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# It is inclusion-reversing: if <math>H_1\subset H_2</math> then <math>E_{H_1}\supset E_{H_2}</math> and if <math>K_1\subset K_2</math> then <math>\operatorname{Gal}(E/K_1)>\operatorname{Gal}(E/K_1)</math>. |
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# It is inclusion-reversing: if <math>H_1\subset H_2</math> then <math>E_{H_1}\supset E_{H_2}</math> and if <math>K_1\subset K_2</math> then <math>\operatorname{Gal}(E/K_1)>\operatorname{Gal}(E/K_1)</math>. |
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# It is degree/index respecting: <math>[E:K]=|\operatorname{Gal}(E/K)|</math> and <math>[K:F]=[\operatorname{Gal}(E/F):\operatorname{Gal}(E/K)]</math>. |
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# It is degree/index respecting: <math>[E:K]=|\operatorname{Gal}(E/K)|</math> and <math>[K:F]=[\operatorname{Gal}(E/F):\operatorname{Gal}(E/K)]</math>. |
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# Splitting fields correspond to normal subgroups: If <math>K</math> in <math>E/K/F</math> is a splitting field then <math>\operatorname{Gal}(E/K)</math> is normal in <math>\operatorname{Gal}(E/F)</math> and <math>\operatorname{Gal}(K/F)\cong\operatorname{Gal}(E/F)/\operatorname{Gal}(E/K)</math>. |
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# Splitting fields correspond to normal subgroups: If <math>K</math> in <math>E/K/F</math> is the splitting field of a polynomial in <math>F[x]</math> then <math>\operatorname{Gal}(E/K)</math> is normal in <math>\operatorname{Gal}(E/F)</math> and <math>\operatorname{Gal}(K/F)\cong\operatorname{Gal}(E/F)/\operatorname{Gal}(E/K)</math>. |
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===Lemmas=== |
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===Lemmas=== |
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Week of...
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Links
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1
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Jan 10
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About, Notes, HW1
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2
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Jan 17
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HW2, Notes
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3
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Jan 24
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HW3, Photo, Notes
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4
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Jan 31
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HW4, Notes
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5
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Feb 7
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HW5, Notes
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6
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Feb 14
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On TT, Notes
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R
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Feb 21
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Reading week
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7
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Feb 28
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Term Test
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8
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Mar 7
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HW6, Notes
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9
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Mar 14
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HW7, Notes
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10
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Mar 21
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HW8, E8, Notes
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11
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Mar 28
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HW9, Notes
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12
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Apr 4
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HW10, Notes
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13
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Apr 11
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Notes, PM
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S
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Apr 16-20
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Study Period
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F
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Apr 24
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Final
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![07-401 Class Photo.jpg](/images/thumb/d/d9/07-401_Class_Photo.jpg/180px-07-401_Class_Photo.jpg) Add your name / see who's in!
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Register of Good Deeds
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In Preparation
The information below is preliminary and cannot be trusted! (v)
The Fundamental Theorem of Galois Theory
It seems we will not have time to prove the Fundamental Theorem of Galois Theory in full. Thus this note is about what we will be missing. The statement appearing here, which is a weak version of the full theorem, is taken from Gallian's book and is meant to match our discussion in class. The proof is taken from Hungerford's book, except modified to fit our notations and conventions and simplified as per our weakened requirements.
Here and everywhere below our base field
will be a field of characteristic 0.
Statement
Theorem. Let
be a splitting field over
. Then there is a correspondence between the set
of intermediate field extensions
lying between
and
and the set
of subgroups
of the Galois group
of the original extension
:
.
The bijection is given by mapping every intermediate extension
to the subgroup
of elements in
that preserve
,
,
and reversely, by mapping every subgroup
of
to its fixed field
:
.
Furthermore, this correspondence has the following further properties:
- It is inclusion-reversing: if
then
and if
then
.
- It is degree/index respecting:
and
.
- Splitting fields correspond to normal subgroups: If
in
is the splitting field of a polynomial in
then
is normal in
and
.
Lemmas
The two lemmas below belong to earlier chapters but we skipped them in class.
The Primitive Element Theorem
The celebrated "Primitive Element Theorem" is just a lemma for us:
Lemma 1. Let
and
be algebraic elements of some extension
of
. Then there exists a single element
of
so that
. (And so by induction, every finite extension of
is "simple", meaning, is generated by a single element, called "a primitive element" for that extension).
Proof. See the proof of Theorem 21.6 on page 375 of Gallian's book.
Splitting Fields are Good at Splitting
Lemma 2. (Compare with Hungerford's Theorem 10.15 on page 355). If
is a splitting field over
and some irreducible polynomial
has a root
in
, then
splits in
.
Proof. Let
be a splitting field of
over
. We need to show that if
is a root of
in
, then
(so all the roots of
are in
and hence
splits in
). Consider the two extensions
and
.
The "smaller fields"
and
in these two extensions are isomorphic as they both arise by adding a root of the same irreducible polynomial (
) to the base field
. The "larger fields"
and
in these two extensions are both the splitting fields of the same polynomial (
) over the respective "small fields", as
is a splitting extension for
and we can use the sub-lemma below. Thus by the uniqueness of splitting extensions, the isomorphism between
and
extends to an isomorphism between
and
, and in particular these two fields are isomorphic and so
. Since all the degrees involved are finite it follows from the last equality and from
that
and therefore
. Therefore
.
Sub-lemma. If
is a splitting extension of some polynomial
and
is an element of some larger extension
of
, then
is also a splitting extension of
.
Proof. Let
be all the roots of
in
. Then they remain roots of
in
, and since
completely splits already in
, these are all the roots of
in
. So
,
and
is obtained by adding all the roots of
to
.
Proof of The Fundamental Theorem
The Bijection
Proof of
. More precisely, we need to show that if
is an intermediate field between
and
, then
. The inclusion
is easy, so we turn to prove the other inclusion. Let
be an element of
which is not in
. We need to show that there is some automorphism
for which
; if such a
exists it follows that
and this implies the other inclusion. So let
be the minimal polynomial of
over
. It is not of degree 1; if it was, we'd have that
contradicting the choice of
. By lemma 2 and using the fact that
is a splitting extension, we know that
splits in
, so
contains all the roots of
. Over a field of characteristic 0 irreducible polynomials cannot have multiple roots and hence
must have at least one other root; call it
. Since
and
have the same minimal polynomial over
, we know that
and
are isomorphic; furthermore, there is an isomorphism
so that
yet
. But
is a splitting field of some polynomial
over
and hence also over
and over
. By the uniqueness of splitting fields, the isomorphism
can be extended to an isomorphism
; i.e., to an automorphism of
. but then
so
, yet
, as required.
Proof of
. More precisely we need to show that if
is a subgroup of the Galois group of
over
, then
. The inclusion
is easy. Note that
is finite since we've proven previously that Galois groups of finite extensions are finite and hence
is finite. We will prove the following sequence of inequalities:
This sequence of course implies that these quantities are all equal and since
it follows that
as required.
The first inequality above follows immediately from the inclusion
.
By the Primitive Element Theorem (Lemma 1) we know that there is some element
so that
. Let
be the minimal polynomial of
over
. Distinct elements of
map
to distinct roots of
, but
has exactly
roots. Hence
, proving the second inequality above.
Let
be an enumeration of all the elements of
, let
(with u as above), and let
be the polynomial
.
Clearly,
. Furthermore, if
, then left multiplication by
permutes the
's (this is always true in groups), and hence the sequence
is a permutation of the sequence
, hence
,
and hence
. Clearly
, so
, so
, proving the third inequality above.
The Further Properties
Proof of Property 1. Easy.
Proof of Property 2. If
, then
as was shown within the proof of
. But every
is
for some
, so
for every
between
and
. The second equality follows from the first and from the multiplicativity of the degree/order/index in towers of extensions and in towers of groups:
Proof of Property 3.