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


The arguments of the following proof relies on computations within $\mathbb Z[x]$ without any reference to bigger rings. 

Let $I=\langle 2x^2+2,3x^3+3,5x^5+5,\dots,px^p+p,\dots\rangle$. Since $$px^p+p=p(x^p+1)=p(x+1)(x^{p-1}-\cdots+1)$$ for any prime $p\ge 3$, we have $$I\subseteq \langle 2x^2+2,x+1\rangle.$$ For $2x^2+2=2(x+1)(x-1)+4$ we get $\langle 2x^2+2,x+1\rangle=\langle 4,x+1\rangle$. Now let's prove that $4,x+1\in I$ and then conclude 

$$I=\langle 4,x+1\rangle.$$

Note that $$3(x^3+1)=3(x+1)(x^2-x+1)$$ $$5(x^5+1)=5(x+1)(x^4-x^3+x^2-x+1)=5(x+1)[x^2(x^2-x+1)-x+1],$$ and therefore $15x^2(x^3+1)-15(x^5+1)=15(x^2-1)$. Since $15\cdot2(x^2+1)-30(x^2-1)=60$ we get $60=4\cdot3\cdot 5\in I$. Repeating the same procedure with $11$ and $13$ instead of $3$ and $5$ we get $4\cdot11\cdot13\in I$, and since $(3\cdot5,11\cdot13)=1$ we find $4\in I$.
It remains to prove $x+1\in I$: since $2x^2+2\in I$ and $4\in I$ we get $2x^2-2\in I$ and summing up $4x^2\in I$. Since $3x^3+3\in I$ and $4x^3\in I$ we get (by subtraction) $x^3-3\in I$, so $x^3+1\in I$. From $5x^5+5=5x^2(x^3+1)-5x^2+5\in I$ we get $5x^2-1=(5x^2-5)+4\in I$. Now recall that $4x^2\in I$, so $x^2-1\in I$ and using $x^3+1\in I$ we get $x+1\in I$.
