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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? –  Gerry Myerson Nov 24 '11 at 1:55
    
In your attempt, what is $\tilde{B}$? –  Dylan Moreland Nov 24 '11 at 18:21
    
Hi Dylan Moreland, it is a polynomial without constant. –  Tashi Nov 24 '11 at 18:25
    
@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. –  Arturo Magidin Nov 24 '11 at 21:50
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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 "$n$." –  Arturo Magidin Nov 24 '11 at 22:42

1 Answer 1

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?

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Can you tell me if my proof is correct? I would be very glad about that. –  Tashi Nov 24 '11 at 17:03
    
Your argument makes as added makes no sense to me whatsoever. You never say what $\widehat{B}$ is, you never relate $\widehat{A}$ to my choice of $a$. You simply assert, with absolutely no justification, that the conclusion follows. So, no, your argument does not seem to be correct. –  Arturo Magidin Nov 24 '11 at 20:21
    
Can you tell me why you use $\widehat{A}$ instead of $\tilde{A}$ and what $\widehat{B}$ is in your case? I believe you choose $a_{0}$ so it is the smallest possible constant? Thank you. –  Tashi Nov 24 '11 at 22:28
    
I used hats because they are more noticeable than the tildes, which are hard to distinguish from overlines. It's simple typographical convenience. Yes, I picked $a_0$ to be the smallest possible positive integer constant in an element of $I$; but you never used this fact in your sentence that you provide as "proof". –  Arturo Magidin Nov 24 '11 at 22:40
    
Im not sure what you want for B or $\widehat{B}$. You want B or $\widehat{B}$ as a linear combination, so $B=(B-qA)$ or in the case $\widehat{B}=(\widehat{B}- t \widehat{A})$ and then you get $AR+BR= AR+(B-qA)R$? So it can be written as linear combination and this part of the theorem follows from it? –  Tashi Nov 24 '11 at 23:35

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