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
[tex]F = \{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 over F is given by:
[tex][F:F]=deg(p(x))[/tex]
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

 

[blue_raver22]

Hello,

Thank you for your question about extension fields. The structure of extension fields can be a bit complex, but I'll do my best to explain it to you.

Firstly, you are correct in your understanding that if we have an irreducible polynomial p(x) in F[x] with a root u in E, then we can construct F(u) as the smallest subfield of E containing F and u. This is known as an algebraic extension, where we are adding a single root to the field.

However, as you mentioned, we can also have multiple roots of p(x) in E. In this case, we can construct a larger extension field, F(u,v), by adding all the roots of p(x) to the field. This is known as a splitting field.

To describe a general element in F(u,v), we can use a similar approach to what you described for F(u). We can write an element as a linear combination of the basis elements \{1,u,v,uv\}, where the coefficients are from F. This allows us to write any element in F(u,v) as a polynomial in u and v with coefficients in F.

The dimension of F(u,v) over F(u) is given by the degree of the minimal polynomial of v over F(u). This can be found by constructing the minimal polynomial using the relation between u and v, and then calculating its degree. In general, the dimension of F(u,v) over F is equal to the product of the degrees of the minimal polynomials of u and v over F.

I hope this helps clarify the structure of extension fields and how to find the dimension of a larger extension field over a smaller one. Please let me know if you have any further questions.