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 08:42, 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.