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Difference between revisions of "Tower law"

(Statement and proof)
 
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Given three [[field]]s, <math>F\subseteq K\subseteq L</math> the '''tower law''' states that if <math>[L:K]</math> and <math>[K:F]</math> are both finite then so is <math>[L:F]</math> and
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Given three [[field]]s, <math>F\subseteq K\subseteq L</math> the '''tower law''' states that if <math>[L:K]</math> and <math>[K:F]</math> are both [[finite]] then so is <math>[L:F]</math> and
 
<cmath>[L:F] = [L:K][K:F].</cmath>
 
<cmath>[L:F] = [L:K][K:F].</cmath>
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And furthermore if either <math>[L:K]</math> or <math>[K:F]</math> is infinite then so is <math>[L:F]</math>.
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Furthermore, if either <math>[L:K]</math> or <math>[K:F]</math> is infinite then so is <math>[L:F]</math>.
  
 
==Proof==
 
==Proof==

Latest revision as of 07:42, 21 August 2009

Given three fields, $F\subseteq K\subseteq L$ the tower law states that if $[L:K]$ and $[K:F]$ are both finite then so is $[L:F]$ and \[[L:F] = [L:K][K:F].\] Furthermore, if either $[L:K]$ or $[K:F]$ is infinite then so is $[L:F]$.

Proof

First consider the case where $[L:K]$ and $[K:F]$ are both finite. Let $[L:K] = m$ and $[K:F] = n$. Let $\{\alpha_1,\alpha_2,\ldots,\alpha_m\}$ be a basis for $L$ over $K$ and $\{\beta_1,\beta_2,\ldots,\beta_n\}$ be a basis for $K$ over $F$. We claim that the set $B=\{\alpha_i\beta_j|1\le i\le m, 1\le j\le n\}$ (which clearly has $mn$ elements) is a basis for $L$ over $F$.

First we show that $B$ spans $L$. Take any $v\in L$. As $\{\alpha_1,\alpha_2,\ldots,\alpha_m\}$ is a basis for $L$ over $K$, we can write $v = \sum_ia_i\alpha_i$, where $a_1,a_2,\ldots,a_n\in K$. And now as $\{\beta_1,\beta_2,\ldots,\beta_n\}$ is a basis for $K$ over $F$ we can write $a_i=\sum_jb_{ij}\beta_j$ where $b_{i1},b_{i2},\ldots,b_{in}\in F$, for each $a_i$. So now \[v = \sum_ia_i\alpha_i = \sum_i\left(\sum_jb_{ij}\beta_j\right)\alpha_i = \sum_{i,j}b_{ij}\alpha_i\beta_j.\] So indeed $B$ spans $L$ over $F$.

Now we show that $B$ is independent. Assume that there are some $\{b_{ij}\}\subseteq F$ such that $\sum_{i,j}b_{ij}\alpha_i\beta_j = 0$. So then we have $\sum_i\left(\sum_jb_{ij}\beta_j\right)\alpha_i = 0.$ So, as $\{\alpha_1,\alpha_2,\ldots,\alpha_m\}$ is independent over $K$ we get that $\sum_jb_{ij}\beta_j = 0.$ For all $i$. And hence as $\{\beta_1,\beta_2,\ldots,\beta_n\}$ is independent over $F$ we get $b_{ij}=0$ for all $i$ and $j$. Therefore $B$ is indeed independent.

Therefore $B$ is indeed a basis, so $[L:F] = |B| = mn$, as desired.

Now we consider the infinite case. By the above argument if $\{\alpha_1,\alpha_2,\ldots,\alpha_m\}\subseteq L$ is independent over $K$ and $\{\beta_1,\beta_2,\ldots,\beta_n\}\subseteq K$ is independent over $F$ then the set $B=\{\alpha_i\beta_j|1\le i\le m, 1\le j\le n\}$ is independent over $F$. Hence if either of $[L:K]$ and $[K:F]$ is infinite then there exisit arbitrarily large independent sets in $L$ over $F$, so $[L:F]$ is infinite as well.

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