So I'm essentially trying to find an explicit description of the smallest subfield generated by a subset of the field. I know that if s is an element of the subset, we must also have its additive and multiplicative inverse. Is there a succint way of describing the subfield set-theoretically? Again, the intuition seems clear, but I'm struggling with a formal description...
Smallest Subfield Generated by a set
3 Answers
It is the set of rational functions over elements of the generating set. I.e., every element has the form: $\frac{p(s_1, \ldots, s_n)}{q(s_1, \ldots, s_n)}$ where $p(X_1, \ldots X_n)$ and $q(X_1, \ldots X_n)$ are polynomials with integer coefficients in the indeterminates $X_1, \ldots, X_n$, for some $n$ and $s_1 \ldots s_n$ range over elements of the generating set such that $q(s_1, \ldots s_n) \neq 0$ .
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0Wouldn't you have to add the condition that $q(s_1,\ldots,s_n)\neq0$ ? – 2015-12-12
There are two ways to consider a subfield $K$ of a field $F$ generated by a subset $S \subset F$. The "building up" approach has already been described, so here is the "top-down" approach (I find this more useful in practice).
We can consider all subfields $L \subset F$ such that $S \subset L \subset F$. Then $K = \bigcap_{L \in \mathcal{L}} L$, where $\mathcal{L} = \{ L | L \subset F \text{ is a subfield with } S \subset L \}$. That is, $K$ is the intersection of all the subfields of $F$ which contain $S$.
To see that $K$ is truly a field, we need to notice that it contains all the right things. Certainly $0,1 \in K$ and if $s \in K$, then $s^{-1}$ and $-s$ are in any field containing $s$, so they are in $K$. Similarly, if $s, t \in K$ then $s+t, st \in K$.
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0Yeah, I prefer this description too. I failed to point out that I already had this version handed to me. This way of thinking works in much greater generality for dealing with "subobjects." – 2012-09-07
I like to think of this by building it up in stages. First take monomials in your subset $S$, so anything you can get by just multiplying elements of $S$ together. Then you can add these monomials up, so you get some finite sum of terms, each of which is a product of elements of $S$. Finally, you can take quotients of these, so for two such expressions $f$ and $g$, where $g\ne0$, you get the quotient $\frac{f}{g}=fg^{-1}$.
So the subfield generated by $S$ is exactly the set of these quotients $\frac{f}{g}$, where $f$ and $g$ are sums of monomials in $S$.