Question

Let Z[x] be the domain of all polynomials with integer coefficients. Consider the constant polynomial 2 and the polynomial x. Let S = {a(x)*2 + b(x)*x: a(x),b(x) w/in Z[x]}. (note: Z is integers symbol)
a)Prove S = {f(x) w/in Z[x]: f(0) w/in 2Z}
b)Prove S is an ideal in Z[x]
c)Prove there is no polynomial d(x) w/in Z[x] such thats S = {q(x)d(x): q(x) w/in Z[x]}. [This Z[x] is not a principle ideal domain]

Answer 1

a)

$$\forall s (x) \in S: s (x) = 2 a (x) + x b (x)$$ $$s (0) = 2 a (0) + 0 = 2 a (0) \in 2 \mathbb {Z} \Rightarrow s (x) \in \left\{f (x) \mid f (0) \in 2 \mathbb {Z} \right\}$$ $$\forall s (x) \in \left\{f (x) \mid f (0) \in 2 \mathbb {Z} \right\} \Rightarrow s (x) = 2 a _ {0} + a _ {1} x + a _ {2} x ^ {2} + \dots + a _ {n} x ^ {n} \in S$$ $$S = \left\{f (x) \mid f (0) \in 2 \mathbb {Z} \right\}$$

b)

S can be expressed in form $S = 2\mathbb{Z}[x] + x\mathbb{Z}[x]$ , so it is finitely generated ideal (by definition) in domain $\mathbb{Z}[x]$ .

c)

If $S = d(x)\mathbb{Z}[x]$ then $d(x)$ have to divide both 2 and $x$ , but only constants are common divisors.

So $d(x) = \text{const}$ , and only possible constants are 1 and -1. So there is no $d(x)$ with this property.

$\mathbf{Z}[\mathbf{x}]$ is not PID.