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The Gaussian integers are defined as the numbers that can be written as $a + bi$ with $a, b$ rational numbers, and for which there is a monic polynomial $P \in \mathbb{Z}[X]$ such that $P(a + bi) = 0$.

I would like to show that these numbers are precisely the $a + bi$ with $a, b$ integers.

The polynomial $P = (X - (a + bi))(X - (a - bi)) = X^{2} - 2aX + (a^{2} + b^{2})$ is in $\mathbb{Q}[X]$ and verifies $P(a + bi) = 0$. If I could show that in fact $P \in \mathbb{Z}[X]$ then I would get that $2a$ and $a^{2} + b^{2}$ are integers, and then I would know how to end the proof.

However, I can't manage to explain why $P$ must be in $\mathbb{Z}[X]$. If we take $b \neq 0$ then $P$ is the polynomial in $\mathbb{Q}[X]$ with smallest degree that verifies $P(a + bi) = 0$. But this does not help me ...

Thank you for your help.

Edit: after some extra research, I found a PDF where it is explicitly said that in order to prove what I talked about; we can first show that $2a$ and $a^{2} + b^{2}$ are in $\mathbb{Z}$, without using anything like Gauss' lemma. So even though I did get an answer, I would be curious to see how to conclude this way: how do we prove that these two numbers are integers ?

JackEight
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    Monic polynomial. If $a,b$ are integers then $P \in \Bbb{Z}[X]{monic}$. Conversely if $P \in \Bbb{Z}[X]{monic},P(a+ib)=0$ then $2a,a^2+b^2\in \Bbb{Z}$ so that... – reuns Dec 19 '20 at 14:42
  • Yes thank you, I had forgotten the essential word monic ... However, I am not sure I understand how this solves my problem. You have show that $P \in \mathbb{Z}[X]_{monic} \iff a, b$ are integers. But $P$ is not necessarily the monic polynomial with the smallest degree such that $P(a + bi) = 0$, right ? – JackEight Dec 19 '20 at 14:52
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    You are right, this is Gauss lemma that the minimal polynomial is in $\Bbb{Z}[X]_{monic}$ – reuns Dec 19 '20 at 14:55
  • Thank you for the idea. But is there not something easier than this ? I mean, I'm satisfied by your answer, but I would prefer something more "natural" – JackEight Dec 19 '20 at 15:06
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    Once you know that any $w\in \Bbb{Q}[i]^*$ is of the form $w=\frac{u}{v}$ with $u,v\in\Bbb{Z}[i]$ and $u$ inversible modulo $v$ (ie. $u\in\Bbb{Z}[i]/v\Bbb{Z}[i]^\times$) then for $P\in \Bbb{Z}[X]_{monic}$ it is immediate that $v^{\deg(P)} P(u/v)\not\equiv 0\bmod (v)$ so $P(u/v)\ne 0$. – reuns Dec 19 '20 at 15:12
  • Thank you. I edited my question because of something I found on the Internet – JackEight Dec 19 '20 at 16:03
  • I meant any $w\in \Bbb{Q}[i],\not \in \Bbb{Z}[i]$ – reuns Dec 19 '20 at 17:01

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