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Let $\mathbb{Q}(\sqrt{d}) = \{a + b \sqrt{d}: a,b \in \mathbb{Q} \}.$ Show that $\mathbb{Q}(\sqrt{d})$ is a field.

Everything seems obvious except for existence of inverses in the multiplicative group of $\mathbb{Q}(\sqrt{d}).$

Suppose $\alpha \in \mathbb{Q}(\sqrt{d}) \Longrightarrow \alpha = a + b \sqrt{d}.$ To prove that $\mathbb{Q}(\sqrt{d})$ contains multiplicative inverses, observe that for a non-square $d$ and $a - b\sqrt{d} \ne 0$ one has $$\frac{1}{\alpha} = \frac{1}{a+b\sqrt{d}} \cdot \frac{a - b \sqrt{d}}{a - b \sqrt{d}} = \frac{a}{a^2-b^2d}- \frac{b}{a^2-b^2d} \sqrt{d}.$$

If $a,b \in \mathbb{Q} \Longrightarrow a^2,b^2 \in \mathbb{Q} \Longrightarrow b^2 d \in \mathbb{Q}$ (since $d \in \mathbb{Z}$) $\Longrightarrow a^2-b^2d \in \mathbb{Q} \Longrightarrow \frac{a}{a^2-b^2d}, \frac{b}{a^2-b^2d} \in \mathbb{Q}.$ Hence $\frac{1}{\alpha} \in \mathbb{Q}(\sqrt{d}).$

Is this correct?

St Vincent
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    We need to make the remark that $a-b\sqrt{d}\ne 0$. So for the calculation to be justified, we need $d$ not a perfect square, and we need to prove that in that case$a-b\sqrt{d}$ can only be $0$ if $a=b=0$, that is, we must show (or remark) that this is true because $\sqrt{d}$ is irrational. – André Nicolas May 27 '15 at 23:14
  • Is the fact that $ d\equiv 2,3 \pmod 4$ a hypothesis? –  May 27 '15 at 23:17
  • @AndréNicolas thanks for the suggestions, I will fix it. – St Vincent May 27 '15 at 23:20
  • You are welcome. Things look fine apart from the little gap I mentioned. – André Nicolas May 27 '15 at 23:21

1 Answers1

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Yes, that is essentially correct.

However, your comment about $d\bmod 4$ is unnecessary (and sort of wrong, because you certainly can consider the field $\mathbb{Q}(\sqrt{d})$ when $d\equiv 1\bmod 4$). Regardless of anything else, $d$ is an integer, so you know $b^2\in\mathbb{Q}$ implies $b^2d\in\mathbb{Q}$.

Zev Chonoles
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