Show that Q[sqrt(2)] is a field

The Mathematician — Wed, 09 Nov 2022 13:00:00 +0000

Show that $\mathbb{Q}[\sqrt{2}]$ is a field

Proof. We have that

\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2} \ | \ a,b \in \mathbb{Q} \}

. So we need to show that all elements in

\mathbb{Q}[\sqrt{2}]

do have an inverse. Indeed, if we take arbitrary

a,b \in \mathbb{Q}

such that

a + b \sqrt{2} \in \mathbb{Q}[\sqrt{2}]

, then

\begin{align*} 1 &= (a + b \sqrt{2}) \frac{a - b \sqrt{2}}{(a + b \sqrt{2})(a - b \sqrt{2})} \\ &= (a + b \sqrt{2}) \frac{a - b \sqrt{2}}{(a^2 - ab\sqrt{2} + ab\sqrt{2} - 2b^2)} \\ &= (a + b \sqrt{2}) \frac{a - b \sqrt{2}}{(a^2 - 2b^2)} \end{align*}

and since

a^2 - 2b^2 \in \mathbb{Q}

and

a - b \sqrt{2} \in \mathbb{Q}[\sqrt{2}]

, this implies that

\frac{a - b \sqrt{2}}{(a^2 - 2b^2)}

is an inverse of

a + b \sqrt{2}

. So this implies that

\mathbb{Q}[\sqrt{2}]

is a field. This completes the proof.

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