Subfield Tests

Theorem

Let \(K\) be a non-empty subset of a field \(\mathbb{F}\). Then \(K\) is a subfield of \(\mathbb{F}\) if and only if

  1. \(K \neq \{0\}\)
  2. \(a - b \in K \quad \forall a, b \in K\)
  3. \(a \times b^{-1} \in K \quad \forall a, b \in K, b \neq 0\)
Proof

The main part of this proof follows from the subgroup tests which we apply to the whole set \(K\) as an additive group, and then to \(K - \{0\}\) as a multiplicative group.

Let \(a \in K\) noting that \(K \neq \varnothing\). Then \(a - a = 0 \in K\) from property (2). Then with \(a = 0\) we have that \(a - b = 0 - b = - b \in K\) for arbitrary \(b \in K\), hence we have closure under inverses in \((K, +)\). As such we have that \(a - (-b) = a + b \in K\) for arbitrary \(a, b \in K\). Hence \((K, +)\) is an abelian group, noting that commutativity is inherited from the parent group.

Now, let \(K^\ast = K - \{0\}\). We will prove that this set is an abelian group under multiplication. Let \(a, b \in K^\ast\), which we note is non-empty by assumption (1). Then from property (3), we have that \(b \times b^{-1} = 1 \in K\). This means that \(1 \times b^{-1} = b^{-1} \in K\) and therefore that \(a \times (b^{-1})^{-1} = a \times b \in K\). Again, commutativity follows from the parent group.

The distributive law then follows from the parent group, and closure as established above.