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.