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I've just learned what the integral closure is.

I would like to find what is the intergral closure of $\mathbb{Z}$ in $\mathbb{Q}[i]$.

Let $\mathcal{R}$ the integral closure of $\mathbb{Z}$ in $\mathbb{Q}[i]$. To determine $\mathcal{R}$ I started to note that $$\mathbb{Z}[i]\subset \mathcal{R}.$$ Indeed if $\alpha = x+iy\in \mathbb{Z}[i]$, $\alpha$ is a root of the monic polynomial $$f(X)=X^2-2xX+x^2+y^2\in\mathbb{Z}[X].$$

And we know that since $\mathbb{Z}$ and $\mathbb{Q}[i]$ are rings so is $\mathcal{R}$.

From there I'm not sure, is it true that there exist no ring $A$ such that $$\mathbb{Z}[i] \subsetneq A \subsetneq \mathbb{Q}[i] \text{ ?}$$ I've assumed that and show that $1/2\notin \mathcal{R}$ for example, otherwise it exists a monic polynomial $f\in \mathbb{Z}[X]$ such that

$$f\left(\frac{1}{2}\right) =\frac{1}{2^n}+a_{n-1}\frac{1}{2^{n-1}}+\cdots+a_1\frac{1}{2}+a_0 =0$$

So

$$2^n f\left(\frac{1}{2}\right) =0 \iff -1=2\underbrace{(a_{n-1}+\cdots +2^{n-2}a_1+2^{n-1}a_0)}_{\in \mathbb{Z}}$$

which is impossible.

In conclusion $\mathbb{Q}[i]\not\subset\mathcal{R}$ so $\mathcal{R}=\mathbb{Z}[i]$.

Is my proof correct ? Is there an other (easier) way to prove that ? And the most important, is there exist a general method to find the integral closure please ?

Thank you for your help.

user26857
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Zanzi
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    To answer one of your questions, do we not have, say, $\mathbb{Z}[i] \subsetneq \mathbb{Z}[i,\frac{1}{2}] \subsetneq \mathbb{Q}[i]$? – Alex Provost Feb 25 '16 at 14:06
  • The usual English terminology would be "integral closure of $\mathbb Z$ in $\mathbb Q(i)$, not "over", and I edited title and text in this regard. – paul garrett Feb 25 '16 at 14:09
  • Your polynomial for $f(X)$ is wrong - it should be $f(X)=X^2-2xX+x^2+y^2$. And, unless absolutely necessary, don't use $x$ and $X$ as variables in the same answer. – Thomas Andrews Feb 25 '16 at 14:25
  • You right @AlexProvost , I feel bad about written that.. – Zanzi Feb 25 '16 at 16:34
  • Yes, you right @ThomasAndrews, I'll edit my post, thanks. – Zanzi Feb 25 '16 at 16:36

1 Answers1

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$X = a+ib \in \mathbb Q[i]$ is integral iff it is a solution of a monic polynomial in $\mathbb Z[X]$ of degree 1 or 2. Calculating $X^2$ we get

$X^2 -2aX +a^2 + b^2=0$.

Thus $-2a$ and $a^2 + b^2$ must be integers. So if $a$ is an integer, $b$ is an integer, giving us $\mathbb Z[i]$. If $a=\frac k 2$ with $k$ odd integer, we get $k^2/4 + b^2 \in \mathbb Z$ and so $4b^2 + k^2 \equiv 0 \pmod 4$, absurd.

There are a lots of methods to find a ring integers, but as far as I know there is not a standard method, unfortunately. For example the ring of integers of $\mathbb Q[\zeta_n]$, with $\zeta_n$ the $n$-th root of unit, is hard to compute (even if it's the expected $\mathbb Z[\zeta_n]$).

Maffred
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    I'm guessing, but not sure, that the OP is wondering about higher degree polynomials (otherwise, $1/2$ is a no-braiiner, right?) It requires some work using unique factorization of polynomials in $\mathbb Q[x]$ and its relation to factorization in $\mathbb Z[x]$. – Thomas Andrews Feb 25 '16 at 14:31
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    I don't get the point. Since the extension's degree is 2 there cannot be integral elememts with mnimum polynomial of dimension higher then 2. – Maffred Feb 25 '16 at 14:36
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    That assumes the OP knows what "extension of degree $2$" means, or a lot of other stuff the OP hasn't gotten to yet. You first have to show that $1/2$ is not the root of a monic integer polynomial, for example, at some point, somewhere. How do you know, if $2a$ or $a^2+b^2$ is not an integer, that $X^2-2aX+(a^2+b^2)$ is not a factor of some higher-degree monic integer polynomial? You have to prove this at some point, and it is long before you have defined degree of extensions. – Thomas Andrews Feb 25 '16 at 14:40
  • I suggest to use the fact that if $A$ is integrally closed, $x$ is integral iff the minimum polynomial is in $A[x]$. All you need to prove this is the fact that the coefficients are symmetric polynomials in the roots. If OP doesn't know anything about Galois Theory and he wants to study ANT I give up! – Maffred Feb 25 '16 at 15:02
  • Hello @Maffred , thank you for your responce. I don't really understand why the degree of the monic polynomial is lower or equal than 2. But I don't know what the "extension's degree" means as well. I'll look it up. Can I have some precisions about how you find $X^2−2aX+a^2+b^2=0$please ? – Zanzi Feb 25 '16 at 16:50
  • @Zanzi I squared both sides of $X=a + ib$ after bringing $a$ on the left side. – Maffred Feb 25 '16 at 16:53
  • oh yes, of course, thank you @Maffred. – Zanzi Feb 25 '16 at 17:02