Integral domain without UF into irriducibles property

[gugo82]

Quite surprisingly, I propose an Algebra exercise...
It can be used also in class, for it is really simple.
Hope Martino will appreciate it.

***

Exercise:

Let:

[tex]$\mathbb{D}:=\left\{ p(X)=\sum_{n=0}^N a_nX^n \in \mathbb{Z}[X]:\ a_1=0 \right\}$[/tex]

(in other words, [tex]$\mathbb{D}$[/tex] is the set of the polynomials over the integers with zero coefficient in the "linear term" [tex]$X$[/tex]).

1. Prove that [tex]$\mathbb{D}$[/tex] is a subring of [tex]$\mathbb{Z} [X]$[/tex] and, more precisely, an integral domain with the operations induced by [tex]$\mathbb{Z} [X]$[/tex].

2. Prove (by providing a suitable counterexample) that [tex]$\mathbb{D}$[/tex] lacks the unique factorization into irriducibles property; in other words, find a polynomial [tex]$q(X)\in \mathbb{D}$[/tex] s.t. it is reducible (i.e. can be represented as product of nonunit lower grade polynomials) and can be expressed in two different forms as product of irriducible polynomials [tex]$\in \mathbb{D}$[/tex].

3. Find two polynomials [tex]$r(X),s(X) \in \mathbb{D}$[/tex] which are irreducible but are not prime.

[gugo82]

"gugo82":Let:

[tex]$\mathbb{D}:=\left\{ p(X)=\sum_{n=0}^N a_nX^n \in \mathbb{Z}[X]:\ a_1=0 \right\}$[/tex]

(in other words, [tex]$\mathbb{D}$[/tex] is the set of the polynomials over the integers with zero coefficient in the "linear term" [tex]$X$[/tex]).

1. Prove that [tex]$\mathbb{D}$[/tex] is a subring of [tex]$\mathbb{Z} [X]$[/tex] and, more precisely, an integral domain with the operations induced by [tex]$\mathbb{Z} [X]$[/tex].

Let [tex]p(X)=\sum_{n=0}^N a_nX^n,\ q(X)=\sum_{n=0}^M b_nX^n \in \mathbb{D}[/tex].
Clearly [tex]$p(X)+q(X)\in \mathbb{D}$[/tex]; on the other hand, if we let [tex]p(X)q(X)=\sum_{n=0}^{N+M} c_nX^n[/tex], then:

[tex]$c_1=a_1b_0+a_0b_1=0$[/tex]

therefore [tex]$p(X)q(X)\in \mathbb{D}$[/tex] and [tex]$\mathbb{D}$[/tex] is a subring of [tex]$\mathbb{Z}[X]$[/tex].
Moreover, if [tex]$p(X)\in \mathbb{D}\setminus \{ 0\}$[/tex] then either:

(1) [tex]$\deg (p(X))=0$[/tex] or [tex]$\deg (p(X))\geq 2$[/tex];

hence degrees' sum rule applies and [tex]$\mathbb{D}$[/tex] is an integral domain.

"gugo82":2. Prove (by providing a suitable counterexample) that [tex]$\mathbb{D}$[/tex] lacks the unique factorization into irriducibles property; in other words, find a polynomial [tex]$q(X)\in \mathbb{D}$[/tex] s.t. it is reducible (i.e. can be represented as product of nonunit lower grade polynomials) and can be expressed in two different forms as product of irriducible polynomials [tex]$\in \mathbb{D}$[/tex].

Let [tex]$q(X)=X^6$[/tex], then:

[tex]$q(X)=X^3\ X^3$[/tex] and [tex]$q(X)=X^2\ X^2\ X^2$[/tex]

are two possible representations of [tex]$q(X)$[/tex] as product of elements [tex]$\in \mathbb{D}$[/tex] (hence [tex]$q(X)$[/tex] is reducible in [tex]$\mathbb{D}$[/tex]), and these two factorizations are in fact distinct, for [tex]$X^2$[/tex] does not divide [tex]$X^3$[/tex] in [tex]$\mathbb{D}$[/tex] (the viceversa is trivial, because of the degrees' sum rule); moreover [tex]$X^2$[/tex] and [tex]$X^3$[/tex] are irriducibles [tex]$\mathbb{D}$[/tex] because of relation (1) and degrees' sum rule.
Therefore [tex]$\mathbb{D}$[/tex] does not have the unique factorization into irriducibles property, as we claimed.

"gugo82":3. Find two polynomials [tex]$r(X),s(X) \in \mathbb{D}$[/tex] which are irreducible but are not prime.

Polynomials [tex]$r(X)=X^2$[/tex] and [tex]$s(X)=X^3$[/tex] are irriducibles, but they are not prime: in fact:

- [tex]$X^2| X^3\ X^3$[/tex] but [tex]$X^2 \not| X^3$[/tex];

- [tex]$X^3 | X^4\ X^2$[/tex] but [tex]$X^3\not| X^2$[/tex] and [tex]$X^3 \not| X^4$[/tex].