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 $K \sqbrk S$ 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 $\map K S$ of $F$ generated by $K \cup S$ is the set of all $x y^{-1} \in F$ with $a, b \in K \sqbrk S$, $b \ne 0$.

$\map K S$ is isomorphic to the field of quotients $Q$ of $K \sqbrk S$.

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

Let $\phi$ be the Evaluation Homomorphism such that $\phi \sqbrk {X_s} = s$.

By Ring Homomorphism Preserves Subrings, $\Img \phi$ is a subring of $F$

This subring contains $\phi \sqbrk {X_s} = s$ for each $s \in S$ and $\map \phi k = k$ 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 $\Img \phi$ is the smallest such subring.

By Universal Property for Field of Quotients, the inclusion $K \sqbrk S \to F$ extends uniquely to a homomorphism $\psi : Q \to F$, given by $\map \psi {a / b} = a b^{-1}$.

We have:


 * $\Img \psi = \set {a b^{-1} \in F: a, b \in K \sqbrk S, \ b \ne 0} \simeq Q$

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

Thus $\Img \phi$ is a subfield of $F$ containing $K \sqbrk S$ and $K \cup S$.

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

Therefore $Q \simeq \Img \phi$ is the smallest such subfield.