My discovery about mathematics

Naoto Meguro :Amateur. MSC2020:13A02.

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The theory of the polynomial ring isn't consistent in ZFC.
ZFC proves every conjecture of the algebraic geometry with its denial.

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Let R be a ring. R[x]=R⊕xR[x]. For ∀p(x)∈R[x], p(x)+0_R[x]=p(x).
xp(x)+x0_R[x]=xp(x). x0_R[x]∈xR[x]. x0_R[x]=0_xR[x}. If x∈R[x], x0_R[x]=0_R[x]. 0_R[x]=0_xR[x].
Theorem 1. The ring R[x] which satisfies R[x]=R⊕xR[x] and x∈R[x] doesn't exist in ZFC.
Proof. 0_R[x]=(0_R,0_xR[x])=(0_R,0_R[x])={{0_R,0_R},{0_R,0_R[x]}}∋{0_R,0_R[x]}∋0_R[x].
This is a contradiction by the axiom of regularity and the axiom of choice. ♦
The theory of R[[x]] and the theory of the integral functions aren't consistent in ZFC too.
Theorem 2. The set theory treating the set N isn't consistent.
Proof. Such a theory defines Z={(m,n)|m,n∈N}/~ ((m,n)~(m",n")⇔m+n"=m"+n)
and treats Z^N and Z[[x]] by the mapping Z^N∋(a,b,c,…)→a+x(b+x(c+…∈Z[[x]]
and the axiom of replacement and leads a contradiction in ZFC like theorem 1. ♦
The proposition including N may be nonsense.
For example, RH=∀n((n∈N)∧(n≥5041)→(∑_d|nd ≤e^γn log log n)) may be nonsense.
If you set up that the object domain includes only natural numbers, you can evolve
the natural number theory without using the set N. So Fermat conjectire isn't
nonsense and may be still unsolved for Taniyama-Weil conjecture may be nonsense.
You need the set N to define the real numbers. So the standard real number theory
isn't consistent by theorem 2. RH is nonsense in it.

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Many conjectures may be back to the starting points or may become nonsense.

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The definition of the mapping by the graph has defect.
The theory of ⊗ isn't consistent if you define the mapping by the graph.

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Let M be a module. Define f: M∋x→x+x∈M. Put F=Z/2Z and i=1_F.
Let G(g)={(x,g(x)) |(x∈A)∧(g(x)∈B)} be the graph of the mapping g: A∋x→g(x)∈B.
G(g⊗i)={(x⊗(1+2Z),y⊗(1+2Z)|(x,y)∈G(g)}.
Theorem 1. If you define the mapping by the graph, h: M∋x→x+x∈Im(f) ⇒ h=f.
Proof. G(f)={(x,x+x) |(x∈M)∧(x+x∈M)}={(x,x+x) | x∈M}={(x,x+x) |(x∈M)∧(x+x∈Im(f))}
=G(h). So h=f.♦
By the axiom of the equality, f=h ⇒ Im(f⊗i)=Im(h⊗i). (You will know that this is wrong.)
Theorem 2. If you define the mapping by the graph,the theory of ⊗ isn't consistent.
Proof. Put M=Z. M⊗F≅F. You may define M⊗F=F and x⊗(y+2Z)=xy+2Z in M⊗F=F.
f⊗i: M⊗F∋x⊗(1+2Z)→(x+x)⊗(1+2Z)∈M⊗F. f⊗i: F∋x+2Z→2x+2Z=0+2Z∈F. Im(f⊗i)≅{0}.
Im(f)⊗F=(2Z)⊗F≅F. You may define Im(f)⊗F=F and 2x⊗(y+2Z)=xy+2Z in Im(f)⊗F=F.
By theorem 1, f: M∋x→x+x∈Im(f). f⊗i:M⊗F∋x⊗(1+2Z)→(x+x)⊗(1+2Z)∈Im(f)⊗F.
f⊗i: F∋x+2Z→2x⊗(1+2Z)=x+2Z∈Im(f)⊗F=(2Z)⊗F=F. Im(f⊗i)=F. F≅{0}.
This is contradiction.♦
Theorem 3. The set theory which assumes the axioms of ZF and existence of the module Z
and defines the mapping by the graph isn't consistent.
Proof. Put M=Z. M, Im(f)=2Z={x+x |x∈M}, 1+2Z={x|(x∈M)∧(x∉2Z)}, F={2Z, 1+2Z}
and G(f)=G(h)={(x, x+x) |x∈M} are sets. So f=h.
The contradiction in the proof of theorem 2 is led when you define M⊗F=F and
x⊗(1+2Z)=x+2Z in M⊗F=F and (2Z)⊗F=F and (2x)⊗(1+2Z)=x+2Z in (2Z)⊗F=F. ♦
Theorem 4. The theory treating the free modules isn't consistent if you define the mapping
by the graph.
Proof. Z is a free module. M⊗N=Z^(M×N)/(0_M⊗0_N). Z^(M×N) and 0_M⊗0_N are free modules.♦
You may not be able to get the tensor products and the projective resolutions of the
modules in the consistent theory.

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You may have to reexamine the theorems proved by using ⊗.
Many conjectures may be back to the starting points or may become nonsense.