My discovery about mathematics

                               1

I indicate defect of the homological algebra and the algebraic geometry by a paradox
of the set theory. Generally, countable union of sets isn't a set. You must not use the
tensor product and the polynomial ring. The algebraic geometry you know isn't consistent.
The standard mathematics may not be consistent. You must reexamine many theorems soon.

                               2

Let X be a countable and consistent axiom system of the standard mathematics (ZFC etc.).
Let d be a free individual symbol. The mathematical axioms in X don't include d.
Theorem 1. If c_m-(c₀∪…∪c_m-1)≠∅ for ∀m∈N,
there isn't a set c for which ∀M((M is a model of X.)⇒∀x((M |= x∈c) ⇔
(M |= x∈c₀)∨(M |= x∈c₁)∨…(infinitely))).
Proof. Assume that such a set c(=c₀∪c₁∪…(infinitely)) exists.
∃b_m(b_m∈c_m-(c₀∪…∪c_m-1)⊂c) The axiom system
X∪{d∈c}∪{d∉c₀}∪…∪{d∉c_m-1} has a model M_m for which M_m |= d=b_m
and is consistent for ∀m(∈N). So X∪{d∈c}∪{d∉c₀}∪{d∉c₁}∪…(infinitely) is consistent
and has a model M. M is a model of X.
So (M |= d∈c)⇔(M |= d∈c₀)∨(M |= d∈c₁)∨…(infinitely).
But (M |= d∉c₀)∧(M |= d∉c₁)∧…(infinitely). This is contradiction.♦
You cannot evolve the measure theory.
Let A be a commutative ring in the mathematics by X.
Theorem 2. For the infinite set S,you must not use A^(S) as a set.
Proof. Put A₀={0}^S. A_m={f|(f∈A^S)∧∃s₁…∃s_m∀s((f(s)≠0)↔(s=s₁)∨…∨(s=s_m))}
is a set. You can set up that b₁∈S, b₂∈S,…(infinitely) (b_i≠b_k for i≠k.).
Define f_m(b_i)=1 (1≤i≤m) f_m(x)=0 (x∈S-({b₁,…b_m}). f_m∈A_m-(A₀∪…∪A_m-1)
Let M be an arbitrary model of X. If A^(S) exists in the mathematics by X,
(M |= f∈A^(S))⇔ (M |= f∈A₀∪A₁∪…)⇔(M |= f∈∪{A₀,A₁,…})⇔ ∃m(M |= f∈A_m)
⇔(M |= f∈A₀)∨(M |= f∈A₁)∨…(infinitely)
Such a set A^(S)=A₀∪A₁∪…(infinitely) doesn't exist by theorem 1.♦
You cannot get the projective resolution of the infinite A module L by A^(L)→L→0 (exact).
Put I=∃g(g:A^(L)∋f→∑_x∈L f(x)x∈L). X |- I ⇒ M |= I M is an arbitrary model of X.
∑_x∈L f(x)x must become a finite sum for every model of X.
Theorem 3. For A modules L and N, L⊗_AN isn't a set if L or N isn't a finite set.
Proof. If L⊗_AN is a set, ∪L⊗_AN=A^(L×N) (or Z^(L×N)) is a set.
This is contradiction by theorem 2 if L×N isn't a finite set.♦
Let g∈L"^L×N be an A-balanced mapping.
Put I"=∃h((h:L⊗_AN∋∑_(x,y)∈L×Nf(x,y)x⊗y →∑_(x,y)∈L×Nf(x,y)g(x,y) ∈L")∧(f∈A^(L×N))).
X |- I" ⇒ M |= I" M is an arbitrary model of X. ∑_(x,y)∈L×Nf(x,y)g(x,y) must become a finite sum for M.
∃h∃k(h=k⊗_ZC) is false by theorem 3 if C is a set. Hodge conjecture is written like
∀w(∃h∃k((h=k⊗_ZC)∧Q(w,h,k))→Q"(w)). This is true if C is a set.
L,N and A must be finite sets to define Tor^A_m(L,N).
Theorem 4. The polynomial ring A[t] isn't a set.
Proof. Put P_m=A+At+…At^m. t^m∈P_m-(P₀∪…∪P_m-1)
Let M be an arbitrary model of X.
(M |= x∈A[t])⇔(M |= x∈P₀)∨(M |= x∈P₁)∨…(infinitely)
A[t] isn't a set by theorem 1.♦
Similarly,A[t₁,…,t_n] isn't a set.
Put 1={∅}, n+1=n∪{n}={∅,1,2,…,n}. These are sets.
Theorem 5. N={1,2,…} isn't a set.
Proof. {n}-({1}∪…∪{n-1})={n}≠∅ Let M be an arbitrary model of X.
(M |= n∈N)⇔(∃m((m∈N)∧(M |= n∈{m}))⇔(M |= n∈{1})∨(M |= n∈{2})∨… N isn't a set by theorem 1.♦
Theorem 6. Z and Q aren't sets.
Proof. If Z=N×N/～ ((a,b)～(c,e)⇔a+e=b+c) is a set, ∪Z=N×N is a set. N is a set then.
This is contradiction by theorem 5. If Q=Z×(Z-{0})/～ ((a,b)～(c,e)⇔ae=bc) is a set,
∪Q=Z×(Z-{0}) is a set. Z is a set then. This is contradiction.♦
Theorem 7. The real number isn't a set.
Proof. If the real number r={a,b} defined by Dedekind cut is a set,∪r=a∪b=Q is a set.
This is contradiction by theorem 6.♦
Theorem 8. R isn't a set.
Proof. If R is a set, Q"={{a,b}|({a,b}∈R)∧∃x∀y((y∈a)↔(y≤x))} is a set.
Q={x|∃c((c∈Q")∧∃a∃b((c={a,b})∧∀y((y∈a)↔(y≤x))))} is a set by the axiom of replacement.
This is contradiction by theorem 6.♦
C isn't a set too.
N,Z,Q,R,C must not appear in the standard mathematics whose object domain includes only sets.
The millennium prize problems except N≠NP problem are nonsense in it. For example,
Riemann hypothesis ∀x∀y((x∈R)∧(y∈R)∧ζ(x+y√-1)=0)→(x=1/2)∨(y=0)) is nonsense.
You cannot construct the mathematics by the set theory.

                         3

Are these troubles fatal or avoidable? I hope to hear your opinions.
The physics isn't consistent by theorem 7. Every phenomenon like
the free energy or the free productivity is possible. This is an answer to the 6th problem
of Hilbert.

Theorem 8. UFD which isn't a field isn't a set.
Proof. Let P be the set of the prime elements of A. Put B₀={e|(e∈A)∧(e|1)},
B_m={x|(x∈A)∧∃q₁…∃q_m((q₁∈P)∧…∧(q_m∈P)∧(x=q₁…q_m))}
Let q(≠0) be a prime element of A. If A is a UFD, q^m∈B_m-({0}∪B₀∪…∪B_m-1)
Let M be an arbitrary model of X. If A is a UFD,
∀x((M |= x∈A)⇔(M |= x∈{0})∨(M |= x∈B₀)∨(M |= x∈B₁)∨(M |= x∈B₂)∨…(infinitely)
A isn't a set by theorem 1.♦ The regular local ring isn't a set?

                                      1

I show a paradox about the zeta function. You can prove and can deny BSD conjecture if L(E,1)=0.
Riemann hypothesis is nonsense too.

                                      2

Put F=Z[[X]],O=Z[X]. Define A+B={a+b|a∈A,b∈B} for A,B∈F/O.
O is the zero of the module F/O.
Theorem 1. O+O+…(infinitely)≠O
Proof. O+O+…(infinitely)={c₁+c₂+…(infinitrly)|c₁∈O,c₂∈O,…}
∋1+X+X²+…(infinitely)∉O ♦
Put D={x|(x∈C)∧(Re(x)≥2)},S={x|(x∈C)∧((Im(x)=0)∨(0≤Re(x)≤1))}.
Let p₁=2,p₂=3,…p_m,… be the prime numbers.
Let g(p,t) be a meromorphic function about t on C for which g(p,t)≠0 for ∀t∈C.
Let f(P,t) be a meromorphic function about t on C for which ∀x((x∈D)→(f(P,x)=∏_p∈Pg(p,x))).
Theorem 2. Theory of ζ(x) isn't consistent.
Proof. Put g(p,t)=1/(1-1/p^t),
V={h(t)|(h(t) is a meromorphic function on C)∧∀x((x∈C)∧(h(x)=0)→(x∈S))}.
Define s(h(t))=O for∀h(t)∈V.
f({2},t)=g(2,t)≠0 for ∀t∈C. f({2},t)∈V s(f({2},t))=O
f({2}∪{3},t)=g(2,t)g(3,t)≠0 for ∀t∈C.
f({2}∪{3},t)∈V s(f({2}∪{3},t))=O=O+O=s(f({2},t))+s(f({3},t))
f({p₁}∪…∪{p_m},t)=g(p₁,t)…g(p_m,t)≠0 for∀t∈C. f({p₁}∪…∪{p_m},t)∈V
s(f({p₁}∪…∪{p_m},t))=O=O+…+O=s(f({p₁},t))+…+s(f({p_m},t)),…
Repeating this infinitely, you get
s(ζ(t))=s(f({p₁}∪{p₂}∪…(infinitely),t))=s(f({p₁},t))+s(f({p₂},t))+…(infinitely)
=O+O+…(infinitely)≠O ζ(t)∉V ∃x((x∈C)∧(ζ(x)=0)∧(x∉S))
This is contradiction by the known theorem.♦ Riemann hypothesis is nonsense.
Theorem 3. You can prove and can deny BSD conjecture if L(E,1)=0.
Proof. assume that L(E,1)=0. You can set up that f({p₁}∪{p₂}∪…(infinitely),t)=L(E,t)
by a suitable g(p,t).
Put V"={h(t)|(h(t) is a meromorphic function on C)∧∀m((m∈N)∧(t=1 is a zero point
of h(t) whose order is m.)→q(m))}, s"(h(t))=O for ∀h(t)∈V".
f({p₁}∪…∪{p_m},1)=g(p₁,1)…g(p_m,1)≠0
f({p₁}∪…∪{p_m},t)∈V" s"(f({p₁}∪…∪{p_m},t)=O=O+…+O=f({p₁},t)+…+f({p_m},t)
So s"(L(E,t))=s"(f({p₁}∪{p₂∪…(infinitely),t))=s"(f({p₁},t))+s"(f({p₂},t))+…(infinitely)
=O+O+…(infinitely)≠O. L(E,t)∉V" ∃m((m∈N)∧(t=1 is a zero point of L(E,t)
whose order is m.)∧(¬q(m))
If you put q(m)=(m≠rank(E(Q))), BSD conjecture is proved.
If you put q(m)=(m=rank(E(Q))), BSD conjecture is denied.♦
Theory of L(E,t) isn't consistent.

                          3

The above means that you must not use infinite symbols like {p₁}∪{p₂}∪…(infinitely)
to express functions. One which isn't consistent isn't only theory of zeta functions but
the standard mathematics. Actually,N={1}∪{2}∪… and {p₁}∪{p₂}∪… aren't sets.
I have written about it in another paper. The physics isn't consistent too.
Every phenomenon like the free productivity is possible. This is an answer to the 6th
problem of Hilbert.