In the ring of polynomials with rational coefficients $\mathbb Q[x]$, let
$$I = \{h(x) \in \mathbb Q[x] : (x^2 - 2) \ \text{ divides }\ \ h(x)\}$$
Show $I$ is an ideal of $\mathbb Q[x]$.
I know I must show that the set is an additive subgroup and that if $s\in I$ and $r\in R$ then $rs$and $sr \in I$, but I am getting confused with the divider property in the first steps. Any help with this proof would help.
Also, how would I be able to show if this was a maximal ideal?
Thank you.
The condition that $x^2 - 2$ divides a polynomial $h(x) \in \mathbb{Q}[x]$ means that there exists $g(x) \in \mathbb{Q}[x]$ such that $h(x) = (x^2 - 2)g(x)$, i.e. $x^2 - 2$ is a factor of $h(x)$. Now if you have two polynomials $h_1(x)$ and $h_2(x)$ in $I$, then $(x^2 -2 )$ is a factor of both, a common factor. So we can write $h_1(x) + h_2(x) = (x^2 - 2)f(x)$ for some $f(x)$ and you can see the sum is in $I$. The other conditions you need to check follow similarly.
To see that $I$ is maximal, you just need to show that $x^2 - 2$ is irreducible (see why?). This follows from Eisenstein's criterion.
Edit: To go from irreducible polynomial to maximal ideal, you can proceed as follows: Suppose $J\subseteq \mathbb{Q}[x]$ is an ideal that properly contains $I$, so $I \subsetneq J$, and let $f \in J \setminus I$. Then $x^2 - 2$ doesn't divide $f$ (since $f \not\in I$) so their GCD is 1. By the Euclidean algorithm, you can find polynomials $s$ and $t$ such that $1 = f\cdot s + (x^2 - 2)\cdot t$. From here (check!) you get that $1 \in J$, hence $J = \mathbb{Q}[x]$. Thus the only ideal properly containing $I$ is the whole ring, so $I$ is maximal.