Let $\mathbb{Q}[\sqrt{2}] = $ {$a + b \sqrt{2}$ | $a, b \in \mathbb{Q} $}. Prove that $\mathbb{Q}[\sqrt{2}]$ is a subfield of $\mathbb{C}$

Asked Jan 22 '20 at 22:10

Active Jan 22 '20 at 22:10

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Let $\mathbb{Q}[\sqrt{2}] = $ {$a + b \sqrt{2}$ | $a, b \in \mathbb{Q} $}. Prove that $\mathbb{Q}[\sqrt{2}]$ is a subfield of $\mathbb{C}$. Hint: First calculate $(a + b\sqrt{2})(a-b\sqrt{2})$

I have that $(a + b\sqrt{2})(a-b\sqrt{2}) = a^2 -2b$

I know that a subfield is a subring that is closed under inverse. However, I do not know how to prove the above.

asked Jan 22 '20 at 22:10

kt046172

What I meant to say was, can you write $\frac{1}{a + b\sqrt{2}}$ as an element of $\Bbb Q[\sqrt{2}]$ by noting that $a^2 - 2b \in \Bbb Q$ – Edward Evans Jan 22 '20 at 22:16
The subring part is straightforward. Use the hint to get inverses (by multiplying top and bottom by the conjugate to rationalize the denominator). – Ned Jan 22 '20 at 22:20

Let $\mathbb{Q}[\sqrt{2}] = $ {$a + b \sqrt{2}$ | $a, b \in \mathbb{Q} $}. Prove that $\mathbb{Q}[\sqrt{2}]$ is a subfield of $\mathbb{C}$

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