Assume that all elements under discussion are algebraic over $F$.

Let the notation "$K=F(A)$" mean that $A\subseteq K$ and there is an injective homomorphism $\sigma:F\to K$, and every element of $K$ can be written as a rational function using elements from the image of $\sigma$ and from $A$.

What I'd like is to prove that the union of two finite field extensions of a base field $F$ is also a finite field extension of $F$. That is, if $K=F(\alpha_i)$ and $L=F(\beta_i)$ for some finite sets $\{\alpha_i\}$, $\{\beta_i\}$, then there is a field $M$ such that $M=F(\gamma_i)$ for some finite set $\{\gamma_i\}$ and there are injective homomorphisms from $K,L\to M$. Note in particular that I don't have a splitting field to work over, since this is part of the proof of the existence of a splitting field.

My original idea was to form the set $R$ of polynomials in $K$ with $n$ variables (where there are $n$ different $\beta_i$ variables), then take the set $R/I$ where $I$ is the ideal generated by all polynomials in $F$ (treated as a subset of $L$) whose evaluation under the map $x_i\mapsto\beta_i$ evaluates to $0$, but then I got confused since this always seems to build a larger field, even if you take $K\cup K$. If I take $K=\Bbb Q(\sqrt[3]{2})$ union with itself three times, do I get a field that splits $x^3-2$? If not, how do I get such a splitting field (for this one polynomial)?

  • 2
    $\begingroup$ Being a finite extension is not equivalent to adjoining a finite set of elements - the elements need to be algebraic, in which case they must already exist within some extension of the base field. Note that "the union" of extensions is ill-defined: they need to exist within a common larger field in order to take the union. Even in that case, taking the union of two fields will not be a field generally (for not closed under the operations). You're speaking of the compositum of extensions, which is the join in some lattice of extensions ordered by inclusion. $\endgroup$ – anon Nov 27 '14 at 1:06
  • $\begingroup$ Also, the set of rational functions (with scalars from a field) is a field, so there will be no proper nonzero ideals to quotient it by. You need to think about quotienting rings of polynomials. $\endgroup$ – anon Nov 27 '14 at 1:09
  • $\begingroup$ @anon I realize that "the union" is ill-defined, and literally taking the union of the base sets won't give me anything close to what I want; the purpose of this question is to find a suitable definition for it. If both fields were subfields of a bigger field, $\overline{A\cup B}$ would do the trick (where the bar means take the closure w.r.t the field operations). $\endgroup$ – Mario Carneiro Nov 27 '14 at 1:27
  • $\begingroup$ @anon Regarding "Being a finite extension is not equivalent to adjoining a finite set of elements", you are right, but these are algebraic field extensions (which are stages in building a complete splitting field on $F$), so we can assume that all the elements in $K,L,M$ are algebraic over $F$. $\endgroup$ – Mario Carneiro Nov 27 '14 at 1:34
  • $\begingroup$ Minor aside: you need to fix the embedding $F \to K$ rather than just require its existence: as an example of the problem, if I let $F = \mathbb{Q}(x)$ be the field of rational functions with rational coefficients, and $K = F(\sqrt{x})$, the algebraic extension of $F$ obtained by adjoining a square root of $x$, then the definition you gave would imply that $K = F(\varnothing)$. $\endgroup$ – Hurkyl Nov 27 '14 at 1:48

It's safe to add one element at a time.

If $K = F(\alpha)$ is a field extension of $F$, then $K \cong F[x] / f(x)$, where $f(x)$ is the minimal polynomial of $\alpha$ over $K$. Thus, if $R$ is any ring over $F$, then the ring homomorphisms $K \to L$ are in one-to-one correspondence with elements $\beta \in R$ satisfying $f(\beta) = 0$.

Now let $K$ and $F(\alpha)$ be field extensions of $F$. You construct the ring $R = K[x] / f(x)$, where $f(x)$ is the minimal polynomial of $\alpha$ (over $F$). It is indeed possible that $R$ is not a field; this happens if and only if $f(x)$ is not irreducible.

So to fix things, what you need is to find an irreducible polynomial $g(x)$ over $K$ such that one of the roots of $g(x)$ is a root of $f(x)$....

An alternative approach, I think, is to recognize that $R$ constructed above is a product of fields.


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