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I am wondering if we can find an irreducible polynomial $g(x)$ in $\mathbb{Z}[x]$ such that

  • The constant term, $c(g)=\pm 1$ and the leading coefficient $\ell(g)=\pm 1$,
  • the ideal generated by $g(x)$ and $5x+7$ is the ring $\mathbb{Z}[x]$, that is, $(g(x),5x+7)=1$, in other words: $g(-7/5)=\pm 1$,
  • the ideal generated by $g(x)$ and $2x-3$ is the ring $\mathbb{Z}[x]$, that is, $(g(x), 2x-3)=1$, in other words: $g(3/2)=\pm 1$.

Thanks.

PS: You can change $5x+7$ and $2x-3$ with any polynomials such that their constant terms and the leading coefficients are not units in $\mathbb{Z}$.

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    If $g \in \mathbb{Z}[x]$ is irreducible, then $\gcd(f,g) = 1$ for all $f \in \mathbb{Z}[x] - \{g, 0\}$.2012-03-07
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    Seems trivial, but how about $x^2 + x + 1$?!2012-03-07
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    @J.D., is 1 in the ideal of ${\bf Z}[x]$ generated by $x^2+x+1$ and $2x-3$? We're not in ${\bf Q}[x]$.2012-03-07
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    @GerryMyerson, it certainly isn't. $\mathbb{Z}[x]/(x^2+x+1)$ is isomorphic to $\mathbb{Z}[\omega]$, the ring of integers of the 3rd cyclotomic field, and $2x-3$ corresponds to $2\omega-3$ under this isomorphism. The group of units in there is torsion, and $2\omega-3$ is not one of them, so the ideal generated by $(2\omega-3)$ cannot be all of $\mathbb{Z}[\omega]$. In fact, the norm of $(2\omega-3)$ down to $\mathbb{Q}$ is 16.2012-03-07
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    I would like to thank you for thinking about this question. First of all, I forgot to write that c(f) is the constant term and l(f) is the leading term in the first bullet and this is not a homework question. I have been trying to find these special irreducible polynomials to give an example in my research. I could find infinitely many irreducible polynomials over the integers if I change 5x+7 with x+7 and 2x-3 with x-3. But I don't know if we have such an irreducible polynomial in this case.2012-03-08
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    @GerryMyerson Thanks for your messages. This is not a homework problem. I don't believe that such a polynomial exists but I don't know how to convince myself. And if it exists it will satisfy a definition given for a two dimensional poset.2012-03-08
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    I'm pretty sure that $g$, if it exists, has degree divisible by 6, but I haven't been able to do better than that. :(2012-03-11
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    Make that divisible by 12.2012-03-11
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    eileendavid82, are you still here? A little bit of engagement with your question would be appreciated, especially given the amount of work Hurkyl put into it.2012-03-19

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