Re: -- Some questions about the definition of a splitting field of a polynomia

On Apr 16, 8:22 am, monodromy88 <monodrom...@xxxxxxxxx> wrote:
Howdy,

Given a polynomial f(x) in F[x], where F is a field,
one defines a splitting field  of f(x) as an extension K of F containing the roots of f; and that K is also generated by the roots of f over F, i.e.
K = F(a_1, .., a_n), where the a_i are the roots of f(x).

Fair enough, but what *is* F(a_1, .., a_n)? Rather,  where does *it* live?
It seems like this definition is just assigning a letter  K to the object F(a_1, .., a_n) and calling it a splitting field of f (and, of course, it doesn't make much sense to talk about the subfield generated by a subset, without reference to an ambient  field containing that subset).

K itself must be (?)  the  subfield of *some field* generated  F and the roots of f,  yes?  

The subfield of what, though? The algebraic closure of F?
This definition seems roundabout to me: it's basically telling me that the splitting field is F(a_1, .., a_n), and leaves it at that.  

When I think of the splitting field of x^2 - 2 over Q, for instance,  I know that the roots live in R; and then I can generate these roots over  Q to get the splitting of x^2 - 2 over Q.

But what's wrong with taking the subfield of the algebraic closure Q* of Q  generated by the roots of
x^2 - 2 over Q and calling *it*  the splitting field of x^2 - 2 over Q?

Are these generated subfields one and the same?   Why?

So what's going on?  When thinking about  the splitting field of some polynomial f(x) in F[x], should I first think of an algebraic closure  F* of F
(since it contains all the roots of f anyway),  and then form the subfield of F*  generated by the roots of f over F, and call it a splitting field of f over F?

I  hope to have been clear  in spite of my confusion!

I look forward to clarification. Thank you!

The general idea is that you can have several algebraic closures which
are set theoretically distinct, that is to say the underlying sets are
different. This does not affect the algebra itself as different
algebraic closures are isomorphic. A particular splitting field of f
over F is the data set consisting of both the set and the field
structure. One of the theorems you prove is that no matter what
splitting field of f over F you take, the mere characterization that
it is a splitting field ensures that two splitting fields of f over F
are isomorphic by a isomorphism which is identity on F, that is that
it takes F to itself pointwise settheoretically. The point of all
these theorems is to show that certain charactizations of fields
ensure that between any two concrete realizations of such a
characterization, there is a well-behaved map that enables you to pass
from one to the other so that doing algebra on one is the same as
doing algebra on the other so that you can do algebra on the abstract
characterization itself. Take for example Q. For any polynomial over
Q, you can find a unique splitting field in the complex field using
the fundamental theorem of algebra. But you can also construct a
splitting field by successively finding roots by mappping a field F
into F[x]/<f(x)> as you are taught that every field can be embedded in
a field in which a polynomial over a root. But no matter how you
construct such an example two such examples are isomorphic by an
isomorphism that is identity on the subfield Q and which has other
desirable properties like that a root alpha of a polynomial p in one
splitting field is taken to a root of p in another field. So the mere
characterization that something is a splitting field enables you to do
algebra on one field without looking for a concrete realization.
.

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