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Let $p(t),q(t)\in \mathbb{Z}[t]$ be non-constant polynomials that have no common roots in $\mathbb{C}$. Prove that any ideal $I\subseteq \mathbb{Z}[t]$ containing $p(t)$ and $q(t)$ must contain a nonzero integer.

A similar question was posted here for several varaible polynomials. But it uses some concepts from normal field extension. Is there any simpler way to solve it from the fact that the ideal generated by $p(t)$ and $q(t)$ in $\mathbb{C}[t]$ will contain a unit.

nkh99
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  • By the dupe $(p,q)=(1)$ in $\Bbb Q[x]$ so scaling its Bezout equation $ap + b q = 1$ by a common denominator yields the sought integer. Please delete this dupe question. – Bill Dubuque Apr 30 '24 at 15:38

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Since there is no common root, the polynomials are coprime over $\mathbb{C}$, and in particular over $\mathbb{Q}$. Hence there are some $u(x), v(x)\in\mathbb{Q}[x]$ such that $u(x)p(x)+v(x)q(x)=1$. Since the coefficients of $u,v$ are rational, there is some $n\in\mathbb{N}$ such that $nu(x), nv(x)\in\mathbb{Z}[x]$. Then:

$n=(nu(x))p(x)+(nv(x))q(x)$

belongs to any ideal of $\mathbb{Z}[x]$ that contains $p$ and $q$.

  • If $p(t),q(t)$ are not coprime over $\mathbb{Q}[x]$ then $I=(p(t))+(q(t))$ is a proper nonzero ideal of $\mathbb{Q}[x].$ Since $\mathbb{Q}$ is a field, $Q[x]$ is a PID. Therefore $I=(s(t))$ for some non-constant polynomial $s(t)$ over $\mathbb{Q}[x]$, which is nothing but the gcd of $p(t),q(t)$. This contradicts that $p(t),q(t)$ have no common zeroes.
nkh99
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Mark
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  • Can you please elaborate how being coprime over $\mathbb{C}$ implies coprime over $\mathbb{Q}$? – nkh99 Apr 30 '24 at 12:54
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    @nkh99 Any common divisor in $\mathbb{Q}[x]$ is in particular a common divisor in $\mathbb{C}[x]$, hence must be constant. – Mark Apr 30 '24 at 12:58
  • This is using the fact that $\mathbb Q[x]$ is a PID. – Kenta S Apr 30 '24 at 14:27
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Apr 30 '24 at 15:38
  • @BillDubuque I know I must be very stupid, but it's totally beyond me how that "duplicate" is the same question as what OP asked. One question is about gcd remaining the same in field extensions, the other one is about why ideal generated by two polynomials in $\mathbb{Z}[x]$ must contain some nonzero integer. Yeah, EXACTLY the same thing. – Mark Apr 30 '24 at 15:47
  • @mark See my closing comment on the question. Where did "EXACTLY" come from? See abstract duplicates. – Bill Dubuque Apr 30 '24 at 15:52