Field Adjoined Set

Theorem
Let $F$ be a field, $S \subseteq F$ a subset and $K \leq F$ a subfield.

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

The subfield $K(S)$ of $F$ generated by $K \cup S$ is the set of all $xy^{-1} \in F$ with $a,b \in K[S]$, $b \neq 0$, and $K(S)$ is isomorphic to the quotient field $Q$ of $K[S]$.

Corollary
Let $A = K[X_1,\ldots,X_n]$ be the ring of polynomial functions in $n$ indeterminates over $K$.

Let $B = K(X_1,\ldots,X_n)$ be the field of rational functions in $n$ indeterminates over $K$.

Let $\alpha_1,\ldots,\alpha_n \in F$. Then:


 * 1. $x \in K[\alpha_1,\ldots,\alpha_n] \Leftrightarrow x = f(\alpha_1,\ldots,\alpha_n)$ for some $f \in A$


 * 2. $x \in K(\alpha_1,\ldots,\alpha_n) \Leftrightarrow x = f(\alpha_1,\ldots,\alpha_n)$ for some $f \in B$


 * 3. $x \in K[S] \Leftrightarrow x \in K[\alpha_1,\ldots,\alpha_n]$ for some $\alpha_1,\ldots,\alpha_n \in S$


 * 4. $x \in K(S) \Leftrightarrow x \in K(\alpha_1,\ldots,\alpha_n)$ for some $\alpha_1,\ldots,\alpha_n \in S$

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

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

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

This subring contains $\phi(X_s) = s$ for each $s \in S$ and $\phi(k) = k$ 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 $\operatorname{im}\phi$ is the smallest such subring.

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

We have:


 * $\operatorname{im}\psi = \{ab^{-1} \in F : a,b \in K[S],\ b\neq 0\} \simeq Q$

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 $K[S]$ and $K \cup S$.

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

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