# Algebraic integers form a ring without irreducible elements, every non-zero prime ideal $\ne(0)$ is maximal, …

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.

• why can I only show that the square root of an algebraic integer is an algebraic integer to solve the problem? – milui Dec 31 '16 at 14:38
• A square cannot be irrreducible! – Bernard Dec 31 '16 at 14:39
• @Bernard a) The coefficients of the OP's polynomials are symmetric in the given roots, so they are in $\mathbb Z[X]$. – user26857 Jan 1 '17 at 0:42
• 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. – Bernard Jan 1 '17 at 1:19
• @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. – reuns Jan 1 '17 at 1:40