1

Given the domain $\mathbb Z[\alpha]$, with $\alpha^2+2\alpha+4=0$, prove that it is normal.

We basically saw two facts about normal domains: one is that any UFD is normal (but $\mathbb Z[\alpha]$ isn't a UFD); the other is the following theorem: let $A$ be a normal domain and $K$ its field of fractions. Let $L$ be an extension of $K$. Then an element of $L$ is integral over $A$ iff the coefficients of its minimal polynomial over $K$ are all in $A$.

However the only way that I see to apply this theorem is with $A=\mathbb Z$, $K=\mathbb Q$ and $L=\mathbb Q[\alpha]$, and it doesn't seem to be useful, since $\mathbb Z[\alpha]$ is not involved. Can you give me a hint on how to start?

user26857
  • 52,094
Dr. Scotti
  • 2,493
  • Maybe you could explain what definition of "normal" you are using, this is a very overloaded term, with heavy dependance on the author. – Captain Lama Nov 06 '21 at 09:30
  • Every UFD is integrally closed. So is "normal" here "integrally closed"? – Dietrich Burde Nov 06 '21 at 09:31
  • Sorry I'm using this definition: a domain $A$ is normal if every element in $\operatorname {Frac} A$ integral over $A$ is in $A$. – Dr. Scotti Nov 06 '21 at 09:33
  • 1
    The roots are $-1\pm \sqrt{-3}$. By the way, the integral closure of $\mathbf{Z}$ in $\mathbf{Q}(\sqrt{-3})$ is the ring $\mathbf{Z}[\frac{1 + \sqrt{-3}}{2}]$, since $-3 \equiv 1 \bmod{4}$. – Dietrich Burde Nov 06 '21 at 09:52
  • 1
    from what @DietrichBurde said, the element $\frac{1+\sqrt{-3}}{2}$ is integral over $\mathbb{Z}[\alpha]$ as it solves the polynomial $x^2-x+1$ but it is not an element of $\mathbb{Z}[\alpha]$. –  Nov 06 '21 at 10:39
  • Sorry there must be a typo in the text of the exercise, also because the second request is to find the minimal natural $n$ such that $\mathbb Z [\alpha, n^{-1}]$ is normal. Even here I don't know what should I do, I tried to prove that $\mathbb Z [\alpha, 2^{-1}]$ is normal but maybe it's false. The only thing I noticed that maybe can be useful is that the field of fractions of $\mathbb Z [\alpha, n^{-1}]$ is again $\mathbb Q[\alpha]=\operatorname {Frac}(\mathbb Z[\alpha])$ – Dr. Scotti Nov 06 '21 at 12:30

1 Answers1

1

$\mathbb Z[\alpha]\simeq\mathbb Z[X]/(X^2+2X+4)$. But $\mathbb Z[X]/(X^2+2X+4)\simeq\mathbb Z[X]/(X^2+3)$ (send $X$ to $X-1$) which is at its turn isomorphic to $\mathbb Z[\sqrt{-3}]$. This is not integrally closed since $\frac{1+\sqrt{-3}}{2}$ is integral over $\mathbb Z[\sqrt{-3}]$ and does not belong to it.

In fact, the integral closure of $\mathbb Z[\sqrt{-3}]$ is $\mathbb Z[\frac{1+\sqrt{-3}}{2}]$. This gives us a hint about the question you mentioned in a comment: the least $n\ge 2$ such that $\mathbb Z[\alpha,n^{-1}]$ is integrally closed is $2$. In order to show that $\mathbb Z[\sqrt{-3},\frac12]$ is integrally closed let $x\in\mathbb Q(\sqrt{-3})$, the field of fractions of all three rings we mentioned in this paragraph, and suppose that $x$ is integral over $\mathbb Z[\sqrt{-3},\frac12]$. Then $2^kx$ is integral over $\mathbb Z[\sqrt{-3}]$ for some non-negative integer $k$. Since it is integral over $\mathbb Z[\frac{1+\sqrt{-3}}{2}]$ which is integrally closed, we have $2^kx\in\mathbb Z[\frac{1+\sqrt{-3}}{2}]$, and this implies that $x\in\mathbb Z[\sqrt{-3},\frac12]$.

user26857
  • 52,094
  • Thank you, the only part that is not clear to me is why the closure of $\mathbb Z[\sqrt 3]$ is $\mathbb Z[\frac{1+\sqrt {-3}}{2}]$; I really couldn't understand the arguement of Dietrich Burde in the comments – Dr. Scotti Nov 07 '21 at 08:45
  • 1
    You probably mean the integral closure of $\mathbb Z[\sqrt{-3}]$. For a proof see here: https://math.stackexchange.com/questions/1237549 – user26857 Nov 07 '21 at 09:04