/IM. M is an A-module and
I is an ideal of A.

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Generally,you cannot get projective resolution of the A module. You will be to treat only Noetherian ring A and
finitely generated A-modules. It may be enough to evolve the commutative algebra.