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I've somewhere seen that whether $i \in\mathbb Q(\sqrt{d})$ is related to the Gaussian integer, but I don't understand why. Could you explain it?

Kanu Kim
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3 Answers3

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Recall that

$$ \mathbb{Q}[\sqrt{-3}] = \lbrace a +b\sqrt{-3}, a,b \in \mathbb{Q} \rbrace $$

So now suppose $\exists a,b \in \mathbb{Q}$

such that $$ \sqrt{-1} = a +b\sqrt{-3}$$

Then it must be the case that $$ -1 = a^2 - 3b^2 + 2ab\sqrt{-3}$$

Since $a^2 - 3b^2 \in \mathbb{Q}$ and $-1$ on the left side is also in $\mathbb{Q}$ it then follows that $2ab\sqrt{-3} \in \mathbb{Q}$ , but $\sqrt{-3} \not\in \mathbb{Q}$ and $\mathbb{Q}$ is closed under multiplication so we deduce that $2ab = 0$. Of course this means either $a = 0$ or $b=0$. Meaning that either:

$$ \sqrt{-1} = a$$

or $$\sqrt{-1} = b\sqrt{-3}$$

The first is ruled out obviously since $\sqrt{-1} \not\in \mathbb{Q}$ now the second case can be analyzed carefully:

$$\sqrt{-1} = b\sqrt{-3} \rightarrow -1 = -3b^2 \rightarrow \frac{1}{3} = b^2$$

But this is also impossible since $\sqrt{3}$ is irrational, so a rational $b$ such that

$$ \frac{1}{3} = b^2$$

Cannot exists, we thus conclude that $$\sqrt{-1} \not \in \mathbb{Q}[\sqrt{-3}]$$

Sidharth Ghoshal
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One quick way to see this: if a field contains both $i$ and $i\sqrt{3}$, then it contains $\sqrt{3}$.

So if $i\in\mathbb{Q}[\sqrt{-3}]$, we have $\sqrt{3}\in\mathbb{Q}[\sqrt{-3}]$, and therefore $\mathbb{Q}[\sqrt{3}]=\mathbb{Q}[\sqrt{-3}]$, as they are both extensions of $\mathbb{Q}$ of degree $2$.

But $\mathbb{Q}[\sqrt{3}]$ embeds into $\mathbb{R}$, while $\mathbb{Q}[\sqrt{-3}]$ doesn't.

Andrew Dudzik
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  • I know this argument, but I confused due to Bourbaki's answer of another question about this: http://math.stackexchange.com/questions/1273256/does-mathbb-q-sqrt-2-contain-a-square-root-of-1 What does mean "occur when d = -1, -3" here? – Kanu Kim Aug 28 '16 at 04:19
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    @KanuKim He observed that the only imaginary quadratic fields with units besides $1$ and $-1$ are $d=-1,-3$. From this observation, we can see that the only imaginary quadratic fields that could possibly contain $i$ are those with $d=-1,-3$. However, he made a mistake in asserting that $d=-3$ actually works. – Andrew Dudzik Aug 28 '16 at 05:31
  • Thanks, I misunderstood the comment. – Kanu Kim Aug 28 '16 at 05:43
  • Could you explain why the only imaginary quadratic fields with units besides (\pm 1) are d = -1, -3? – Kanu Kim Aug 28 '16 at 06:00
  • @KanuKim If $u$ is a unit in the ring of integers of a quadratic field, then either $u$ is rational (i.e. $u=\pm 1$) or $u$ satisfies a quadratic equation $X^2 + aX \pm 1$ with $a\in\mathbb{Z}$. Then the discriminant is $a^2 \mp 4$, which is always positive unless $a\in {-1,0,1}$. – Andrew Dudzik Aug 28 '16 at 06:25
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I'm not sure what you're thinking of, but a general result here is that for squarefree integers $a$ and $b$, different from $0$ and $1$, we have $$\sqrt{a}\in\mathbb{Q}(\sqrt{b})\iff\mathbb{Q}(\sqrt{a})=\mathbb{Q}(\sqrt{b})\iff a=b$$ In particular, it is certainly false that $\sqrt{-1}\in\mathbb{Q}(\sqrt{-3})$.

Zev Chonoles
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