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.