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Question on generating subfields (field extensions).

Let F⊂L F \subset L be a field extension and A⊂L A \subset L be a subset we define the intermediate field as the subfield of L L where we "add" the elements of A A and denote this subfield as K(A) K(A).

How does one figure out how the elements of such a field look like?

Example

Suppose we take the extension Q⊂C \mathbb{Q} \subset \Bbb{C} then we let the subset of C \mathbb{C} be {i,e,π} \{i,e,\pi \} for example what how can we describe the elements of Q(i,e,π) \mathbb{Q}(i,e, \pi).

What I mean by that is R(i)={a+bi:a,b∈R}=C \mathbb{R}(i)=\{a+bi : a,b \in \mathbb{R}\}=\Bbb{C} or Q(i)={a+bi:a,b∈Q} \mathbb{Q}(i)=\{a+bi : a,b \in \mathbb{Q}\}.

I believe these are called the Gaussian integers.But when the subset has a few elements more it's confusing me as to what they elements look like.

Hopefully what I've posted makes some sense at least to describe what I don't get. Any algebra legends out there?
For nicer field extensions (algebraic extensions) things aren't bad. Let K be a field and let f(X) be an irreducible degree d polynomial over K. Then we can build a field extension (slight abuse of `=' follows)

L=K[X]/(f(X))={ad−1Xd−1+…+a1X+a0:ai∈K} L = K[X]/(f(X)) = \{ a_{d-1}X^{d-1} + \ldots + a_1X +a_0 : a_i\in K\}

where X in L behaves like a root of f; so we write L=K(α) L = K(\alpha) where α\alpha is a root of f (usually in C, but really whatever the algebraic closure of K is).

To be super concrete,

Q(23)=Q[X]/(X3−2)={a2232+a123+a0:ai∈Q}\mathbb{Q}(\sqrt[3]{2}) = \mathbb{Q}[X]/(X^3-2) = \{a_2\sqrt[3]{2}^2 +a_1\sqrt[3]{2} +a_0 : a_i\in\mathbb{Q}\}

What about adjoining many roots? Good news about separable extensions (which all algebraic extensions over Q are), they all have the form K(α)K(\alpha)! This is the primitive element theorem. For example Q(2,3)=Q(2+3)\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2}+\sqrt{3}).

When you want to add transcendental elements writing down general elements isn't so nice. For instance Q(e)≅Q(X)\mathbb{Q}(e) \cong \mathbb{Q}(X), the field of rational functions over Q, because e is transcendental over Q - therefore expressions cannot be simplified as e does not satisfy any polynomial over Q.

This can answer your more general question.Q(i,π,e)≅Q(i)(X,Y)\mathbb{Q}(i,\pi, e) \cong \mathbb{Q}(i)(X,Y), the field of rational functions in two variables over Q(i)\mathbb{Q}(i).
(edited 7 years ago)

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