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Wingeer
#1
Mar6-12, 10:09 PM
P: 79
Hello,
I have a quick question about extension fields.
We know that if E is an extension field of F and if we have got an irreducible polynomial p(x) in F[x] with a root u in E, then we can construct F(u) which is the smallest subfield of E containing F and u. This by defining a homomorphism:

[tex]\Phi : F[x] \to E[/tex]
by
[tex]\Phi (f(x)) = f(u)[/tex].

Then, since the ideal generated by p(x) in F[x] is maximal, and
[tex]Ker \Phi = p(x)[/tex]
we have, using the fundamental theorem of homomorphisms, that:
[tex]F[x] / (p(x))[/tex]
is isomorphic to
[tex]F[u] = \{a_0 + a_0u + \cdots + a_mu^m | a_0 + a_0x + \cdots + a_mx^m \in F[x] \}[/tex]
Which in fact is equal to F(u), the smallest subfield of E containing F and u.

The dimension of F[u] over F is given by:
[tex][F[u]:F]=deg(p(x))[/tex]
Also in drawing that conclusion we need the fact that the set [tex]\{1,u,u^2, \cdots, u^{n-1}\}[/tex] is a basis for F[u].

This is all fine, but what stumped me is the concept of expanding over this field again by adding, say another root v in E of p(x) (assuming such v exists, of course). We also assume that v is not algebraic in F(u), so that we need another extension to cover the roots of p(x).
I reckon that [tex]F(u,v)[/tex] is an alternative, but I would like to describe a general element in this field, like one could for an element in F(u). Is there also a nice way to find the dimension of F(u,v) over F(u)? Is it simply 2?

I hope that I made myself clear, and if there are any uncertainties be kind to ask.
Thanks in advance.
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