My discovery about mathematics

Put S={x|(x∈C)∧((Im x=0)∨(Re x=1/2))}, D={x|(x∈C)∧(Re x≥2)},
Z={f|∃h((h is a meromorphic function onC.)∧(h|_D=f)∧∀x((h(x)=0)→(x∈S)))}.
If f,g∈Z,∃h,∃k, h and k are meromorphic functions and h|_D=f and k|_D=g.
h(x)=0⇒x∈S,k(x)=0⇒x∈S. hk is a meromorphic function and hk|_D=fg. h(x)k(x)=0⇒
h(x)=0 or k(x)=0 ⇒x∈S
So fg∈Z. Define c(f)=O for ∀f∈Z. For f,g∈Z,c(fg)=O=O+O=c(f)+c(g).
Put h(p,x)=1/(1-1/p^x)=e^{(log p)x}/(e^{(log p)x}-1) (p is a prime number.)
h(p,x) is a meromorphic function. h(p,x)≠0 for ∀x∈C. So f(p,x)=h(p,x)|_D ∈Z.
If Riemann hypothesis is true, ζ(x)|_D=f(2,x)f(3,x)f(5,x)…(infinitely)∈Z. c(ζ(x)|_D)=O then.
c(f(2,x)f(3,x))=c(f(2,x))+c(f(3,x)),c(f(2,x)f(3,x)f(5,x))=c(f(2,x))+c(f(3,x))+c(f(5,x)),…,
c(f(2,x)…f(p,x))=c(f(2,x))+…+c(f(p,x)),… Repeating this infinitely,you get
c(f(2,x)f(3,x)…(infinitely))=O+O+…(infinitely)≠O This is contradiction.
Riemann hypothesis is wrong? The hypothesis is nonsense by theorem 6,though.

Let F be a flat Z module. F is a finite set to define tensor products. F≅⊕_1≤i≤mF(n_i) (F(n)≅Z/nZ).
For n=n_i, 0→nF(n²)≅F(n)→F(n²)→F(n²)/nF(n²)≅F(n)→0 (exact).
0→F(n)⊗F→F(n²)⊗F→F(n)⊗F→0 (exact)
0→F(n)⊗F(n)→F(n²)⊗F(n)→F(n)⊗F(n)→0 (exact) 0→F(n)→F(n)→F(n)→0 (exact)
0→F(n)→0→F(n)→F(n)→0 (exact) F(n)=0 F=0
Z is torsionfree as a Z module but isn't a flat Z module though Z is a commutative semihereditary
ring. Z isn't a set from the first,though.

You may assume the axioms ∀m∀I((m∉I)→(∑_i∈I∪{m} x_i=(∑_i∈I x_i)+x_m)) and ∀m(∑_i∈{m} x_i=x_m).
Put N={1,2,3,…}.
Theorem 1. ∑_i∈N x_i=x₁+x₂+…(infinitely)
Theorem 2. ∑_i∈N x=∑_i∈N-{1} x
Proof. ∑_i∈N-{1} x_i=x₂+x₃+…(infinitely)
So ∑_i∈N-{1} x=x+x+…(infinitely)=∑_i∈N x. ♦
Let L be a module.
Theorem 3. If x_i,y_i∈L for ∀i∈N, ∑_i∈N x_i+∑_i∈N y_i=∑_i∈N (x_i+y_i).
Let 0" be the zero of L. Assume that 0"≠b∈L. Put a=∑_i∈N0", d=∑_i∈N b,
d"=∑_i∈N (-b).
Theorem 4. a≠0"
Theorem 5. The real number theory isn't consistent.
Proof. Set up that L=R={r+∑_i∈I 1/2ⁱ |(r∈Z)∧(I⊂N)},b=0+∑_i∈N 1/2ⁱ. 0"=0+∑_i∈∅ 1/2ⁱ=0
0"=0"b=0(0+∑_i∈N 1/2ⁱ)=00+∑_i∈N 0/2ⁱ=∑_i∈N 0=∑_i∈N 0"=a This is contradiction by theorem 4.♦