As shown in the title: How to find all the algebraic integer in $\mathbb{Q}[\sqrt{-1}]$ and $\mathbb{Q}[\sqrt{-3}]$
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Have you shown/seen/read that the sum and product of two algebraic integers is again algebraic? – Servaes Jan 05 '17 at 13:35
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Yes, I know it! – Yanner Jan 05 '17 at 13:36
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Well then, what is your definition of $\Bbb{Q}[\sqrt{-1}]$? – Servaes Jan 05 '17 at 13:38
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1In this case you can do it explicitly. Pick an element of your field and compute its minimal polynomial over $\mathbf Q$. See what the element must look like in order for the minimal polynomial to have integral coefficients. – Alex Macedo Jan 05 '17 at 13:39
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1Maybe he meant $\mathbb Q(\sqrt{-1})$ rather than $\mathbb Q[\sqrt{-1}]$. – Robert Soupe Jan 05 '17 at 13:39
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@Robert Soupe What's the difference? – Servaes Jan 05 '17 at 13:40
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@RobertSoupe adjoining an algebraic element to a field always yields a field – Alex Macedo Jan 05 '17 at 13:40
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It's a subfield of $\mathbb{C}$, which is generated by $\mathbb{Q}$ and $\sqrt{-1}$. – Yanner Jan 05 '17 at 13:41
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1In general, for these quadratics, the algebraic integers include at least numbers of the form $a + b \sqrt d$ where $a$ and $b$ are usual integers. It maybe also include other numbers depending on whether $d \equiv 1 \pmod 4$. If no one else writes up a more thorough answer, I'll do so tonight. – Robert Soupe Jan 05 '17 at 13:42
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@Alex We like to get hung up on notation on this website. – Robert Soupe Jan 05 '17 at 13:43
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I think no different, $\mathbb{Q}[x]$ and $\mathbb{Q}(x)$ has the same elements as a set. – Yanner Jan 05 '17 at 13:43
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@ Alex Macedo Thanks for your hint! – Yanner Jan 05 '17 at 13:51
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1This is a duplicate question, but I think we've failed to identify the correct original. – Mr. Brooks Jan 05 '17 at 22:15