# Difference between revisions of "Tower law"

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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 | + | 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> | ||

− | + | 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, the **tower law** states that if and are both finite then so is and
Furthermore, if either or is infinite then so is .

## Proof

First consider the case where and are both finite. Let and . Let be a basis for over and be a basis for over . We claim that the set (which clearly has elements) is a basis for over .

First we show that spans . Take any . As is a basis for over , we can write , where . And now as is a basis for over we can write where , for each . So now So indeed spans over .

Now we show that is independent. Assume that there are some such that . So then we have So, as is independent over we get that For all . And hence as is independent over we get for all and . Therefore is indeed independent.

Therefore is indeed a basis, so , as desired.

Now we consider the infinite case. By the above argument if is independent over and is independent over then the set is independent over . Hence if either of and is infinite then there exisit arbitrarily large independent sets in over , so is infinite as well.