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Let $K⊆M⊆L$, such that $[L:K]=n$ for some $n∈mathbbN$. Prove $[M:K]$ is well-defined.
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I wanted to delve into Galois theory as soon as possible, so I started reading a book on the topic without previous knowledge of other areas of mathematics which are often needed to understand Galois. Because of this I do not understand the usual definition of a field extension (which makes use of the concept of a Vector Space), but rather employ a different (I believe equivalent) definition:
Let $L, K$ be subfields of $mathbbC$ such that $K⊆L$
Let $A = $ $x_1, ..., x_n$ $⊆L$ be a set of linearly independent elements over $K$ such that $L = $ k_i∈K$, then $[L:K]=n$. We call $A$ a basis for $L:K$.
From such definition, is it possible to prove that if $K⊆M⊆L$, such that $[L:K]=n$ for some $n∈mathbbN$ then $[M:K]$ is well defined. In other words, that there exists a set $B⊆M$ which is a basis for $M:K$.
I'm aware my definition only speaks of finite field extensions.
I would really appreciate any help/thoughts.
abstract-algebra field-theory galois-theory extension-field
edited Jul 19 at 21:34
Jyrki Lahtonen
105k12161355
asked Jul 19 at 20:29
Leo
689416
 |Â
show 1 more comment
up vote
1
down vote
favorite
I wanted to delve into Galois theory as soon as possible, so I started reading a book on the topic without previous knowledge of other areas of mathematics which are often needed to understand Galois. Because of this I do not understand the usual definition of a field extension (which makes use of the concept of a Vector Space), but rather employ a different (I believe equivalent) definition:
Let $L, K$ be subfields of $mathbbC$ such that $K⊆L$
Let $A = $ $x_1, ..., x_n$ $⊆L$ be a set of linearly independent elements over $K$ such that $L = $ k_i∈K$, then $[L:K]=n$. We call $A$ a basis for $L:K$.
From such definition, is it possible to prove that if $K⊆M⊆L$, such that $[L:K]=n$ for some $n∈mathbbN$ then $[M:K]$ is well defined. In other words, that there exists a set $B⊆M$ which is a basis for $M:K$.
I'm aware my definition only speaks of finite field extensions.
I would really appreciate any help/thoughts.
abstract-algebra field-theory galois-theory extension-field
edited Jul 19 at 21:34
Jyrki Lahtonen
105k12161355
asked Jul 19 at 20:29
Leo
689416
1
The dimension of a vector space is well defined, and it is the cardinality of any of its basis.
– Berci
Jul 19 at 20:36
All subfields of $BbbC$ contain $BbbQ$, so they have infinitely many elements. Therefore finite-fields was an inappropriate tag.
– Jyrki Lahtonen
Jul 19 at 21:35
Anyway, you may benefit from reviewing a bit of linear algebra. $L$ is a finitely generated vector space over $K$, and $M$ is a subspace. Therefore $M$ is also finitely generated.
– Jyrki Lahtonen
Jul 19 at 21:40
@Berci. I'm aware that by using the concept of a Vector Space my question can be easily answered. Yet I was wondering if there was a way to prove the statement without making use of such concept.
– Leo
Jul 19 at 22:39
Well, looking for (the existence of) a 'basis' is typically a linear algebra task, you might use different terminology, seemingly avoiding vector spaces, but in the end, it's just the theorem you're going to use that any basis of a subspace extends to the ambient vector space.
– Berci
Jul 20 at 6:25
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I wanted to delve into Galois theory as soon as possible, so I started reading a book on the topic without previous knowledge of other areas of mathematics which are often needed to understand Galois. Because of this I do not understand the usual definition of a field extension (which makes use of the concept of a Vector Space), but rather employ a different (I believe equivalent) definition:
Let $L, K$ be subfields of $mathbbC$ such that $K⊆L$
Let $A = $ $x_1, ..., x_n$ $⊆L$ be a set of linearly independent elements over $K$ such that $L = $ k_i∈K$, then $[L:K]=n$. We call $A$ a basis for $L:K$.
From such definition, is it possible to prove that if $K⊆M⊆L$, such that $[L:K]=n$ for some $n∈mathbbN$ then $[M:K]$ is well defined. In other words, that there exists a set $B⊆M$ which is a basis for $M:K$.
I'm aware my definition only speaks of finite field extensions.
I would really appreciate any help/thoughts.
abstract-algebra field-theory galois-theory extension-field
edited Jul 19 at 21:34
Jyrki Lahtonen
105k12161355
asked Jul 19 at 20:29
Leo
689416
I wanted to delve into Galois theory as soon as possible, so I started reading a book on the topic without previous knowledge of other areas of mathematics which are often needed to understand Galois. Because of this I do not understand the usual definition of a field extension (which makes use of the concept of a Vector Space), but rather employ a different (I believe equivalent) definition:
Let $L, K$ be subfields of $mathbbC$ such that $K⊆L$
Let $A = $ $x_1, ..., x_n$ $⊆L$ be a set of linearly independent elements over $K$ such that $L = $ k_i∈K$, then $[L:K]=n$. We call $A$ a basis for $L:K$.
From such definition, is it possible to prove that if $K⊆M⊆L$, such that $[L:K]=n$ for some $n∈mathbbN$ then $[M:K]$ is well defined. In other words, that there exists a set $B⊆M$ which is a basis for $M:K$.
I'm aware my definition only speaks of finite field extensions.
I would really appreciate any help/thoughts.
abstract-algebra field-theory galois-theory extension-field
edited Jul 19 at 21:34
Jyrki Lahtonen
105k12161355
asked Jul 19 at 20:29
Leo
689416
edited Jul 19 at 21:34
Jyrki Lahtonen
105k12161355
edited Jul 19 at 21:34
Jyrki Lahtonen
105k12161355
edited Jul 19 at 21:34
Jyrki Lahtonen
105k12161355
105k12161355
asked Jul 19 at 20:29
Leo
689416
asked Jul 19 at 20:29
Leo
689416
asked Jul 19 at 20:29
Leo
689416
689416
1
The dimension of a vector space is well defined, and it is the cardinality of any of its basis.
– Berci
Jul 19 at 20:36
All subfields of $BbbC$ contain $BbbQ$, so they have infinitely many elements. Therefore finite-fields was an inappropriate tag.
– Jyrki Lahtonen
Jul 19 at 21:35
Anyway, you may benefit from reviewing a bit of linear algebra. $L$ is a finitely generated vector space over $K$, and $M$ is a subspace. Therefore $M$ is also finitely generated.
– Jyrki Lahtonen
Jul 19 at 21:40
@Berci. I'm aware that by using the concept of a Vector Space my question can be easily answered. Yet I was wondering if there was a way to prove the statement without making use of such concept.
– Leo
Jul 19 at 22:39
Well, looking for (the existence of) a 'basis' is typically a linear algebra task, you might use different terminology, seemingly avoiding vector spaces, but in the end, it's just the theorem you're going to use that any basis of a subspace extends to the ambient vector space.
– Berci
Jul 20 at 6:25
 |Â
show 1 more comment
1
The dimension of a vector space is well defined, and it is the cardinality of any of its basis.
– Berci
Jul 19 at 20:36
All subfields of $BbbC$ contain $BbbQ$, so they have infinitely many elements. Therefore finite-fields was an inappropriate tag.
– Jyrki Lahtonen
Jul 19 at 21:35
Anyway, you may benefit from reviewing a bit of linear algebra. $L$ is a finitely generated vector space over $K$, and $M$ is a subspace. Therefore $M$ is also finitely generated.
– Jyrki Lahtonen
Jul 19 at 21:40
@Berci. I'm aware that by using the concept of a Vector Space my question can be easily answered. Yet I was wondering if there was a way to prove the statement without making use of such concept.
– Leo
Jul 19 at 22:39
Well, looking for (the existence of) a 'basis' is typically a linear algebra task, you might use different terminology, seemingly avoiding vector spaces, but in the end, it's just the theorem you're going to use that any basis of a subspace extends to the ambient vector space.
– Berci
Jul 20 at 6:25
1
1
The dimension of a vector space is well defined, and it is the cardinality of any of its basis.
– Berci
Jul 19 at 20:36
The dimension of a vector space is well defined, and it is the cardinality of any of its basis.
– Berci
Jul 19 at 20:36
All subfields of $BbbC$ contain $BbbQ$, so they have infinitely many elements. Therefore finite-fields was an inappropriate tag.
– Jyrki Lahtonen
Jul 19 at 21:35
All subfields of $BbbC$ contain $BbbQ$, so they have infinitely many elements. Therefore finite-fields was an inappropriate tag.
– Jyrki Lahtonen
Jul 19 at 21:35
Anyway, you may benefit from reviewing a bit of linear algebra. $L$ is a finitely generated vector space over $K$, and $M$ is a subspace. Therefore $M$ is also finitely generated.
– Jyrki Lahtonen
Jul 19 at 21:40
Anyway, you may benefit from reviewing a bit of linear algebra. $L$ is a finitely generated vector space over $K$, and $M$ is a subspace. Therefore $M$ is also finitely generated.
– Jyrki Lahtonen
Jul 19 at 21:40
@Berci. I'm aware that by using the concept of a Vector Space my question can be easily answered. Yet I was wondering if there was a way to prove the statement without making use of such concept.
– Leo
Jul 19 at 22:39
@Berci. I'm aware that by using the concept of a Vector Space my question can be easily answered. Yet I was wondering if there was a way to prove the statement without making use of such concept.
– Leo
Jul 19 at 22:39
Well, looking for (the existence of) a 'basis' is typically a linear algebra task, you might use different terminology, seemingly avoiding vector spaces, but in the end, it's just the theorem you're going to use that any basis of a subspace extends to the ambient vector space.
– Berci
Jul 20 at 6:25
Well, looking for (the existence of) a 'basis' is typically a linear algebra task, you might use different terminology, seemingly avoiding vector spaces, but in the end, it's just the theorem you're going to use that any basis of a subspace extends to the ambient vector space.
– Berci
Jul 20 at 6:25
 |Â
show 1 more comment
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1
The dimension of a vector space is well defined, and it is the cardinality of any of its basis.
– Berci
Jul 19 at 20:36
All subfields of $BbbC$ contain $BbbQ$, so they have infinitely many elements. Therefore finite-fields was an inappropriate tag.
– Jyrki Lahtonen
Jul 19 at 21:35
Anyway, you may benefit from reviewing a bit of linear algebra. $L$ is a finitely generated vector space over $K$, and $M$ is a subspace. Therefore $M$ is also finitely generated.
– Jyrki Lahtonen
Jul 19 at 21:40
@Berci. I'm aware that by using the concept of a Vector Space my question can be easily answered. Yet I was wondering if there was a way to prove the statement without making use of such concept.
– Leo
Jul 19 at 22:39
Well, looking for (the existence of) a 'basis' is typically a linear algebra task, you might use different terminology, seemingly avoiding vector spaces, but in the end, it's just the theorem you're going to use that any basis of a subspace extends to the ambient vector space.
– Berci
Jul 20 at 6:25