Subfield Test/Three-Step

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Theorem

Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.

Let $K$ be a subset of $F$.


$\struct {K, +, \times}$ is a subfield of $\struct {F, +, \times}$ if and only if these all hold:

$(1): \quad K^* \ne \O$
$(2): \quad \forall x, y \in K: x - y \in K$
$(3): \quad \forall x, y \in K: \dfrac x y \in K$

where $K^*$ denotes $K \setminus \set {0_F}$.


Proof

Necessary Condition

Let $\struct {K, +, \times}$ be a subfield of $\struct {F, +, \times}$.

Then the conditions $(1)$ to $(3)$ all hold by virtue of the field axioms.

$\Box$


Sufficient Condition

Suppose the conditions $(1)$ to $(3)$ hold.

As $K^* \ne \O$, it follows that $K \ne \O$.

From the One-Step Subgroup Test, it follows that $\struct {K, +}$ is a subgroup of $\struct {F, +}$.

Also from the One-Step Subgroup Test, it follows that $\struct {K^*, \times}$ is a subgroup of $\struct {F^*, \times}$.


As $\struct {F, +, \times}$ is a field, then $\times$ is commutative on all of $F$.

From Restriction of Commutative Operation is Commutative, $\times$ is commutative also on $K$.


As $\struct {F, +, \times}$ is a field, then $\times$ is distributive over $+$ on all of $F$.

From Restriction of Operation Distributivity $\times$ is distributive over $+$ also on $K$.


Thus $\struct {K, +, \times}$ is a field.

$\blacksquare$


Sources