Field Adjoined Set

Theorem
Let $F$ be a field.

Let $S \subseteq F$ be a subset of $F$.

Let $K \le F$ be a subfield of $F$.

The subring $K \left[{S}\right]$ of $F$ generated by $K \cup S$ is the set of all finite linear combinations of powers of elements of $S$ with coefficients in $K$.

The subfield $K \left({S}\right)$ of $F$ generated by $K \cup S$ is the set of all $x y^{-1} \in F$ with $a, b \in K \left[{S}\right]$, $b \ne 0$.

$K \left({S}\right)$ is isomorphic to the quotient field $Q$ of $K \left[{S}\right]$.

Proof
Let $\left\{{X_s: s \in S}\right\}$ be a family of indeterminates indexed by $S$.

Let $\phi$ be the Evaluation Homomorphism such that $\phi \left({X_s}\right) = s$.

By Ring Homomorphism Preserves Subrings, $\operatorname{Im} \phi$ is a subring of $F$

This subring contains $\phi \left({X_s}\right) = s$ for each $s \in S$ and $\phi \left({k}\right) = k$ for all $k \in K$.

Moreover since it is obtained by evaluating polynomials on $S$ it is the set of all finite linear combinations of powers elements of $S$ with coefficients in $K$ as claimed.

All these linear combinations must belong to any subring of $F$ that contains $K$ and $S$ (otherwise it is not closed), so $\operatorname{Im} \phi$ is the smallest such subring.

By Universal Property for Quotient Field, the inclusion $K \left[{S}\right] \to F$ extends uniquely to a homomorphism $\psi : Q \to F$, given by $\psi \left({a / b}\right) = a b^{-1}$.

We have:


 * $\operatorname{Im} \psi = \left\{{a b^{-1} \in F: a, b \in K \left[{S}\right], \ b \ne 0}\right\} \simeq Q$

The isomorphism comes from the fact that a field homomorphism is injective and the First Isomorphism Theorem.

Thus $\operatorname{Im} \phi$ is a subfield of $F$ containing $K \left[{S}\right]$ and $K \cup S$.

Any subfield of $F$ containing $K$ and $S$ must contain $K \cup S$, $K \left[{S}\right]$ and all $a b^{-1}$ with $a, b \in K \left[{S}\right]$ (otherwise it would not be closed).

Therefore $Q \simeq \operatorname{Im} \phi$ is the smallest such subfield.