Ideal In Polynomial intersection of Ideal and integers Let $I = AR + BR $ be an ideal  of $R= \mathbb{Z}[x]$. 
i) Can you show that $I \cap \mathbb{Z} = t\mathbb{Z} ; t\in \ \mathbb{Z}$ ?  
ii) Can you show that if $t\ge 1$, then it holds that $I=\tilde{A}R+\tilde{B}R$ where $\tilde{A}= a_{0}+a_{1}x+a_{2}x^{2}+\cdots$; $\tilde{B}=b_{1}x+b_{2}x^{2}+\cdots$ and $a_{0}\ge 1$ divides $t$ ? 

Reading Arturo Magidin's commentary: "Consider the collection of all positive integers that are constant terms of elements of $I$. This collection is not empty, since it contains $t$, and therefore contains a smallest element $a_0$. Let $\widehat{a}(x)$ be a polynomial in $I$ with constant term $a_0$. Then $a_0\leq t$, and since $t\in I$, by taking integral linear combinations of $\widehat{a}(x)$ and $t$ we can obtain a polynomial with constant term $\gcd(a_0,t)$. As this will be positive and less than or equal to $a_0$, it follows that it must equal $a_0$, hence $a_0|t$. Since it is positive by construction, $a_0\geq 1$."
What he/you has done is that he chose $\tilde{A}$ with the smallest possible positive constant $a_{0}$. Now because $AR+BR=AR+(B-tA)R$ it follows hat $I=\tilde{A}R+\tilde{B}R$.
Is this the way he/you meant it? Or does it follow directly? 
Thank you very much for your help. 
 A: Your argument for (i) does not make any sense to me; you have to show that there exists a $t$ with the desired property, but you don't know you have one yet. So you cannot prove that $I\cap\mathbb{Z}=t\mathbb{Z}$ by double inclusion, because you don't have two sets to compare; all you have is $I\cap\mathbb{Z}$.
Also, it is extremely sloppy to use $x$ as an arbitrary element of an ideal when you are working in a ring that already uses $x$ in a very specific capacity: you are working in the ring of polynomials in $x$! Don't use $x$ for anything except the indeterminate.
For (i): Let $I$ be an arbitrary ideal of $\mathbb{Z}[x]$. I claim that $I\cap\mathbb{Z}$ is an ideal of $\mathbb{Z}$.
Indeed, since $I$ and $\mathbb{Z}$ are both additive subgroups of $\mathbb{Z}[x]$, then $I\cap \mathbb{Z}[x]$ is an additive subgroup of $\mathbb{Z}[x]$ that is contained in $\mathbb{Z}$, hence an additive subgroup of $\mathbb{Z}$. And if $a\in I\cap \mathbb{Z}$ and $r\in \mathbb{Z}$, theN $ra\in I$ (because $I$ is an ideal of $\mathbb{Z}[x]$), and since $r,a\in\mathbb{Z}$ and $\mathbb{Z}$ is closed under products, $ra\in\mathbb{Z}$. Thus, $ra\in I\cap\mathbb{Z}$. Hence $I\cap\mathbb{Z}$ is an ideal of $\mathbb{Z}$. 
Since all ideals of $\mathbb{Z}$ are of the form $t\mathbb{Z}$ for some $t\geq 0$, (i) follows.
I'm assuming that in (ii), $d$ should be $t$, so you are considering the case in which your ideal intersects $\mathbb{Z}$ nontrivially. You are trying to show that you can also generate the ideal with polynomials of the given form (again, very bad form to introduce $A$ and $B$ and never say what they were).
Consider the collection of all positive integers that are constant terms of elements of $I$. This collection is not empty, since it contains $t$, and therefore contains a smallest element $a_0$. Let $\widehat{a}(x)$ be a polynomial in $I$ with constant term $a_0$. Then $a_0\leq t$, and since $t\in I$, by taking integral linear combinations of $\widehat{a}(x)$ and $t$ we can obtain a polynomial with constant term $\gcd(a_0,t)$. As this will be positive and less than or equal to $a_0$, it follows that it must equal $a_0$, hence $a_0|t$. Since it is positive by construction, $a_0\geq 1$.
That seems like a wise choice of $\widehat{a}(x)$ to me. Can you take it from here?
