Prove that $I$ is a finitely-generated ideal of $\mathbb{Z}[x]$ by finding a finite set of generators for $I$.

Question

Let $I = \{f(x) \in \mathbb{Z}[x] : \text{$f(n) \equiv 0 \pmod2$ for all $n \in \mathbb{Z}$}\}$. Prove that $I$ is a finitely-generated ideal of $\mathbb{Z}[x]$ by finding a finite set of generators for $I$.

I have no idea how to proceed with this problem!

Answer 1

Note that for a polynomial $f=a_m x^m+\cdots +a_0$ being in $I$ is equivalent to say that the constant term is even and all coefficients add up to an even number. Now I claim that with $J:=(x^2+x,x^2-x,2)$ we have $J=I$: let $g$ be the polynomial with those coefficients $b_i=a_i$ if $a_i$ is even and $b_i=a_i-1$ if $a_i$ is odd. Then $g\in J$ and if we can show $f-g\in J$, then so is $f$. But this is clear since $x^l+x^k \in J$ for all $l,k \in \mathbb{N}$ and $f-g$ is a sum of these. This shows $I\subseteq J$ and the other inclusion holds also.

Answer 2

$n\equiv n' \pmod 2$ implies $f(n) \equiv f(n') \pmod 2$. So the condition is $f(0) \equiv f(1) \equiv 0 \pmod 2$.

Consider an arbitrary polynomial $f(x)$ and divide by $x(x-1)$ with remainder. We get $$f(x) = x(x-1) q(x) + a x + b$$

Now $f(0) = b$ and $f(1) = a+b$. Therefore $f$ is in the ideal if and only if $a \equiv b \equiv 0 \pmod 2$.

We conclude $I = (x(x-1),2)$

A similar proof shows that if $p$ is prime the ideal $$I_p= \{ f \in \mathbb{Z}[x]\ | \ f(\mathbb{Z})\subset p \mathbb{Z}\}$$ is generated by $x(x-1)\ldots (x-p+1)$ and $p$. This implies in particular (use Fermat's little theorem ) that the polynomial $(x^p-x)- x(x-1)\ldots (x-p+1)$ has all the coefficients divisible by $p$. For instance $(x^5 - x)- x(x-1)(x-2)(x-3)(x-4)=5 x (x - 1) (2 x^2 - 5 x + 5)$.

Note that if $p$ is not prime, the ideal $I_p$ may not be generated by $x(x-1)\ldots (x-p+1)$ and $p$. For instance, $x(x-1)(x-2)$ is in $I_6$.

Note that every ideal of $\mathbb{Z}[x]$ is finitely generated, that is, $\mathbb{Z}[x]$ is noetherian.