Undergrad Free Abelian Groups .... Aluffi Proposition 5.6

[lavinia]

So what is the structure of Ras an abelian group under addition?

 

[fresh_42]

So what is the structure of Ras an abelian group under addition?

The factor group of the free abelian group generated by all real numbers by the subgroup generated by all relations $(p,q) = \{p \text{ and } q \text{ are represented by the same Dedekind cut}\}$.
Edit: Maybe not a satisfactory algebraic explanation but we have to define the relations which makes to real numbers equal.

 

[lavinia]

The factor group of the free abelian group generated by all real numbers by the subgroup generated by all relations $(p,q) = \{p \text{ and } q \text{ are represented by the same Dedekind cut}\}$.
Edit: Maybe not a satisfactory algebraic explanation but we have to define the relations which makes to real numbers equal.

I do not see why this is true.

 

[fresh_42]

Let $C = ℝ$ be the continuum and basis of the free abelian group $F$. Further let $ι : C → F$ be the embedding (of the alphabet) and $id: C → ℝ_+$ the identity. Then there is a unique group homomorphism $π: F → ℝ_+$ which extends $id$, i.e. $id = π \, ι$ because $F$ is free. $π$ is onto. The question is what kernel $π$ has or what makes $4 + 5 = 9$ since $4 + 5$ and $9$ are different elements in $F$. The Dedekind cuts were what occurs to me. Maybe there is a more algebraic way to define the equivalence classes.

 

[mathwonk]

mike is emil's son. (artin)

I will go out on alimb in such fine company and offer my free course notes as well, math 843 on this page:

http://alpha.math.uga.edu/~roy/

Of course Iearned from the sources you have in hand and even cribbed from some of them.

 

[fresh_42]

mike is emil's son. (artin)

... and a prof of mine once said: A mathematician's talent is transmitted to his son-in-law

 

[lavinia]

Let $C = ℝ$ be the continuum and basis of the free abelian group $F$. Further let $ι : C → F$ be the embedding (of the alphabet) and $id: C → ℝ_+$ the identity. Then there is a unique group homomorphism $π: F → ℝ_+$ which extends $id$, i.e. $id = π \, ι$ because $F$ is free. $π$ is onto. The question is what kernel $π$ has or what makes $4 + 5 = 9$ since $4 + 5$ and $9$ are different elements in $F$. The Dedekind cuts were what occurs to me. Maybe there is a more algebraic way to define the equivalence classes.

It seems to me that Dedekind cuts assume that you already know what addition is.

 

[fresh_42]

It seems to me that Dedekind cuts assume that you already know what addition is.

I thought one needs the order in ℝ and the embedding of ℚ. But you are right, this breaks the definition requirements only down on ℚ.
So we are left with the interesting question: What is addition? Why are 4+5 and 9 equivalent? I have to think about it. I never really lost a thought on it. (PF is an ever lasting fount of challenges ...)

Edit: How about defining $(a,0) \sim a$ for all $a \in ℝ$ and then $(a,b) \sim c ⇔ \nexists d : ( (a,b) \sim d)$ proceeding by transfinite induction on the well-ordering of $ℝ$? I'm almost certain that it cant't be done without the axiom and the ordering. But I have to admit that I'm no logician.

Edit2: Perhaps some more care with the definition is needed which I consider a technical issue (well definition, commutativity, inverse).

 

[mathwonk]

@fresh. well John Tate was Emil Artin's son in law, having married Karin Artin, so maybe.

 

[micromass]

As for the addition structure on $\mathbb{R}$. It might help to look at $\mathbb{R}$ as a $\mathbb{Q}$-vector space of dimension $2^{\aleph_0}$. So we could see it as something like $\bigoplus_{2^{\aleph_0}~\text{factors}} \mathbb{Q}$

 

[lavinia]

As for the addition structure on $\mathbb{R}$. It might help to look at $\mathbb{R}$ as a $\mathbb{Q}$-vector space of dimension $2^{\aleph_0}$. So we could see it as something like $\bigoplus_{2^{\aleph_0}~\text{factors}} \mathbb{Q}$

Right. Then one needs the structure of the rationals.