Finding generators for an ideal of $\Bbb{Z}[x]$ We know that $\Bbb{Z}$ is Noetherian. Hence, we can conclude that $\Bbb{Z}[x]$ is Noetherian, too. Consider the ideal generated by $\langle 2x^2+2,3x^3+3,5x^5+5,…,px^p+p,…\rangle$ for all prime natural numbers $p$. How can I determine a finite number of elements which generate this ideal? Clearly the Euclidean algorithm is not valid in $\Bbb{Z}[x]$.
Thank you

Hint Use the Euclidian Algorithm in $\mathbb{Q}[X]$, then multiply whatever what you get by the common denominator. If needed.
Edit By the extended EA we have
$$3x^3+3=\frac{3}{2}x (2x^2+2)+3-3x$$
$$2x^2+2=-\frac{2}{3}(-3x+3)+4$$
Therefore 
$$4=2x^2+2+\frac{2}{3}(3x^3+3-\frac{3}{2}x (2x^2+2))\\
=(2x^2+2)(x-1)+\frac{2}{3}(3x^3+3)$$
This shows that $12 \in I$.
Now, for each $p \neq 2,3$ we can write $1=12m+pn$, and hence
$$x^p+1=12(x^p+1)m+n(px^p+1) \in I$$
Using again the Euclidian algorithm in $Q[x]$ we get
$$x^7+1=x^2(x^5+1)+(-x^2+1)$$
$$x^5+1=(-x^2+1)(-x^3-x)+x+1$$
$$-x^2+1=(x+1)(-x+1)$$
Thus, $x+1$ is a linear combination of $x^5+1$ and $x^7+1$ with all the coefficients in $Z[X]$. Moreover, as it divides $px^p+p$ for all $p \neq 2$, it is trivial to show
$$I=(2x^2+2, x+1) \,.$$
Now, use again the Euclidean algorithm,  for those two polynomials to get
$$4=2x^2+2-2(x-1)(x+1)$$
