Is this ideal principal?

Question

Let $I = \{ f(x)∈ \mathbb{Z}[x] | f(0) \text{ is an even integer}\}$

This is an ideal of the ring $R = \mathbb{Z}[x]$. Is it principal?

I have the definition of a principal ideal but I'm unsure if I've applied it properly.

If I choose $a = 2$, where $a$ is meant to generate $I$ (i.e. $I=aR$).

Is that enough to show that $I$ is a principal ideal?

Answer 1

The ideal is not the same as $2R$, because the polynomial $x$ belongs to $I$, but is not of the form $2g(x)$, for any $g(x)\in\mathbb{Z}[x]$.

Hint: this is not a principal ideal. Suppose $h(x)$ is a generator; then $2=h(x)p(x)$ and $x=h(x)q(x)$. Can you find a contradiction?

Answer 2

The ideal $I$ is not principal. You started the right way, by picking a candidate of generator, $a=2$. This is a reasonable candidate, because all even integers (even degree $0$ polynomials and $0$) are in your ideal. In particular, for degree reasons, if the ideal was principal this would have to be a generator. So the question is whether $2R$ is the whole ideal $I$ or not (it is easily seen that $2R\subseteq I$). The claim now is that this is a strict inclusion, i.e. $$ 2R\subsetneq I $$ To show this you need to find a polynomial in $I$ but not in $2R$, i.e. a polynomial in $I$ with not all coefficients even.

Hint: $0$ is an even number. What polynomials have $f(0)=0$? Can you find one such polynomial (say of degree $1$) with an odd coefficient?

Answer 3

I claim that in fact

$I = \langle 2, x \rangle, \tag 1$

so it is not principal. We show that (1) binds:

We have

$I = \{f(x) \in \Bbb Z[x] \mid f(0) \; \text{is even} \}; \tag 2$

thus

$f(x) = \displaystyle \sum_0^n f_i x^i \in I \Longleftrightarrow f_0 = f(0) = 2m, \; m \in \Bbb Z; \tag 3$

we may thus write

$f(x) = \displaystyle \sum_0^n f_i x^i = f_0 + \sum_1^n f_i x^i = 2m + x \sum_1^n f_i x^{i - 1} \in \langle 2, x \rangle, \tag 4$

which shows

$I \subset \langle 2, x \rangle; \tag 5$

likewise if

$f(x) \in \langle 2, x \rangle, \tag 6$

we have

$\exists h(x), g(x) \in \Bbb Z[x], \; f(x) = 2h(x) + xg(x); \tag 7$

if we write

$h(x) = \displaystyle \sum_0^{\deg h} h_i x^i \tag 8$

then

$2h(x) = \displaystyle \sum_0^{\deg h} 2h_i x^i = 2h_0 + \sum_1^{\deg h} 2h_i x^i; \tag 9$

thus

$f(0) = 2h(0) + 0(g(0)) = 2h_0, \tag{10}$

and so

$f(x) \in I; \tag{11}$

therefore (1) is established.

Now if $I$ is principle, we have some $d(x) \in \Bbb Z[x]$ such that

$I = \langle 2, x \rangle = \langle d(x) \rangle; \tag{12}$

therefore,

$d(x) \mid 2, \; d(x) \mid x; \tag{13}$

now

$d(x) \mid 2 \Longrightarrow d(x) = \pm 1, \pm 2, \tag{14}$

and we cannot have

$d(x) = \pm 1, \tag{15}$

since this implies

$I = \langle d(x) \rangle = \Bbb Z[x], \tag{16}$

which we have already seen is impossible by virtue of the fact that $I$ contains no polynomial with odd constant term; and if

$d(x) = \pm 2, \tag{17}$

then we have

$2 \mid x, \tag{18}$

which is also obviously impossible. Therefore we conclude that

$\not \exists d(x) \in \Bbb Z[x] \mid I = \langle d(x) \rangle, \tag{19}$

that is, that $I$ is not a principal ideal in $\Bbb Z[x]$.