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I am trying to find an integral closure of $\mathbb{Z}$ in $\mathbb{Q}[I]$

Wikipedia says that integral closure of $\mathbb{Z}$ in $\mathbb{Q}$ is a ring of integers $O_{\mathbb{Q}}$ and I think that the answer should be $O_{\mathbb{Q}}[I]$

But here it's said that the answer is Z[i], because Z is a UFD, and UFD is closed in its field of fractions.

Here the answer is finitely generated as well.

I see, that $x^n-2=0$ provides $\sqrt[n]{2}$ which lie in algebraic closure of $\mathbb{Z}$. I am afraid that I have messed up with terminology, so could you explain me what's wrong?

FFjet
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Lada Dudnikova
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    Is $I$ the imaginary unit for you? In other words, a solution of $x^2+1=0$? – Jyrki Lahtonen Aug 30 '19 at 14:50
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    Anyway, if I understood it correctly, you took a wrong turn when you forgot about the meaning of integral closure of $\Bbb{Z}$ in the field $K$. That last part of that phrase means that only the elements of $K$ are considered. So when you need to determine the integral closure of $\Bbb{Z}$ in the field $\Bbb{Q}(i)$, you only need to decide which numbers of the form $a+bi$, $a,b$ both rational, are integral over $\Bbb{Z}$. – Jyrki Lahtonen Aug 30 '19 at 14:55
  • (cont'd) The number $\root n\of2$ is not of that form, so is left out of the reckoning this time. But it would be included if the question were about the integral closure in the field of real numbers $\Bbb{R}$ (or some other suitable field). – Jyrki Lahtonen Aug 30 '19 at 14:56
  • The answer is, indeed, $\Bbb{Z}[i]$. Do you see why? – Jyrki Lahtonen Aug 30 '19 at 14:57
  • @JurkiLantonen No, I don't see why. Do you mean that the minimal polynomial of $\sqrtn$ is not $x^n-2$, or something else? – Lada Dudnikova Aug 30 '19 at 15:02
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    You already know from your other questions : we consider the integral closure of a ring $R$ in a field $K$ containing it. So what is your $R$ and $K$. – reuns Aug 30 '19 at 15:03
  • @JurkiLantonen I want to find an integral closure in the ring $\mathbb{Q}[i]$, not a field $\mathbb{Q}(I)$ – Lada Dudnikova Aug 30 '19 at 15:03
  • @JurkyLahtonen so, do you mean that we don't consider the integral closure in the ring, but in a field? – Lada Dudnikova Aug 30 '19 at 15:06
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    $\Bbb{Q}(i)$ and $\Bbb{Q}[i]$ are actually the same thing, but, yeah, nothings stops you from considering the integral closure inside a ring either. – Jyrki Lahtonen Aug 30 '19 at 15:06
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    The number $z=\root n\of2$ is not an element of the integral closure of $\Bbb{Z}$ in $\Bbb{Q}[i]$ because $z\notin \Bbb{Q}[i]$. – Jyrki Lahtonen Aug 30 '19 at 15:08
  • @JurkiLahtonen oh, I see. – Lada Dudnikova Aug 30 '19 at 15:11
  • @reuns Is it right that for a finite extension of $\mathbb{Q}$ the integral closure is finitely generated, but it's not finitely generated for infinite extensions? – Lada Dudnikova Aug 30 '19 at 15:30
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    $K/F$ is an Infinite extension means $K$ is algebraic over $F$ and it is an infinite dimensional $F$-vector space. With $R$ a subring of $F$ with $Frac(R)=F$ and $S$ the integral closure of $R$ in $K$ then $Frac(S)= K$ so if it is finitely generated then $K$ is finite dimensional. – reuns Aug 30 '19 at 15:34
  • @reuns and is it right that if you have an integral closure $A'$ of the ring $A$ in some field $K$ then the integral closure $A'_{ext}$ in the finite field extension $L$ is finitely generated as $A'$-algebra? – Lada Dudnikova Aug 30 '19 at 15:37
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    If the ideals of your base ring are finite generated so if the base ring is noetherian or something like that. For number fields and for algebraic varieties those things follow easily. – reuns Aug 30 '19 at 15:44

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