Proving the Dimension Bound for Intermediate Field Extensions

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In summary, the conversation discusses the relationship between finite extensions $K$ and $F$, and intermediate extensions $L_1$ and $L_2$. It is shown that there is a basis of $L_1L_2$ over $L_1$ consisting of elements of $L_2$, and it is proven that $[L_1L_2:F]\leq [L_1:F]\cdot [L_2:F]$. The conversation also explores the concept of basis reduction and how it can be used to show that the dimension of $L_1L_2$ as a $L_1$-vector space is less than or equal to the dimension of $L_
  • #1
mathmari
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Hey! :eek:

Let $K$ be a finite extension of $F$ and let $L_1, L_2$ be intermediate extensions.
  1. Show that there is a basis of $L_1L_2$ over $L_1$ that consists of elements of $L_2$.
  2. Prove that $[L_1L_2:F]\leq [L_1:F]\cdot [L_2:F]$.
I have done the following:

  1. Let $[L_1:F]=n$, $[L_2:F]=m$.

    Let $a_1, a_2, \ldots , a_n$ be a basis for $L_1$ over $F$ and let $b_1, b_2, \ldots , b_m$ be a basis for $L_2$ over $F$.

    $L_1L_2$ is the smallest field that contains $L_1$ and $L_2$.

    So, we have that $$L_1L_2=F(a_1, a_2, \ldots , a_n, b_1, b_2, \ldots , b_m)=L_1(b_1, b_2, \ldots , b_m)$$

    Does this mean that $b_1, b_2, \ldots , b_m$ is a basis for $L_1L_2$ over $L_1$ that consists of elements of $L_2$ ? (Wondering)

    $ $
  2. We have that $F\subseteq L_1\subseteq L_1L_2$ so we get that $[L_1L_2:F]=[L_1L_2:L_1][L_1:F]$.

    We have that $[L_1L_2:L_1]=[L_1(b_1, b_2, \ldots , b_m):L_1]$. We have to show that this is smaller or equal to $[L_2:F]$, right?

    How could we show that? (Wondering)
 
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  • #2
mathmari said:
Hey! :eek:

Let $K$ be a finite extension of $F$ and let $L_1, L_2$ be intermediate extensions.
  1. Show that there is a basis of $L_1L_2$ over $L_1$ that consists of elements of $L_2$.
  2. Prove that $[L_1L_2:F]\leq [L_1:F]\cdot [L_2:F]$.
I have done the following:

  1. Let $[L_1:F]=n$, $[L_2:F]=m$.

    Let $a_1, a_2, \ldots , a_n$ be a basis for $L_1$ over $F$ and let $b_1, b_2, \ldots , b_m$ be a basis for $L_2$ over $F$.

    $L_1L_2$ is the smallest field that contains $L_1$ and $L_2$.

    So, we have that $$L_1L_2=F(a_1, a_2, \ldots , a_n, b_1, b_2, \ldots , b_m)=L_1(b_1, b_2, \ldots , b_m)$$

    Does this mean that $b_1, b_2, \ldots , b_m$ is a basis for $L_1L_2$ over $L_1$ that consists of elements of $L_2$ ? (Wondering)

    $ $
  2. We have that $F\subseteq L_1\subseteq L_1L_2$ so we get that $[L_1L_2:F]=[L_1L_2:L_1][L_1:F]$.

    We have that $[L_1L_2:L_1]=[L_1(b_1, b_2, \ldots , b_m):L_1]$. We have to show that this is smaller or equal to $[L_2:F]$, right?

    How could we show that? (Wondering)
Hi mathmari,

It is true that $L_1L_2=L_1(b_1,\dots,b_m)$, but the $b_i$ are not necessarily independent over $L1$; we only know that they are independent over $F$.

In a vector space, any spanning set contains a basis. By removing redundant elements, you can extract form the $b_i$ a basis of $L_1L_2$ over $L_1$. This shows that $[L_1L_2:L_1]\le m$, and the result follows.
 
  • #3
castor28 said:
It is true that $L_1L_2=L_1(b_1,\dots,b_m)$, but the $b_i$ are not necessarily independent over $L1$; we only know that they are independent over $F$.

In a vector space, any spanning set contains a basis. By removing redundant elements, you can extract form the $b_i$ a basis of $L_1L_2$ over $L_1$.

Since $L_1L_2=L_1(b_1, b_2, \ldots , b_m)$ it follows that $L_2$ spans $L_1L_2$ as a vector space over $L_1$. From the Basis Reduction Theorem we get that we can remove some elements of $L_2$ to obtain a basis of $L_1L_2$ over $L_1$.

Have I understood that correctly? (Wondering)
castor28 said:
This shows that $[L_1L_2:L_1]\le m$, and the result follows.

We have that $[L_1L_2:L_1]$= dimension of $L_1L_2$ as a $L_1$-vector space = $\dim_{L_1}(L_1L_2)$ = number of elements of the basis of $L_1L_2$ over $L_1$.

Since the basis consists of some elements of $L_2$, we get that $[L_1L_2:L_1]\leq $ dimension of $L_2$. Is this correct?

Is this same as to say that $[L_1L_2:L_1]\leq $ dimension of $L_2$ as a $F$-vector space?

(Wondering)
 
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  • #4
Hi mathmari

All that is correct. For the second part, we could be a little more precise and modify the second paragraph as:

Since the basis consists of some elements of the basis of $L_2$ over $F$, we get that $[L_1L_2:L_1]\leq$ dimension of $L_2$ over $F$.
 
  • #5
castor28 said:
All that is correct. For the second part, we could be a little more precise and modify the second paragraph as:

Since the basis consists of some elements of the basis of $L_2$ over $F$, we get that $[L_1L_2:L_1]\leq$ dimension of $L_2$ over $F$.

Ah ok! Thank you so much! (Smile)
 

FAQ: Proving the Dimension Bound for Intermediate Field Extensions

What is an intermediate field extension?

An intermediate field extension is a field extension that lies between two other field extensions. In other words, if we have fields F 1 and F 2, and F 1 ⊆ K ⊆ F 2, then K is an intermediate field extension.

How is an intermediate field extension different from a simple field extension?

A simple field extension is a field extension that is obtained by adjoining a single element to a base field. In contrast, an intermediate field extension is obtained by adjoining multiple elements to a base field.

Can an intermediate field extension be infinite?

Yes, an intermediate field extension can be infinite. For example, if we have the field extension ℚ(√2) ⊆ ℚ(√2,√3) ⊆ ℚ(√2,√3,√5), then the intermediate field extension ℚ(√2,√3) is infinite since it contains infinitely many elements of the form a + b√2 + c√3 + d√6, where a, b, c, d are rational numbers.

What is the degree of an intermediate field extension?

The degree of an intermediate field extension is the degree of the field extension obtained by adjoining all the elements in the intermediate field extension to the base field. In other words, if the intermediate field extension is obtained by adjoining n elements to the base field, then the degree of the intermediate field extension is n.

How are intermediate field extensions useful in field theory?

Intermediate field extensions play an important role in Galois theory and the study of field automorphisms. They also help us understand the structure of field extensions and their relationship to each other. Additionally, intermediate field extensions are used in many practical applications, such as cryptography and coding theory.

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