Field on R^3 and isomorphism between C and R

In summary: Thus, C is an algebraic closed field.2) No, they are not isomorphic. The field on R^n is just isomorphic to the field on C, as we showed.3) Yes, R^n is a finite extension of R.
  • #1
GreenBeret
5
0
Hey Guys ;

I need to discuss this problem with you. 1st of all , I'm going to post some posts about some questions with answers .
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Q) Could we define a multiplication operation on [tex]\mathbb R^3[/tex] to have a field on it ??

A) Yes we could define it as follows :

take a group isomorphism [tex]f:\mathbb R^3 \rightarrow \mathbb R[/tex] and define [tex]\cdot[/tex] for each [tex]a,b \in \mathbb R^3[/tex] as [tex]a\cdot b = f^{-1} (f(a)\cdot f(b))[/tex] .Thus, we get [tex]\mathbb R^3[/tex] equipped with a multiplication, and in fact this is a field.

So , in the same steps we could carry this for [tex]\mathbb R^n[/tex].

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Q) Is [tex] f[/tex] exist ?

Yes, we can show that as [tex](\mathbb C , +)[/tex] and [tex](\mathbb R ,+)[/tex] are isomorphic depending on axiom of choice.

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Q) How we can show that [tex](\mathbb C , +)[/tex] and [tex](\mathbb R ,+)[/tex] are isomorphic ?

A) Both are vector spaces over the field [tex]\mathbb Q[/tex], and they have the same dimension.So they are isomorphic (use the base of R over Q, and C over Q, that exists by the axiom of choice; you cannot write it out explicitly...)

Need the procedure of that.

Q) The answer :

Let V be a vector space (over some field K)
Use Zorn's lemma [there is no procedure for finding bases in general vector spaces]:
the set of all linearly independent sets is such that if we have a chain of them, the union
is an upperbound. So Zorn applies; there is a maximal linearly independent subset and
this can be shown to be a base for the vector space.
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Q) I know the dimension of [tex] \mathbb C , \mathbb R [/tex] are not equal ..? and also how could we prove that [tex] \mathbb R^3 , \mathbb R [/tex] are isomorphic ?
Need more details ...

A) This requires some set theory, especially cardinality.

Dimension doesn't only depend on the vector space [tex]V[/tex], but also from the base field [tex]K[/tex]. For example, as a complex vector space [tex]\mathbb C[/tex] has dimension [tex]1[/tex], as a real vector space it has dimension [tex]2[/tex] and as a rational vector space it has dimension [tex]\infty[/tex].
The effect that [tex]\mathbb R \cong \mathbb R^3[/tex] as [tex]\mathbb Q[/tex]-vector spaces (not as [tex]\mathbb R[/tex]-vector spaces!) is similar to the "fact" there are equally many prime numbers than integers (there are simply countable infinite many).

Let [tex]V,W[/tex] be vector spaces over a field [tex]K[/tex] and having bases [tex]\mathcal V = \{v_i\}[/tex] and [tex]\mathcal W = \{w_j\}[/tex] (both of them possibly infinite). If there is a bijection [tex]b: \mathcal V \to \mathcal W[/tex], then this gives an isomorphism [tex]\bar b[/tex] of [tex]K[/tex]-vector spaces, defined by [tex]\bar b \left( \sum k_i v_i \right) = \sum k_i b(v_i)[/tex] for [tex]k_i \in K[/tex] (where the sum is finite/all but finitely many summands are [tex]0[/tex]; note that a basis is defined to reach any elements as a finite sum).
So vector spaces are isomorphic if a set of bases is isomorphic (which for sets means: there is a bijection). Now let [tex]K[/tex] be an infinite field and [tex]V,W[/tex] be [tex]K[/tex]-vector spaces such that [tex]|K| < |V| = |W|[/tex] (here [tex]|\cdot |[/tex] means cardinality, so this means: there is a bijection [tex]V \to W[/tex] as sets [not necessarily as vector spaces!] and there is an injection [tex]K \to V[/tex] but not an injection [tex]V \to K[/tex] [again as sets, not spaces!]). Then by the axiom of choice and some set theory, any [tex]K[/tex]-basis [tex]\mathcal V[/tex] of [tex]V[/tex] has the same cardinality than [tex]V[/tex] itself, so [tex]|\mathcal V | = |V|[/tex]. Similar, [tex]|\mathcal W | = |W|[/tex], thus [tex]|\mathcal V | = |V| = |W| = |\mathcal W |[/tex], meaning that there is a bijection between the bases. By the previous said, this means that [tex]V[/tex] and [tex]W[/tex] are isomorphic as [tex]K[/tex]-vector spaces.
This applies for [tex]K = \mathbb Q[/tex], [tex]V = \mathbb R^m[/tex], [tex]W = \mathbb R^n[/tex] as well.

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Q) But you show that both of two vector spaces are isomorphic ? not as groups with additive?

A)V and W are isomorphic as additive abelian groups (a bijective linear map
is clearly a bijective homomorphism of the corresponding additive groups.

====================================================================

My Questions :

1) I saw some of post here that [tex] dim_{\mathbb Q} (\mathbb R)[/tex] is not countable , How can show that [tex] dim_{\mathbb C} ( \mathbb C)=1,dim_{\mathbb R} ( \mathbb C)=2[/tex].

2) Now if we show that we can define a field on [tex] \mathbb R^n[/tex] as above , we get a contrary to the fact [tex] \mathbb C [/tex] is an algebraic closed ? or the field on [tex] \mathbb R^n[/tex] is just isomorphic to [tex]\mathbb C[/tex] since we show above :

[tex] \mathbb R^n \cong \mathbb R \cong \mathb C[/tex].

3) Is [tex] \mathbb R^n [/tex] is a finite extension for [tex] \mathbb R[/tex].

I'm waiting your replays .

:smile:
 
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  • #2
Your answers:

1) By exhibiting bases. 1 is a basis of C over C, and 1,i are bases of C over R.

2) What you've done is to show that if we have a set S that is in bijection with R (as a set) we can define a field structure on S. But then that just makes S isomorphic as a field to R. This is an empty statement - it is just saying 'give elements of R different labels'.

3) You've not defined R^n as a field extension of R, so this question isn't valid.
 
  • #3
matt grime said:
Your answers:

1) By exhibiting bases. 1 is a basis of C over C, and 1,i are bases of C over R.

2) What you've done is to show that if we have a set S that is in bijection with R (as a set) we can define a field structure on S. But then that just makes S isomorphic as a field to R. This is an empty statement - it is just saying 'give elements of R different labels'.

3) You've not defined R^n as a field extension of R, so this question isn't valid.

For 2 :So, we have showed there a field on [tex] \mathbb R^n[/tex]??for me I need more details here ...

3) We know every finite extension of [tex]\mathbb R[/tex] is [tex]\mathbb R[/tex] itself or it's isomorphic to [tex] \mathbb C[/tex] .

The field [tex](\mathbb R^n, +,\cdot)[/tex] is an extension of [tex] (\mathbb R,+,\cdot)[/tex] or not ?? that what I meant :smile:
 
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  • #4
No, you have not 'shown there is a field structure on R^n' in any meaningful way.

You did not prove that your attempt to make R^n into a field by using an 'isomorphism' (in what category?) actually made R^n into a field. And even if it were successful, all it does is define an abstract field structure on the underlying set of R^n that makes it just a copy of R.
 
  • #5
What do you think ?

Could we define a field on [tex] \mathbb R^n[/tex] ??

If so , how ?
 
  • #6
Let me explain what I think you're trying to do.

Let F be a field. Let G be an abelian group. Suppose that G is isomorphic to The additive group of F, then by identifying G with the additive subgroup of F I can attempt to make G into a field in 'the obvious way'.

*If* that is succesful, then all you've done is create a field that is isomorphic to the one you started with.
 

1. What is a field on R^3?

A field on R^3 is a mathematical structure that consists of a set of numbers (such as real or complex numbers) and two operations, addition and multiplication, that satisfy certain properties. In the case of R^3, these operations are defined on three-dimensional vectors.

2. How is R^3 related to C?

R^3 and C (the set of complex numbers) are both examples of fields. R^3 is a field of real numbers, while C is a field of complex numbers. They are related because they both have similar algebraic structures and properties.

3. What is an isomorphism?

An isomorphism is a mathematical concept that describes a one-to-one correspondence between two mathematical structures. In the case of fields, an isomorphism between two fields means that they have the same algebraic structure and properties.

4. How is C isomorphic to R?

C and R are isomorphic because they both have the same algebraic structure of being fields. This means that they have similar properties, such as closure under addition and multiplication, and the existence of multiplicative inverses for non-zero elements.

5. Why is the isomorphism between C and R important?

The isomorphism between C and R is important in mathematics because it allows us to use the powerful tools and techniques developed for one field (such as complex analysis for C) in the other field. It also helps to deepen our understanding of both fields and their properties.

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