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

On Apr 15, 8:22 pm, 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!

Hello,

Your question is very clearly described. I will not be able to live
up to that standard.

If f(x) splits into linear factors over F, there is no problem.
Otherwise, f(x) is divisible by a polynomial g(x) with coefficients in
F (g(x) could be f(x)).

Now look at the polynomial ring F[x], and call two polynomials in F{x}
equivalent if their difference is divisible by g(x). On the
equivalence classes, define addition and multiplication in the natural
way. We obtain a field, usually called F[x]/(g(x)). This field does
not contain F, but it the equivalence classes of constant polynomials
form a subfield of F[x]/(g(x)) naturally isomorphic to F. So we may
replace this copy of F with F itself, and thus obtain an extension
field F' of F. It is easy to show that the equation g(x)=0$ has a
solution in F'. If the orinal polynomial f(x) does not split into
linear factors over F', continue. After a while we reach a field K
which contains F and is a splitting field of f(x).

But we sloppily talk about "the" splitting field of f(x). This is
because it can be shown that if K, K' are any splitting fields of f
(x), then K and K' are isomorphic, via an isomorphism that is the
identity when restricted to F.

So the answer is that "the" splitting field of f(x) is a gadget that
in principle can be produced by an elaborate construction using
equivalence classes of polynomials. But the details of the
construction are more or less irrelevant (except for the crucial fact
that they work). Usually the only important thing about "the"
splitting field is its defining property, that over it the polynomial
splits, and that the polynomial splits over no proper subfield.

There are deliberate ambiguities built into common usage. It is very
correct to point them out, but then to behave as if they do not matter.
.

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