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Question on generating subfields (field extensions).

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

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


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

    What I mean by that is  \mathbb{R}(i)=\{a+bi : a,b \in \mathbb{R}\}=\Bbb{C} or  \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)) = \{ 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(\alpha) where \alpha is a root of f (usually in C, but really whatever the algebraic closure of K is).

    To be super concrete,

    \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(\alpha)! This is the primitive element theorem. For example \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 \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.\mathbb{Q}(i,\pi, e) \cong \mathbb{Q}(i)(X,Y), the field of rational functions in two variables over \mathbb{Q}(i).
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