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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? – 2017-01-05
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    Yes, I know it! – 2017-01-05
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    Well then, what is your definition of $\Bbb{Q}[\sqrt{-1}]$? – 2017-01-05
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    In 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. – 2017-01-05
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    Maybe he meant $\mathbb Q(\sqrt{-1})$ rather than $\mathbb Q[\sqrt{-1}]$. – 2017-01-05
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    @Robert Soupe What's the difference? – 2017-01-05
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    @RobertSoupe adjoining an algebraic element to a field always yields a field – 2017-01-05
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    It's a subfield of $\mathbb{C}$, which is generated by $\mathbb{Q}$ and $\sqrt{-1}$. – 2017-01-05
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    In 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. – 2017-01-05
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    @Alex We like to get hung up on notation on this website. – 2017-01-05
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    I think no different, $\mathbb{Q}[x]$ and $\mathbb{Q}(x)$ has the same elements as a set. – 2017-01-05
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    @ Alex Macedo Thanks for your hint! – 2017-01-05
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    This is a duplicate question, but I think we've failed to identify the correct original. – 2017-01-05

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