Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations

Let $F$ be a field (Field), regarded as a commutative ring by Every field is a commutative ring with $1 \ne 0$; it is an integral domain, and it is a commutative division ring. A subset $K \subseteq F$ is a subfield of $F$ when

• (K1) $K$ is a subring of $F$ (Subring: a subset containing $1_R$ and closed under addition, additive inverses and multiplication);
• (K2) $x^{-1} \in K$ for every $x \in K$ with $x \ne 0_F$.

Equivalently, by Subring criterion: $S \subseteq R$ is a subring if and only if $1_R \in S$ and $a - b \in S$ and $ab \in S$ for all $a, b \in S$; and an intersection of subrings is a subring, $K$ is a subfield exactly when $1_F \in K$, $a - b \in K$ and $ab \in K$ for all $a, b \in K$, and $x^{-1} \in K$ for every nonzero $x \in K$.

Why $K$ is then a field, and with the same $0$ and $1$. By (K1) and Subring: a subset containing $1_R$ and closed under addition, additive inverses and multiplication, $K$ with the restricted operations is a ring whose zero is $0_F$ and whose identity is $1_F$; its multiplication is commutative, being the restriction of a commutative one (Commutative ring). Since $1_F \ne 0_F$ in $F$ and both lie in $K$, we have $1_K \ne 0_K$. Let $x \in K$ with $x \ne 0_K$; then $x \ne 0_F$, so $x^{-1} \in F$ exists and lies in $K$ by (K2), and $x x^{-1} = 1_F = 1_K = x^{-1} x$. So every nonzero element of $K$ is a unit of the ring $K$, and $K$ is a commutative division ring (Division ring: a ring with $1 \ne 0$ in which every nonzero element is a unit); by Every commutative division ring is a field, so "field" and "commutative division ring" name the same structures and the published definition and the ring-theoretic one agree it is a field. Moreover the inverse of $x$ computed in $K$ is its inverse computed in $F$, since $x^{-1}$ already satisfies the defining equation inside $K$.

In particular

$0_K = 0_F, \qquad 1_K = 1_F, \qquad (-x)_K = (-x)_F, \qquad (x^{-1})_K = (x^{-1})_F \quad (x \in K,\ x \ne 0_F).$

A subfield of an ordered field inherits the order. Let $(F,P)$ be an ordered field (Ordered field) and $K$ a subfield. Put $P_K := P \cap K$. Then (O1) holds in $K$: for $x \in K$ we have $-x \in K$ by (K1), and exactly one of $x \in P$, $x = 0_F$, $-x \in P$ holds in $F$, so exactly one of $x \in P_K$, $x = 0_K$, $-x \in P_K$ holds. And (O2) holds: if $x, y \in P_K$ then $x + y$ and $xy$ lie in $P$ by (O2) in $F$ and in $K$ by (K1), hence in $P_K$. So $(K,P_K)$ is an ordered field, and its order is the restriction of the order of $F$, because $a < b$ means $b - a \in P$ on both sides and $b - a$ is the same element in $K$ as in $F$.

• The inverse-closure clause is not implied by (K1). The integers sit inside the rationals as a subring that is not a subfield, since $2$ is nonzero there and $2^{-1}$ is not an integer; the companion page records that witness. So (K2) is doing work.

• The agreement of the two zeros and the two identities is the load-bearing part. A later page restricts the scalars of a vector space along a subfield inclusion, and every axiom checked there uses that $1_K$ acts as $1_F$ does. Nothing would go through if a "subfield" were merely a subset that happens to be a field under some operations of its own.

• Why the definition goes through subrings rather than restating the field axioms. All of (A), (M) and (D) except the existence of inverses are already guaranteed by Subring: a subset containing $1_R$ and closed under addition, additive inverses and multiplication, and the two bridge lemmas of this page convert the resulting commutative division ring back into a field. Restating the axioms would create a second definition of a field on this page, which is exactly what Every field is a commutative ring with $1 \ne 0$; it is an integral domain, and it is a commutative division ring and Every commutative division ring is a field, so "field" and "commutative division ring" name the same structures and the published definition and the ring-theoretic one agree exist to prevent.