MHB Proving the Dimension Bound for Intermediate Field Extensions

[mathmari]

Hey!

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)

 

[castor28]

Hey!

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.

 

[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 $L_1$.

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)

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)

 

[castor28]

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 $F$.

 

[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 $F$.

Ah ok! Thank you so much! (Smile)