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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.

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    $d$ makes a brief, unannounced, appearance in ii) and then vanishes, never to be seen again. What was it there for? In your proof of i), you are being sloppy with the notation. Is $x$ an element, or is it a set of elements? what do you mean by "an ideal with constant $t{\bf Z}$? how do you "cut" a constant with itself? how does cutting a constant with itself give an ideal?2011-11-24
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    In your attempt, what is $\tilde{B}$?2011-11-24
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    Hi Dylan Moreland, it is a polynomial without constant.2011-11-24
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    @Tashi: That's not the question. The question is: *explicitly*, in terms of $\widehat{A}$, $t$, $A$, and $B$, *what is* $\tilde{B}$? You *want it* to be a polynomial without a constant. You haven't said *what* it is.2011-11-24
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    @Arturo Magidin , Sorry, but it is : $\widehat{B}= b_{1}x + b_{2}x^{2}....+b_{n}x^{n} $.2011-11-24
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    @Gerry Myerson Sorry, it is not cut it is called intersection.2011-11-24
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    @Tashi: That's an empty response. It doesn't have any meaning. You don't say what $b_1$, $b_2$, etc. are. Let me put it simply: if I were to give you an *explicit* $A$ and $B$, where in your argument do you tell me how to find $\tilde{B}$? Nowhere. Just because you write out "Ah, $\tilde(B)$ is $b_1x+\cdots+b_nx^n$", you aren't telling me anything other than the fact that you are going to call its coefficients $b_1$, $b_2,\ldots,b_n$. Your answer is as if I asked you "How many toes do you have in your right foot", and you answered "$n$."2011-11-24

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