- #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 \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.
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 \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.
====================================================================
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 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.
I'm waiting your replays .
I need to discuss this problem with you. 1st of all , I'm going to post some posts about some questions with answers .
=======================================================================
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.
=======================================================================
Q) Is f exist ?
Yes, we can show that as (\mathbb C , +) and (\mathbb R ,+) are isomorphic depending on axiom of choice.
=======================================================================
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.
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 \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.
====================================================================
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 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.
I'm waiting your replays .
