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Field on R^3 and isomorphism between C and R

  1. Jul 2, 2008 #1
    Hey Guys ;

    I need to discuss this problem with you. 1st of all , I'm gonna post some posts about some questions with answers .

    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].

    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.


    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.

    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.

    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 .

  2. jcsd
  3. Jul 2, 2008 #2

    matt grime

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    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.
  4. Jul 2, 2008 #3
    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:
    Last edited: Jul 2, 2008
  5. Jul 2, 2008 #4

    matt grime

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    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.
  6. Jul 2, 2008 #5
    What do you think ?

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

    If so , how ???
  7. Jul 2, 2008 #6

    matt grime

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    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.
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