The direct sum

2017-07-31 04:55:33 | Mathematics

I indicate that theory of the direct sum isn't consistent by Gödel's theorem.
You must reexamine the homological algebra.

Assume that theory of the direct sum is consistent. Put A=Z/2Z={0",1"}.
Let X be a countable and consistent axiom system of theory of the direct sum.
Let d be a free individual symbol. The mathematical axioms in X don't include d.
You may set up that 0"≠1" ∈X and m≠n ∈X for m∈N, n∈N, m≠n and
∀x({(y,1")|y∈x}∈Ax) ∈X and ∀S∀f((f∈A(S))→(∑m∈S f(m)∈A)) ∈X.
Theorem 1. Theory of the direct sum isn't consistent.
Proof. X∪{Ad=A(d)}∪{1∈d}∪…∪{n∈d} has a model Mn for which
Mn |= d={1,2,…n} and is consistent for ∀n∈N.
So X∪{Ad=A(d)}∪{1∈d}∪{2∈d}∪…(infinitely) is consistent and has a countable model M.
Let D be the object domain of M.Define c∈c"⇔ M |= c∈c" for c∈D and c"∈D.
c=c" ⇔ c∈{c"}. So c=c" ⇔ M |= c=c" for c∈D and c"∈D. d(⊃N) becomes a countable set.
Put e={(x,1")|x∈d}. d∈D. e∈D. M |= e∈Ad=A(d). M |= ∑m∈d e(m)=∑m∈d 1" ∈A.
m∈d 1" ∈A. Put p=1"+1"+…(infinitely)=∑m∈c 1". c is an arbitrary countable set. If c"∈c,
p=1"+∑m∈c-{c"} 1"=1"+p. If p=1",1"=1"+1"=0". So p≠1". Similarly p≠0". p∉A.But p=∑m∈d 1" ∈A.
This is contradiction.♦

You must not use the direct sum. Generally, you cannot get the projective resolution of the module.
You cannot define Ext and Tor then. You cannot define the tensor product by direct sums. You must
rewrite the proofs of theorems got by using ⊕ or ⊗ like Fermat conjecture, Mordel conjecture,
Serre's problem etc.. Conjectures treating ⊕ or ⊗ like Tate conjecture etc. are provable nonsense.
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