I'm really stuck in this exercise. Hope you can help me somehow. Any advice would be very nice.

Let $$\overline{\mathbb{Z}}:=\{z \in \mathbb{C}\mid f(z)=0 \text{ for a monic polynomial } f\in \mathbb{Z}[X]\}\subset\overline{\mathbb{Q}}\subset\mathbb{C}.$$

Show that:

a) $\overline{\mathbb{Z}}$ is a ring with no irreducible elements.

b) Every prime ideal $I \ne \{ 0 \}$ in $\overline{\mathbb{Z}}$ is maximal.

c) Prove whether $ \frac{-1+\sqrt{3}}{2}$ and $ \frac{-1+\sqrt{-3}}{2}$ are in $\overline{\mathbb{Z}}$.

At first, for a) "Showing that $\overline{\mathbb{Z}}$ is a ring" I would start with, showing that for $\alpha, \beta \in \overline{\mathbb{Z}}$, $-\alpha, \alpha +\beta$ and $\alpha \beta$ is in $\overline{\mathbb{Z}}$. I have already proved it for the easy first case : $-\alpha$. Now for $\alpha + \beta$: Suppose that $\alpha$ is the root of a polynomial, named $f$ and $\beta$ is the root of a polynomial, named $g$. Let $\alpha_1,..., \alpha_n$ be the set of the entire roots from $f$ and $\beta_1,...,\beta_m$ the roots from $g$. Now I would consider the polynomial:

$$h(X)=\prod \limits_{i=1}^{n}\prod \limits_{j=1}^{m} (X-(\alpha_i+\beta_j)).$$

At that point I don't know how to prove that the coefficients of $h$ are in $\mathbb{Z}$.

For $\alpha\beta$ it should work with $$\prod \limits_{i=1}^{n}\prod \limits_{j=1}^{m} (X-\alpha_i \beta_j),$$ but here I have a similar problem.

For the rest of a) and particularly in b) I have no clue.

I also tried to find some polynomials in c) but always failed. What comes to in my mind is that if we showed a, then we can use the fact that $\overline{\mathbb{Z}}$ is a ring and maybe first proof that $ \frac{-1}{2}$ is in $\overline{\mathbb{Z}}$ and then $\frac{\sqrt{3}}{2}$, but it's not so easy for me because those polynomials should be monic...

As you see I have many problems. Thank you very much, even if you can help a little.


Hint for a and c:

Prove that $\alpha\in \overline{\mathbf Z}\;$ if and only the ring $\mathbf Z[\alpha]$ is a finitely generated $\mathbf Z$-module.

No irreducible elements: just show the square root of an algebraic integer is an algebraic integer.

b) Show that any element $\alpha\notin \mathfrak p$ (a non-zero prime ideal of $\overline{\mathbf Z}$) is a unit modulo $\mathfrak p$. For that, show that $\mathfrak p\cap\mathbf Z[\alpha] $ is a maximal ideal of $\mathbf Z[\alpha] $.

Sub-hint: Set $\mathfrak p\cap\mathbf Z=p\mathbf Z$ and show the quotient $\mathbf Z[\alpha]/\mathfrak p\cap\mathbf Z[\alpha]$ is a finite-dimensional $\mathbf Z/p\mathbf Z$-vector space.

c) Compute the minimal polynomials of each element.

  • $\begingroup$ why can I only show that the square root of an algebraic integer is an algebraic integer to solve the problem? $\endgroup$ – milui Dec 31 '16 at 14:38
  • 2
    $\begingroup$ A square cannot be irrreducible! $\endgroup$ – Bernard Dec 31 '16 at 14:39
  • $\begingroup$ @Bernard a) The coefficients of the OP's polynomials are symmetric in the given roots, so they are in $\mathbb Z[X]$. $\endgroup$ – user26857 Jan 1 '17 at 0:42
  • $\begingroup$ Are you sure it's obvious? They're symmetric by groups (the $\alpha_i$s and the $\beta_j$s) for sure, but globally? Further, it uses a sophisticated theorem, and I think the argument with finitely generated modules (abelian groups in the present case) is simpler. $\endgroup$ – Bernard Jan 1 '17 at 1:19
  • $\begingroup$ @Bernard And for the finitely generated modules, I like to replace $\alpha$ by $A$ the companion matrix of its minimal polynomial, $\mathrm{Z}[A]$ being easier to visualize in my opinion, and it is very natural to apply the adjugate matrix argument for showing all its elements are algebraic integers. $\endgroup$ – reuns Jan 1 '17 at 1:40

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