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I'm trying to find a degree of the extension $ \mathbb{Q} \subset \mathbb{Q}(\sqrt{2} + i) $. Once I'm done with that, I'd like to find a basis of $ \mathbb{Q} (\sqrt{2} + i) $ as a $ \mathbb{Q} $ -linear space

My attempt: $ (\sqrt{2} + i)^2 = 1 + 2\sqrt{2}i $. This is a root of the following polynomial: $P(x) = x^2 - 2x + 9 \in \mathbb{Q}[x]$ (that can be found out by guessing - maybe there is a more efficient / universal way?)

Therefore $ a = \sqrt{2} + i$ is a root of $ R(x) = P(x^2) = x^4 - 2x^2 + 9 \in \mathbb{Q}[x] $

If I manage to prove that $R(x) $ is irreducible in $ \mathbb{Q}[x] $, it'll mean that the degree is 4. Eisenstein's criterion is of no use. Other way is to express $ R(x) $ as $ R(x) = (x - a_1)(x-a_2)(x-a_3)(x- a_4) $, where $ a_i $ are in $ \mathbb{C} \setminus \mathbb{R} $ and show that all products of form $ (x- a_i)(x-a_j) $ are not from $ \mathbb{Q}[x] $ - but that's not very elegant. Is there a better way?

And aboout finding a basis - I don't really have a good idea how to do that.

So my questions is: is the following argument all right? And how do I find a basis of this extension?

I'd appreciate some help

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    $\begingroup$ You can consider the polynomial $R(x+k)$ where $k\in \mathbb{Z}$. This polynomial is irreducible if and only if $R(x)$ is. Try to find a $k$, by a little brute forcing, that gives $R(x+k)$ the proper form to apply Eisenstein. $\endgroup$ – Nicolas Bourbaki Apr 13 '15 at 17:32
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    $\begingroup$ Isn't it more natural to prove that $\mathbb{Q}(\sqrt{2} + i) = \mathbb{Q}(\sqrt{2}, i) $ ? $\endgroup$ – krirkrirk Apr 13 '15 at 17:34
  • $\begingroup$ A basis of $\mathbb{Q}(\alpha)$ is given by the powers of $\alpha$ (up to the degree of the extension minus one). $\endgroup$ – Joel Cohen Apr 13 '15 at 17:35
  • $\begingroup$ +1 I love this question. It's so thoughtful. It's clear you're new to this content but equally clear you're really interested in making sense of it for yourself. $\endgroup$ – Ben Blum-Smith Apr 13 '15 at 17:36
  • $\begingroup$ @krirkrirk - I am guessing that the OP didn't realize this. $\endgroup$ – Ben Blum-Smith Apr 13 '15 at 17:37
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Note: I don't address your attempts in this answer, and just dive toward an alternate solution, so look away if you aren't interested. I did leave some stuff for you to do, though.

Well, let's first try to prove the claim: $$\Bbb Q(\sqrt{2} + i) = \Bbb Q(\sqrt{2}, i)$$ If this is true, then both will have the same basis and degree.

  1. Notice that $\sqrt{2} + i \in \Bbb Q(\sqrt{2} + i)$, so $(\sqrt{2} + i)^{2} \in \Bbb Q(\sqrt{2} + i)$. But $(\sqrt{2} + i)^{2} = 2 + 2\sqrt{2}i - 1 = 1 + 2\sqrt{2}i$. This means $1 + 2\sqrt{2}i \in \Bbb Q(\sqrt{2} + i)$.

  2. Then by subtracting $1$ and subsequently dividing by $2$, we get that $\sqrt{2}i \in \Bbb Q(\sqrt{2} + i)$.

  3. But $\Bbb Q(\sqrt{2} + i)$ is closed under multiplication, so $(\sqrt{2} + i)\sqrt{2}i \in \Bbb Q(\sqrt{2} + i)$. And $(\sqrt{2} + i)\sqrt{2}i = 2i - \sqrt{2}$.

  4. Then $2i - \sqrt{2} + (\sqrt{2} + i) = 3i \in \Bbb Q(\sqrt{2} + i)$, and dividing by $3$ gives us that $i \in \Bbb Q(\sqrt{2} + i)$.

  5. Similarly, if we add $(2i - \sqrt{2})$ with $-2(\sqrt{2} + i)$, we get $-2\sqrt{2}$, which implies $\sqrt{2} \in \Bbb Q (\sqrt{2} + i)$.

Thus, since $\sqrt{2}, i \in Q (\sqrt{2} + i)$, it follows that $\Bbb Q(\sqrt{2}, i) \subseteq Q (\sqrt{2} + i)$.

It should be clear that $Q(\sqrt{2}, i) \supseteq Q (\sqrt{2} + i)$, so the two are equal.

This helps us because now, we can find the degree and the basis. Try to figure both out and let me know if you need more help.

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  • $\begingroup$ Also, I encourage you to use a similar proof technique to prove that for any natural numbers $m, n$, $\Bbb Q(\sqrt{m} + \sqrt{n}) = \Bbb Q(\sqrt{n}, \sqrt{m})$. $\endgroup$ – layman Apr 13 '15 at 17:41
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    $\begingroup$ A easier way (I think) is to write $a = \sqrt{2} +i$ and to develop the identites $(a-i)^{2} = 2$ and $(a - \sqrt{2})^{2} = -1$ $\endgroup$ – krirkrirk Apr 13 '15 at 17:41
  • $\begingroup$ @krirkrirk Sure, that looks much cleaner than my work. Typing calculations like this are a pain, and after all that hard work, it doesn't even look nice in the end. $\endgroup$ – layman Apr 13 '15 at 17:44
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Hint: Notice that $\mathbb Q (\sqrt 2 + i) = \mathbb Q (\sqrt 2 , i)$. In fact,

$(\supseteq)$ Show that $$\sqrt 2 - i = \frac{(\sqrt 2 - i)(\sqrt 2 + i)}{\sqrt 2+ i}$$ is in $\mathbb Q (\sqrt 2 + i)$

$(\subseteq)$ It's clear.

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Since $\;P(x)\;$ is real its complex non-real roots appear as conjugate pairs, then also $\;\sqrt2-i\;$ is a root, and thus $\;x^4-2x^2+9\;$ is divisible by

$$\;(x-(\sqrt2-i))(x-(\sqrt2+i))=x^2-2\sqrt2 x+3\;$$

Dividing $\;P(x)\;$ by the above quadratic we get

$$x^4-2x^2+9=\left(x^2-2\sqrt2x+3\right)\left(x^2+2\sqrt2x+3\right)$$

Since $\;P(x)\;$ has no rational roots (why?) the only possibility to factor it over the rationals is as product of two quadratics. But the above shows this is possible only with two real non-rational quadratics, and since $\;\Bbb R[x]\;$ is a UFD this shows $\;P(x)\;$ is really irreducible.

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