# Field on R^3 and isomorphism between C and R

Hey Guys ;

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

A) Yes we could define it as follows :

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

So , in the same steps we could carry this for $$\mathbb R^n$$.

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Q) Is $$f$$ exist ?

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

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Q) How we can show that $$(\mathbb C , +)$$ and $$(\mathbb R ,+)$$ are isomorphic ?

A) Both are vector spaces over the field $$\mathbb Q$$, 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.

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 $$\mathbb C , \mathbb R$$ are not equal ..? and also how could we prove that $$\mathbb R^3 , \mathbb R$$ are isomorphic ???
Need more details ...

A) This requires some set theory, especially cardinality.

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

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

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My Questions :

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

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

$$\mathbb R^n \cong \mathbb R \cong \mathb C$$.

3) Is $$\mathbb R^n$$ is a finite extension for $$\mathbb R$$.

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matt grime
Homework Helper

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.

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 $$\mathbb R^n$$??for me I need more details here ...

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

The field $$(\mathbb R^n, +,\cdot)$$ is an extension of $$(\mathbb R,+,\cdot)$$ or not ?? that what I meant

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matt grime
Homework Helper
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.

What do you think ?

Could we define a field on $$\mathbb R^n$$ ??

If so , how ???

matt grime