Hi, vasucatga:
I interpret the Clarification to mean you only want the facts, quickly
but without proof.
Let a be the real cube root of 2, and let w be a complex cube root of 1.
Note that w satisfies w^2 + w + 1 = 0, and hence is "quadratic" in nature.
The splitting field of X^3  2 over Q is then Q(a,w).
1) A basis for Q(a,w) over Q is {1,a,a^2,w,wa,wa^2}.
2) The degree of the field extension is therefore 6 (the same as the
dimension of the extension Q(a,w) considered as a vector space over
Q).
3) The Galois group for Q(a,w) over Q is S_3, a nonabelian but
solvable group of order 6 which is called the symmetric group of
degree 3 (permutations on 3 symbols).
I hope this is helpful. Let me know if further Clarification is wanted.
regards, mathtalkga 
Clarification of Answer by
mathtalkga
on
06 May 2004 11:28 PDT
Hi, vasucatga:
Let me point out that your original Question add these further aspects:
"and also determine all subgroups of the Galois group and give all
intermediate subfields of the splitting field of the polynomial."
In order to provide the information that you seem to need quickly, I
not only omitted any proof of the facts stated, but I also limited my
Answer to "the three things I asked for you.1) basis, 2)degree and
3)determine the galois group for the given polynomials."
But I would be happy to elaborate on the Galois connection (the
correspondance between the subgroups of the automorphism group and the
intermediate subfields) as well as to fill in any gaps in the theory
for you.
regards, mathtalkga

Request for Answer Clarification by
vasucatga
on
06 May 2004 12:09 PDT
hi mathtalkga
could you please work out those things clearly with proofs for me.
waiting for your answer ASAP
thanks,
vasucatga

Request for Answer Clarification by
vasucatga
on
06 May 2004 12:18 PDT
hi mathtalkga,
I need the proofs of the things you provided so that I can
understand clearly and also elaborate on the Galois connection(the
correspondence between the subgroups of the automorphism group and the
intermediate subfields)
and also I request you to prove the other question I posted with all the details.
thanks,
vasucatga
waiting for you answer
srinivas kalla

Clarification of Answer by
mathtalkga
on
06 May 2004 12:31 PDT
Hi, vasucatga:
I see that you have now requested a proof of the three "facts"
requested as well as an explanation of the intermediate subfields.
In order to provide this as quickly as possible, I need you to respond
to my original request for clarification, describing your background
in algebra so that I do not unnecessarily belabor details which are
well known to you.
For example, I asked about whether a proof that X^3  2 was
irreducible in Q[X] as well as in Z[X], to which you replied that it
was not necessary "to prove that they belong to Z[X] or Q[X]." But
that is not my point. A polynomial is irreducible if it has no
nontrivial factorization (one with both factors nonunits).
Other aspects of an undergraduate background in algebra that would be
relevant here are abstract vector spaces and permutation groups. My
Answer to your extended Question would be much longer if it is
necessary to fill in the background on those topics.
So once more, I respectfully ask that you explain your background and
interest in an amount of detail that is consist with the care you
expect me to put into answering your inquiry. I believe that you will
find my explanations much more useful if you will only cooperate in
this small request.
regards, mathtalkga

Request for Answer Clarification by
vasucatga
on
06 May 2004 12:52 PDT
hi mathtalkga,
I am graduate student in math. Frankly speaking I am not good at
algebra. From my undergraduate I used to have problem with algebra. I
hope you understand my problem
wiating for ur reply
thanks,
vasucatga

Clarification of Answer by
mathtalkga
on
07 May 2004 15:09 PDT
Let's try it this way. An undergraduate treatment of abstract algebra
would include the definitions and basic properties of groups, rings,
fields, and vector spaces, esp. the "structure preserving maps" on
these called homomorphisms.
I will approach the material with this sort of background in mind, but
if you find that my explanation is unclear, post a reply asking for a
Clarification of that material whenever necessary.
So we may begin with the notion of a field extension, which means that
field F is contained in field G (G is an extension of field F, F a
subfield of G).
A mapping from field G onto itself that preserves sums and products is
a special case of a (ring) homomorphism that is called a field
automorphism. There is always at least one field automorphism, namely
the identity mapping f(x) = x.
The Galois group of a field extension G over F is defined, when G is a
field extension of F, to be the collection of all field automorphisms
of G which, when restricted to F, are the identity automorphism on F.
This collection of mappings is a group under the operation of function
composition. (The inverse function of an automorphism is also an
automorphism, etc.) Often we label this group:
Gal(G/F) "the Galois group of G over F"
In your case we are dealing with a base field F = Q the rationals.
This turns out to be an especially nice case, so called
"characteristic zero". As you may know there are finite fields like
Z/pZ for prime number p, which contain only finitely many elements. Q
is however an infinite field and contains no smaller fields than
itself. The statement of many results in Galois theory are simpler
for fields of characteristic 0 (fields whose smallest subfield is Q)
than in the general case (including fields of characteristic p for p a
finite prime).
The ring of polynomials in one variable over a field F[X] plays an
important role in both undergraduate and graduate algebra. This ring
is a Euclidean domain (whenever the coefficients F are taken from a
field), and therefore a principle ideal domain (PID).
Now the nice thing about a PID is that all ideals are principle, i.e.
generated by a single element. Specifically, if F is a field, then an
ideal I of F[X] would necessarily be of the form all multiples of p(X)
from some polynomial p(X) in the ring F[X]. (One way to identify such
a generator p(X) is to look for a nonzero polynomial of smallest
degree in the ideal if any exists.)
Now there's a significant but perhaps subtle notational difference in
talking about F[X], which is a ring of polynomials (a domain but not a
field since X has no inverse), and F(a), which will be our notation
for a "simple field extension", that is a field extension of F which
is the "smallest possible" one that contains both F and a. Such a
definition would make sense in the context of element a belonging to
G, a field extension of F, for we could then take the intersection of
all subfields of G that contain both F and a, and define F(a) to be
the result. This would be a "natural" approach to defining F(a), but
it's also possible to construct the simple field extension in a more
abstract way, as a homomorphic image of F[X].
Define as an ideal I of F[X] all polynomials p(X) which are zero when
a is substituted for X (imagining the evaluation to be done in G,
which is a field containing both F and a, so all the arithmetic can be
carried out). Then:
F(a) is isomorphic to F[X]/I by the "evaluation map" p(X) :> p(a)
Now while F[X] was an infinite dimensional vector space over F, with
countable basis {1,X,X^2,X^3,...}, the simple field extension F(a) may
turn out to be finite dimensional. I say "may" because it might
happen that I = {0}, i.e. that the only polynomial in X that evaluates
to 0 when a is substituted for X is the zero polynomial. Such would
be the case when a is transcendental over F, e.g. when F = Q and a is
pi.
But if a is algebraic, meaning that there are nonzero polynomials over
F that are satisfied by a, then F(a) will turn out to be a finite
dimensional vector space over F. In fact the dimension of F(a) as a
vector space over F is just the same as the minimal degree of a
nonzero polynomial "satisfied" by a, which is the same as the
generator of I. For then the minimal polynomial, if say of degree n,
tells us that a^n can be expressed as a linear combination of lower
order powers of a and 1:
p(a) = 0 implies a^n belongs to F[1,a,...,a^(n1)]
Therefore, using the minimality of degree for polynomial p(X), one can prove:
{1,a,...,a^(n1)} is a basis for the extension F(a) over F
By the definition a "splitting field" for a polynomial p(X) in F[X] is
a smallest extension of F in which "all roots" of p(X) can be found,
or to be more precise, in which p(X) can be expressed as a product of
first degree factors.
If we restrict ourselves (without loss of generality because F is a
field) to monic polynomials (leading coefficient 1), then we are
saying the splitting field for p(X) would be a field extension G of F
in which:
p(X) = (X  a_1)...(X  a_n) in G[X]
where the degree of p(X) must match the number of "linear" factors shown.
One of the amazing results of field extensions is that any splitting
field of a polynomial over Q (the rationals) is a simple field
extension of Q. Now this is a little confusing because in general the
minimal polynomial of the "primitive element" b that gives us the
splitting field as Q(b) is not the same as the polynomial being split.
Take as an example the problem we studied above. You asked for the
splitting field over Q of p(X) = X^3  2. To get all roots of this
cubic polynomial requires an extension Q(a,w) that turns out to be of
dimension six over Q, so if we sought a simple extension Q(b) =
Q(a,w), then the minimal polynomial for b would be of degree six
(compared with the degree of the split polynomial p(X) which is merely
degree three).
Hopefully this has served to begin the discussion on a familiar and
somewhat elementary footing. I will pause to see if you need
Clarification of the definitions and basic results that have been
described to this point.
regards, mathtalkga

Clarification of Answer by
mathtalkga
on
07 May 2004 19:39 PDT
It might be useful, to judge whether the material above needs further
Clarification, to ask yourself how it applies in the problem you are
studying.
Of course we have already noted that the base field F = Q the rational
numbers, and the extension G which we are most interested in is the
splitting field G of the cubic polynomial p(X) = X^3  2.
After we are clear together on the three basic facts that I addressed
in my Answer (the basis for the splitting field, its degree, and the
group of automorphisms), then we can take up the Galois correspondance
between intermediate subfields of the extension G/F and the subgroups
of Gal(G/F).
regards, mathtalkga

Clarification of Answer by
mathtalkga
on
08 May 2004 08:29 PDT
Hi, vasucatga:
Let's proceed now with the details of constructing a basis for the
splitting field of X^3  2 over Q.
Now the splitting field for X^3  2 over Q will be the smallest field
extension of Q that contains its three roots. Let a be the real cube
root of 2, and note that the other two complex cube roots of 2 can be
expressed as aw and aw^2 where w is a complex cube root of 1. Taken
together these are the only roots of X^3  2, and:
X^3  2 = (X  a)(X  aw)(X  aw^2)
Therefore the splitting field of this polynomial over Q may be expressed as:
Q(a,aw,aw^2)
which would mean the result of a succession of three "simple" field
extensions (first a, then aw, finally aw^2 being adjoined).
But in fact a + aw + aw^2 = 0, so already after adjoining a and aw, we
would have the third root aw^2 as:
aw^2 = (a + aw)
In any case we previously identified the splitting field in this way:
Q(a,w)
and a moments reflection shows that if a and aw are in a field, so
must w be in the field (because a is nonzero and has an inverse).
In defining a basis for Q(a,w) it is probably helpful to think of a
"tower composition" of field extensions:
Q contained in Q(a) contained in Q(a,w)
It will be shown that Q(a) is dimension 3 over Q, and that Q(a,w) is
dimension 2 over Q(a). We then show that the basis {1,a,a^2} for Q(a)
over Q may be combined with the basis {1,w} of Q(a,w) over Q(a) to
yield the combined basis that we cited earlier for Q(a,w) over Q:
{1,a,a^2,w,wa,wa^2}
This basis contains six elements, the products of elements from the
two "composed" bases, so Q(a,w) is dimension 6 over Q.
Let's back up and consider the two "steps" in our tower. First Q(a)
is a field extension of Q, and because a is by definition a root of
X^3  2 = 0, we know:
a^3 = 2
So in the spanning set {1,a,a^2,a^3,...} for Q(a) over Q, we have a
linear dependence relation for a^3 and all higher powers of a. a^3
can be expressed as 2, so it follows that all the higher powers can be
expressed in terms of {1,a,a^2} as well.
But why is {1,a,a^2} a linearly independent set (and therefore a
basis, since by the above it spans)? If there were a dependence
relation:
c_0*1 + c_1*a + c_2*a^2 = 0
with coefficients c_0,c_1,c_2 (not all zero) in base field Q, then it
would mean that the minimal polynomial for a was of degree at most 2,
certainly less than 3.
But p(X) = X^3  2 is irreducible (for example, by Eisenstein's
criterion), and the minimal polynomial for a must divide p(X) because
a is a root of p(X). Then p(X) is the minimal polynomial for a
(irreducible means no nonunit divisors) and hence the existence of a
lower degree relation among the powers of a is impossible.
Next consider the extension Q(a,w) over Q(a). Here although w is a
complex cube root of 1, it is actually a quadratic extension. That is
because of the factorization:
X^3  1 = (X  1)(X^2 + X + 1)
The minimal polynomial for w turns out to be the second, quadratic
factor. Note that the complex conjugate of w is just as good a choice
for the extension, as in fact:
conjugate(w) = w^2 = 1/w
By reasoning as before the spanning set {1,w,w^2,...} for Q(a,w) over
Q(a) can be chopped down to {1,w} in view of the dependence relation
w^2 + w + 1 = 0, and the linear independence of {1,w} is clear from
the simple fact that the complex number w does not belong to Q(a),
which consists only of real values.
It might be a good insight to note that complex conjugation is an
automorphism on Q(a,w) which "fixes" Q(a). That is, since Q(a)
consists only of real values, the complex conjugation automorphism
maps each element of Q(a) to itself. This insight will prove helpful
later in our discussion of the intermediate fields between Q and
Q(a,w) and the associated subgroups of Gal(Q(a,w)/Q).
Now it can be concluded that where {1,a,a^2} is a basis for Q(a) over
Q, and {1,w} is a basis for Q(a,w) over Q(a), that {1,a,a^2,w,wa,wa^2}
is a basis for Q(a,w) over Q.
The "easy" part of proving this conclusion is that {1,a,a^2,w,wa,wa^2}
is a spanning set. For given any element of Q(a,w), it can be
expressed as a linear combination of 1 and w using coefficients from
Q(a), and these coefficients in Q(a) can in turn be expressed using
the basis {1,a,a^2} for Q(a) over Q with rational coefficients.
Expanding the expressions out gives the desired combination of terms
from {1,a,a^2,w,wa,wa^2} with rational coefficients.
The linear independence of {1,a,a^2,w,wa,wa^2} is only slightly hard
to show. For suppose that we have a dependence relation among these
terms over Q, not all coefficients therein being zero. If a nonzero
coefficient involves one of the last three terms, containing w, then
we have a linear dependence relation for w over Q(a), contradicting
what we showed before. On the other hand if none of the terms involve
w, we would have a dependence relation among {1,a,a^2}, again
contradicting our previous conclusion about the linear independence of
these over Q.
The construction here of a "composite" basis using the intermediate
bases from "steps" in a tower of field extensions is a general one,
though we have argued only the facts for the particular example in
order to illustrate the ideas clearly.
As before if some aspects of this discussion need Clarification, please ask.
regards, mathtalkga

Clarification of Answer by
mathtalkga
on
12 May 2004 17:44 PDT
Hi, vasucatga:
Please feel free to ask even very basic questions about the material
we are covering. Without your feedback I'm unable to determine what
progress if any is being made toward a clear understanding of the
Galois theory for field extensions.
Here's a link that contains a fairly concise summary of definitions:
[Mathworld  Extension Field]
http://mathworld.wolfram.com/ExtensionField.html
and for the specific topic of the minimal polynomial of an algebraic
element in an extension field:
[Mathworld  Extension Field Minimal Polynomial]
http://mathworld.wolfram.com/ExtensionFieldMinimalPolynomial.html
It may be helpful to review these notions in connection with the
examples just worked. First we found that Q(a) where a is the (real)
cube root of 2 is a threedimensional vector space over Q, and that
the minimal polynomial of a over Q is the cubic polynomial:
X^3  2
Second we found that Q(a,w) where w is the (complex) cube root of 1 is
a twodimensional vector space over Q(a), and that the minimal
polynomial of w over Q(a) is the quadratic polynomial:
X^2 + X + 1
because X^3  1 factors as (X  1)(X^2 + X + 1) over Q(a). [This is
also the minimal polynomial for w over Q, since its coefficients are
in Q.]
You might guess from these examples (and even more so from the
techniques of proof we used) that the dimension of a simple field
extension is equal to the degree of the minimal polynomial for the
"adjoined" element. Your guess would be correct in this case.
We also argued in the particular case:
Q contained in Q(a) contained in Q(a,w)
that the dimensions of a "tower" of field extensions like these will
multiply, so that Q(a,w) is dimension 6 = 3 * 2 over Q. This too is a
general theorem, and the construction we gave of a basis for the top
of tower over the base field, by taking products of elements from the
intermediate extension bases, is also valid in general.
We now come to the first "tricky" part of the theory, which begins
with a definition. Let us recall that for a field extension F
contained in G, the automorphism group Gal(G/F) means those
automorphisms of G which "fix" F (that is, map each element of F to
itself; the restriction of these automorphisms of G to subfield F is
always the identity map on F).
It is not hard to show that the subset of G which is fixed by all
automorphisms in Gal(G/F) is a subfield of G which contains F.
We say (define) that G/F is a Galois extension iff that subfield of G
fixed by all of Gal(G/F) equals F.
We can use our earlier examples to illustrate a field extension that
is _not_ a Galois extension as well as one that is. For instance
Gal(Q(a)/Q) is just the trivial group (containing only the identity
automorphism), from which it follws that the subfield fixed by
Gal(Q(a)/Q) is not Q but actually all of Q(a).
To see why this is, consider an automorphism f in Gal(Q(a)/Q). That
is, it must be an automorphism f:Q(a) > Q(a) such that f is the
identity on Q. But what might f(a) be? Since a is a "root" of
minimal polynomial X^3  2, and since the coefficients of this
polynomial are in Q, it follows that:
(f(a))^3  2 = 0
i.e. that f(a) is also a root of X^3  2. But Q(a) only contains one
(real) root of this polynomial, namely a itself. So perforce f(a) =
a, and from this it follows that f is the identity on all of Q(a)
(since every element in Q(a) is expressible as a polynomial in a with
coefficients in Q).
Thus Q(a)/Q is _not_ a Galois extension.
On the other hand Q(w)/Q is a Galois extension, as is Q(a,w)/Q(a).
The reason is that 1/w is the complex conjugate of w, so Gal(Q(w)/Q)
contains a nontrivial automorphism, ie. complex conjugation. This
means Gal(Q(w)/Q) will not fix all of Q(w), but only a proper subfield
of Q(w) which contains Q. Since the dimension of Q(w) over Q is 2,
there's no "wiggle" room. The subfield of Q(w) fixed by Gal(Q(w)/Q)
is equal to Q.
What makes the difference in these examples? Well, in a nutshell the
difference is what we call a splitting field. Q(w) is a splitting
field over Q of the polynomial X^2 + X + 1, meaning that it contains
"all roots" of the polynomial so that the polynomial "splits" into
first degree factors:
X^2 + X + 1 = (X  w)(X  1/w)
Here w = 1/2 + i * sqrt(3)/2, and you can verify that the
factorization works (or that the roots are correct by the quadratic
formula).
But in the extension Q(a)/Q we did not have a splitting field. We
know that in particular the minimal polynomial X^3  2 does not split
in Q(a), because we get only one root there (the real one) and not
either of the complex conjugate pair of additional roots (that we'd
find in field C, for example). It turns out that if a simple field
extension is a splitting field for one choice of adjoined element (and
its minimal polynomial), it is for any other choice as well.
In fact the splitting field for X^3  2 turns out to be all of Q(a,w),
which as we've seen is a six dimensional space over Q.
So with these illustrations of the Galois group and a Galois extension
in mind, we are ready to next discuss the final aspects of your
Question, namely the Galois group Gal(Q(a,w)/Q) and its subgroups and
the corresponding intermediate subfields between Q and Q(a,w).
regards, mathtalkga
