Notice the following tower of fields.
Base field is K=Q, intermediate field is L=Q(√3) and top field is M=Q(√7).
Field L=Q(√3)={a+b√3|a,b∈Q}.
Field M=Q(√3)={a+b√3+c√7+d√7x3=21|a,b,c,d∈Q}.
Note that the field M is a vector space over L and K. Indeed, from the expression M=Q(√3)={a+b√3+c√7+d√7x3=21|a,b,c,d∈Q} one can see that M as vector space over K has dimension 4.
That is [M:K]=4.
There is another way of expression M as a vector space over L.
Letting u1,u2 be elements of this form u1=a+b√3 and u2=c+d√3, one can express elements of M as u1+√7(u2).
That is [M:L]=2. Similarly [L:K]=2. Thus,
[M:K]=[M:L][L:K]=2∗2=4.
First theorem of Field extensions formalizes these observations.
Theorem 1:Let K,L,M be fields with K⊂L⊂M. Then [M:K] is finite if and only if both [M:L] and [L:K] are finite and in that case [M:K]=[M:L][L:K].
Proof is a generalization of above observations.
Assume that [M:K] is finite, then consider L as subspace of M over K. As a subspace of finite vector space [L:K] is finite. Any basis that spans M over K (for example, {a,b,c,d} in above fields), also spans M over L; hence, [M:L] is also finite.
Assume the vector space M over L has [M:L]=m and L over K has [L:K]=n, then we need to show that [M:K]=mn.
Let z∈M. z can be expressed as z=∑iuihi where hi∈L. And hi=∑jcijvj. Putting all this together, z=∑i∑jcijuivj. i,j range over m and n. This is the spanning set for M over K.
To show this is independent, we need to show that whenever ∑ijcijuivj=0 we need to have cij=0.
∑ijcijuivj can be rewritten as ∑ihiui where hi∈L. Since ui form a basis for M over L, we must have each hi=0 and since this means ∑cijvj=0 and independence of vj leads to each cij=0.
One consequence of this thoerem is that if the order of M/L is a prime, then this precludes any fields between M and L.
Field L=Q(√3)={a+b√3|a,b∈Q}.
Field M=Q(√3)={a+b√3+c√7+d√7x3=21|a,b,c,d∈Q}.
Note that the field M is a vector space over L and K. Indeed, from the expression M=Q(√3)={a+b√3+c√7+d√7x3=21|a,b,c,d∈Q} one can see that M as vector space over K has dimension 4.
That is [M:K]=4.
There is another way of expression M as a vector space over L.
Letting u1,u2 be elements of this form u1=a+b√3 and u2=c+d√3, one can express elements of M as u1+√7(u2).
That is [M:L]=2. Similarly [L:K]=2. Thus,
[M:K]=[M:L][L:K]=2∗2=4.
First theorem of Field extensions formalizes these observations.
Theorem 1:Let K,L,M be fields with K⊂L⊂M. Then [M:K] is finite if and only if both [M:L] and [L:K] are finite and in that case [M:K]=[M:L][L:K].
Proof is a generalization of above observations.
Assume that [M:K] is finite, then consider L as subspace of M over K. As a subspace of finite vector space [L:K] is finite. Any basis that spans M over K (for example, {a,b,c,d} in above fields), also spans M over L; hence, [M:L] is also finite.
Assume the vector space M over L has [M:L]=m and L over K has [L:K]=n, then we need to show that [M:K]=mn.
Let z∈M. z can be expressed as z=∑iuihi where hi∈L. And hi=∑jcijvj. Putting all this together, z=∑i∑jcijuivj. i,j range over m and n. This is the spanning set for M over K.
To show this is independent, we need to show that whenever ∑ijcijuivj=0 we need to have cij=0.
∑ijcijuivj can be rewritten as ∑ihiui where hi∈L. Since ui form a basis for M over L, we must have each hi=0 and since this means ∑cijvj=0 and independence of vj leads to each cij=0.
One consequence of this thoerem is that if the order of M/L is a prime, then this precludes any fields between M and L.
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