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.

Naoto Meguro. Amateur.
MSC 2010. Primary 08A65 ;Secondary 26E99 .
Key Words and Phrases. Countable sums, the real number theory.
The abstract. The real number theory by countable sums isn't consistent.

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The usual algebra doesn't treat infinite sums of elements of the module. I try to think them and lead
contradiction of the real number theory etc..

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Put N"={0}∪N. Define ∑_n∈N" a(n)=a(0)+a(1)+…(infinitely) =…((a(0)+a(1))+a(2))+a(3))+…(infinitely)
and (∑_n∈N" a(n))+(∑_n∈N" b(n))=∑_n∈N" (a(n)+b(n)).
Let M be a Z-module. Put E(M)={∑_n∈N" c(n)|c(n)∈M} and 0"=∑_n∈N"0 (0∈M).
E(M) is a module whose zero is 0".
Theorem 1. c∈M⇒ 0"=c+0+0+0+…(infinitely).
Proof. E(M)∋∑_n∈N" c=…((c+c)+c)+c)+…=…((2c+c)+c)+…=2c+c+c+….
(∑_n∈N" -c)+(∑_n∈N" c)=(-c+c)+(-c+c)+…=(-c+2c)+(-c+c)+(-c+c)+…=0"=c+0+0+…(infinitely). ♦
Define (∑_n∈N" a(n))(∑_n∈N" b(n))=∑_n∈N"(∑_0≤k≤n a(k)b(n-k)).
Let A be a commutative ring. E(A) is a ring then.
Theorem 2. E(A)={0"}
Proof. By theorem 1, 0"=1+0+0+…(infinitely). If x∈E(A), x=(1+0+0+…(infinitely))x=0" x=0". ♦
You may define R={∑_n∈N" c(n)|(c(0)∈Z)∧∀n((n∈N)→(c(n)=0)∨(c(n)=1/2ⁿ))}. By theorem 2,
{0"}=E(Q)⊃R. R={0"}? 0"=E(k[x₁,…,x_m])⊃k[[x₁…,x_m]]={0"}?
{0"}=E(Z)⊃Z_p. Z_p={0"}? {0"}=E(C)∋∑_n∈N" 1/(n+1)^s (s∈C). ζ(s)=0" for ∀s∈C?

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What I do is only indication of contradiction. The mathematics using R or k[[x₁,…,x_m]] or Z_p is doubtful
until someone corrects the above defect of it. If an irrational number exists in the real world, the laws of nature allow
0.000…=1.000…=2.000…=…. I proposed an experiment about it to European Nuclear Society etc..