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my question: I want to prove that there are two elements in $R = \{f(x) \in \mathbb{Q}[x] | f(0) \in \mathbb{Z}\}$ that do not have a greatest common divisor.

my attempt: For $x \in R$, $\forall k \in \mathbb{Z}$, it seems that $\frac{x}{k} | x$, and $\frac{x}{k_{1}} | \frac{x}{k_{2}} \Leftrightarrow k_{2} | k_{1}$, so $\gcd(x, x)$ doesn't exist. Is this correct? Thank you very much for any prompt.

jhzg
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    There always is a gcd of $x$ and $x$, which is $x$. Note that $\mathrm{gcd}(x,x)$ is ambiguous as gcd isn't unique in a general context. – Maxence1402 May 23 '25 at 12:09
  • @Maxence1402 What does “gcd(x, x) is ambiguous” mean? Are there other examples that show there can be two elements without a gcd? – jhzg May 23 '25 at 12:15
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    here is another example. – lulu May 23 '25 at 12:33
  • @Lulu Sorry, Sorry, I meant in the ring $R$ in my question. – jhzg May 23 '25 at 13:27
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    $\gcd(\color{#0a0}{a,a}) \approx a,$ since $,d\mid \color{#0a0}{a,a}\iff d\mid a\ \ $ – Bill Dubuque May 23 '25 at 13:30
  • But $\frac{x}{2}, \frac{x}{3} | x$, and $\frac{x}{2} \nmid \frac{x}{3}$. – jhzg May 23 '25 at 13:46
  • That is not what you should be interested in. A gcd of $x$ and $x$ is an element such that every $d$ that divides both $x$ and $x$ also divides the gcd (and the other way round). – Tzimmo May 23 '25 at 13:50
  • @BillDubuque I understand what you mean now, and you're right. Then, are there concrete examples in this ring where the gcd does not exist? – jhzg May 23 '25 at 14:07

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I want to prove that there are two elements in $=\{()∈ℚ[]|(0)∈ℤ\}$ that do not have a greatest common divisor.

As per the wiki article the construction $\mathbb Z+x\mathbb Q[x]$ (which amounts to the ring you are talking about) yields a GCD domain that is not a UFD. So, any two elements are going to have a GCD.

(I believe the title question has already been dealt with in the comments.)

If you still need a ring with two elements not having a gcd, we already have solutions for that.

rschwieb
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