My discovery about mathematics

グロタンディーク先生が「収穫と蒔いた種と」で、ほのめかしている代数幾何学やホモロジー代数
やカテゴリーの理論の欠陥について素人なりに推測してみました。あの本を読みこなす手がかりに
なるかも知れません。

                                    1

Noetherian domains which don't coexist exist. Many theorems about Noetherian domains are false.

                                    2

Let X be a countable axiom system of the standard mathematics. Let d and N" be the individual symbols
which X doesn't include. Let a and A be the individual symbols for which
X⊃{a≠0, a∈A, ∀a"((a"∈A)→(a"a≠1)).}. Set up that (1≠0)∈X,(1+1≠0)∈X, (1+1+1≠0)∈X,…(infinitely).
You can set up a logical formula T(D,I) ⇔ D is an integral domain and I is an ideal of D.
T(D,I)=∀x∀y∀z((x∈D)∧(y∈D)∧(z∈D)→(x+y∈D)∧(xy∈D)∧(x+y=y+x)∧(xy=yx)∧((x+y)+z=x+(y+z))∧
((xy)z=x(yz))∧(x(y+z)=xy+xz)∧(0∈D)∧(1∈D)∧(x+0=x)∧(x1=x)∧(-x∈D)∧(x+(-x)=0)∧
((x≠0)∧(y≠0)→(xy≠0))∧((x∈I)∧(y∈I)→(x+y∈I)∧(-x∈I)∧(zx∈I)))∧(D⊃I)∧(D≠I).
Set up that T(A,{0})∧∀U((U≠∅)→(∀I((I∈U)→T(A,I))→∃I"((I"∈U)∧∀I((I∈U)∧(I⊃I")→(I=I")))))∈X.
A is Noetherian domain then. Set up the axiom of replacement and ¬∃n(n∈∅) in X.
Set up that ∀x∀y((y∈Ax)↔∃a"((a"∈A)∧(y=a"x)) ∈X. X|- ∀x∀w∃y(∀z((z∈y)↔∃n((z=A(w/aⁿ))∧(n∈x))) then.
A(d/aⁿ⁺¹)∋ad/aⁿ⁺¹=d/aⁿ. So A(d/aⁿ⁺¹)⊃A(d/aⁿ).
If A(d/aⁿ)=A(d/aⁿ⁺¹) and d∈A-{0},∃a"((a"∈A)∧(d/aⁿ⁺¹=a"d/aⁿ)). a" ad=d .a" a=1.
Theorem 1. Noetherian domain which doesn't coexist with the set N and isn't a field exists if X is consistent.
Proof. If X is consistent, X∪{d∈A-{0}, T(A,A(d/a¹)), T(A,A(d/a²)),…T(A,A(d/aⁿ))} has a model in which d=aⁿ⁺¹
and is consistent for ∀n∈N. So X"=X∪{d∈A-{0}, T(A,A(d/a¹)), T(A,A(d/a²)),…(infinitely)} is consistent and
has a model M. Let S be the object domain of M. Assume that N∈S. You may set up that N"=N.
By the axiom of replacement, M |= ∃U"(∀I((I∈U")↔∃n((I=A(d/aⁿ))∧(n∈N")))). M |=(N"=N≠∅). So M |=(U"≠∅).
M |= I∈U" ⇒ ∃n(M |= I=A(d/aⁿ), M |= n∈N") ⇒ M |= T(A,A(d/aⁿ)).
M |= ∀I((I∈U")→T(A,I)). M |= ∃T((T∈U")∧∀I((I∈U")∧(I⊃T)→(T=I))). M |= ∃m"((T=A(d/a^m"))∧(m"∈N")).
M |= m"+1∈N". M |= A(d/a^m"+1)⊃A(d/a^m")=T. M |= A(d/a^m"+1)∈U".
M |= T=A(d/a^m")=A(d/a^m"+1). M |= ∃a"((a"∈A)∧(a" a=1)). This is contradiction. So N∉S.
A in M coexists with the only elements of S. A in M is Noetherian domain which doesn't coexist with N.♦
Let A" be Noetherian domain which doesn't coexist with N and isn't a field and ch(A")=0 like A in M.
If X is consistent,A" exists. Put k={ n/n" |(n∈A")∧(n"∈A"-{0})} and C=k[x₁,…,x_m].
Theorem 2. A" and C don't coexist if X is consistent.
Proof. If C exists as a set, P={f |(f∈C)∧∃n(f=x₁ⁿ)} and N={n | x₁ⁿ∈P.}-{0} exist as sets.
If A" and C coexist, A" and N coexist.♦
Noetherian domains A" and F[x₁] (F is a field) don't coexist too.
Theorem 3. If C exists, X isn't consistent.
Proof. If C exists, k exists by the axiom of replacement and a surjection C∋f(x₁,…,x_m)→f(0,…,0)∈k.
A"={n/n" | (n/n"∈k)∧(n"=1)} exists. By theorem 2, X isn't consistent.♦
If C/p exists for p∈Spec(C), C=∪(C/p) exists. The coordinate rings do not always exist if X is consistent.
Theorem 4. For the infinite set J, A" and B=⊕_i∈JR (R≠{0}) don't coexist if X is consistent.
Proof. If B exists as a set, N={n | (n=Card({i | r_i≠0}))∧((r_i)_i∈J∈B)}-{0} exists as a set.♦
A" is an integral domain and isn't a field and is an infinite set and is a Z-module. But you cannot set up that
⊕_a"∈A" Z→A"→0 (exact). Put W=A"⊗_ZA". If W exists,∃W"(W=B"/W"). B"=⊕_{(x,y)∈A"×A"} Z. B"=∪W exists.
By W∋∑a_i⊗a"_i → ∑a_ia"_i ∈A", A" exists. This is contradiction by theorem 4.
Projective resolutions and tensor products of modules do not always exist.
For 0→A"→A"→0→0 (exact), 0→A"⊗_ZA"→A"⊗_ZA"→0→0 (exact) is false.
A" isn't a flat Z-module though na"≠0 for ∀n∈N and ∀a"∈A"-{0}.

                              3

You must confirm coexistence of modules befor you think a diagram including the modules.
The theory treating arbitrary modules isn't consistent for the modules may not coexist.

Theorem 1. There isn't a countable and consistent axiom system S(A) which define
Noetherian domain A including a(≠0) which isn't a unit of A.
Proof. Assume that such S(A) and A and a exist. Let d be a free individual symbol.
S(A) doesn't include d. S(A)∪{d≠0, d∈A, d/a∈A,d/a²∈A,…,d/aⁿ∈A} has a model
in which d=aⁿ and is consistent for ∀n∈N. So S(A)∪{d∈A, d/a∈A, d/a²∈A,…(infinitely)}
is consistent and has a model M in which M |= Ad⊂A(d/a)⊂A(d/a²)⊂…⊂A. M is a model of S(A).
A is a Noetherian domain in M. ∃m((m∈N)∧(M |= A(d/a^m)=A(d/a^m+1)). ∃a"(M |= a"(d/a^m)=d/a^m+1)
M is a model of S(A). A is a domain in M. M |= a"a=1. S(A) |- ∀b((b∈A)→(ba≠1)). M |= a"a≠1.
This is contradiction. ♦