How Do You Calculate the Dimension of F(u,v) Over F(u)?

[Wingeer]

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:

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

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

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

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 F(u,v) 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.

 

[mathwonk]

you have to factor the polynomial p(x) for u that was formerly irreducible over F, and see what its irreducible factors are now over F(u). One of them will be the irreducible polynomial for v over F(u). Its degree will be the dimension of the extension F(u,v) over F(u).

The general element of F(u,v) is of course a polynomial in u and v with coefficients in F. (with certain bounded degrees in u and in v.)

 

[Wingeer]

you have to factor the polynomial p(x) for u that was formerly irreducible over F, and see what its irreducible factors are now over F(u). One of them will be the irreducible polynomial for v over F(u). Its degree will be the dimension of the extension F(u,v) over F(u).

The general element of F(u,v) is of course a polynomial in u and v with coefficients in F. (with certain bounded degrees in u and in v.)

Could you kindly elaborate a bit on the bold text in the quote? Or check if I got it right. Say, for instance that we are working with
p(x)=x^3-2 \in Q[x]
the roots of this polynomial are:
x=\sqrt[3]{2},\omega \sqrt[3]{2},\omega^2 \sqrt[3]{2}
Where,
\omega = e^{\frac{2\pi i}{3}}
First let us construct an extension: Q(\sqrt[3]{2}) By the theory in the opening post we can conclude that
[Q(\sqrt[3]{2}):Q]=deg(x^3-2)=3
And that
Q(\sqrt[3]{2})=\{ a_0 + a_1 \sqrt[3]{2} + a_2 \sqrt[3]{2}^2 | a_0 + a_1x + a_2x^2 \in Q[x] \}
We now consider p(x) over the new extension field. By factoring out the root we get that:
p(x)=x^3-2 = (x-\sqrt[3]{2})(x^2 +\sqrt[3]{2}x + \sqrt[3]{4})
So the new irreducible polynomial is now:
p_1(x) = (x^2 +\sqrt[3]{2}x + \sqrt[3]{4}
We see that
v=\omega \sqrt[3]{2}
is a root for this polynomial. We therefore construct a new extension (and here we do not actually need "whole" v, as the cube root of 2 is already in our extension field?), namely:
Q(\sqrt[3]{2}, \omega)
Where:
[Q(\sqrt[3]{2}, \omega):Q(\sqrt[3]{2})]=deg(p_1(x))=2

Does this final extension field contain all the roots of p(x)? What is the basis of this extension?
Also, what did you mean with the comment about "certain bounded degrees"?

Thank You so much for the help!

 

[mathwonk]

read section 19 of these notes for the example of the splitting field of X^4-2 over Q.

http://www.math.uga.edu/%7Eroy/843-2.pdf