A problem with polynomial rings an ideals Define $R=\{\sum_{i=0}^m a_iX^i\in \mathbb{Q}[X], a_0\in \mathbb{Z}\}$ and $I=\{\sum_{i=0}^m a_iX^i\in \mathbb{Q}[X], a_0=0\}$.
Prove
Part (a) $R$ is a subring of $\mathbb{Q}[x]$ and $I$ ideal of $R$.
To see that $R$ is a subring: 
(1) It is obvious $R\subset\mathbb{Q}[x]$,
(2) $\left(\sum_{i=0}^m a_iX^i\right) + \left(\sum_{i=0}^m b_iX^i\right)=\sum_{i=0}^m (a_i+b_i)X^i\in R,$
(3) $\left(\sum_{i=0}^m a_iX^i\right)\left(\sum_{i=0}^m b_iX^i\right)=a_0b_0+a_0b_1x^{2}+a_0b_2x^{3}+\dots + a_0b_nx^{n^2}+\dots+a_1b_nx^{(n+1)}+\dots+a_nb_0x^n+a_nb_1x^{n+1}+a_1b_2x^{3}+\dots + a_nb_1x^{n^2}+a_1b_nx^{(n+1)}+\dots+a_nb_nx^{n^2}\in R$.
(4) Additive inverse $\sum_{i=0}^m (-a_i)X^i$
(5) Multiplicative inverse Not sure of a closed form for it. I know it exists and can be found by means of the Euclidean algorithm.
To see that $I$ is ideal
(1) Because of the condition $a_0=0$ is $I\subset R$
(2) Because of the calculation above for (3), since the elements of $I$ have a similar form, both products $ir, ri$, where $i\in I, r\in R$, are in $R$.
Part (b) If $p_1, \dots, p_t\in I$ and are not zero, show that $p_i=\frac{m_i}{n_i}X^{k_i}+\text{ terms of higher degree}$. Also prove $\frac{1}{2n_1\dots n_t}X \in I$ but is not in $\sum_{i}^t Rp_i$
I'm lost on how to do (b).
 A: Clearly $\frac{1}{2n_1\dots n_t}X \in I$ suppose $\frac{1}{2n_1\dots n_t}X \in \sum_{i}^t Rp_i$ then you will get elements from $\{a_1,a_2,...,a_t\} \in R$ such that $\sum_{i}^t a_ip_i = \frac{1}{2n_1\dots n_t}X$. Now from here on I think we will make this into cases 
If all $k_1's$ are strictly bigger than one then there is no way we can obtain any multiple of $X$. similarly, if any $k_i$ is strictly bigger than 1 then $a_i$ must be zero.
If some of $k_i's$ are 1 ($k_i$ cant be zero), then we get $\sum_{i}^t a_ip_i = \frac{1}{2n_1\dots n_t}X$  and we can reduce it further to see that actually $a_i \in \mathbb{Z}$ for all i. If it happens then we can not get $\frac{1}{2n_1\dots n_t}$.
I think I made it right this time.
A: Things mostly look fine (or at least on the right track) so I will just comment in the places it is apparently needed...

(5) Multiplicative inverse Not sure of a closed form for it. I know it exists and can be found by means of the Euclidean algorithm.

Why are you looking for multiplicative inverses when confirming this is a ring? That isn't an axiom. In fact only $\pm 1$ will have inverses in this ring. If you were looking for the multiplicative identity, then $1$ is it.

I'm lost on how to do (b).

You should just observe that $R/I\cong \mathbb Z$, which is a domain. That means the kernel, $I$, is a prime ideal.
Well, $I$ is just $X\mathbb Q[X]$, so naturally if you collect the $p_i$, you can factor out their common $X$'s to the front, until each one is of the form $X^e q_i(X)$ where $q_i(X)$ is a polynomial with nonzero constant term. Gathering up all the $X$'s in front, and multiplying all the nonzero constant terms together, you have that the lowest degree term is the product of those $X$'s and that nonzero constant.
$\frac{X}{n}\in I$ for every $n\in\mathbb Z\setminus\{0\}$, so I'm not sure what the holdup is there.
Thinking about that last half of the last half...
