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Let $S$ be a local integral domain and $S[X]$ be a polynomial ring.

Choose $f, g$ from $S[X]$ as follows:

$f:= X^n + c_{n-1}X^{n-1} + ... + c_1X + c_0$

$g:= a_mX^m + ... + a_0$,

where $a_0, a_1,...,a_m$ all lie in the unique maximal ideal $m_S$ of $S$.

If $g$ is irreducible, does the ideal $(f,g)$ contain a non-trivial element of $S$ other than zero?

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    $\begingroup$ Why not if, for example $\deg f=0$? $\endgroup$
    – SashaP
    Commented Sep 21, 2016 at 7:52
  • $\begingroup$ If $g$ is prime, then the answer is yes. One shows that multiplication by $g$ on $S[X]/(f)$ is injective and thus has a nontrivial determinant. And a matrix times its adjoint is the determinant times the identity matrix. $\endgroup$ Commented Sep 21, 2016 at 9:38
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    $\begingroup$ @PierreMATSUMI It's really sad that after more than 3 years you did not accept any answer to your questions. $\endgroup$
    – user26857
    Commented Oct 11, 2016 at 7:24

1 Answer 1

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No. Take for $S$ the ring of real power series in $t$ for which the coefficients of $t$ and $t^3$ both vanish. Take $f:=X^2-t^6$, $g:=t^5+t^2X$. We must check that there is no $h$ so that $gh$ is a nonzero constant modulo $f$. This is straightforward.

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  • $\begingroup$ Because g is irreducible, this means (g) is prime. Then as you say, the answer is Yes, isn't it? What's the relation of your example of real formal power series ring? Pierre $\endgroup$ Commented Sep 21, 2016 at 10:11
  • $\begingroup$ No, that is false. In my answer $g$ is irreducible, but $(g)$ is not prime. Could you rephrase your question about `the relation' ? $\endgroup$ Commented Sep 21, 2016 at 13:18
  • $\begingroup$ Dear Wilberd van der Kallen, thanks a lot. Pierre Matsumi $\endgroup$ Commented Sep 25, 2016 at 12:17
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    $\begingroup$ @Pierre MATSUMI.This is not the way it is done. If you agree, you should accept the answer. $\endgroup$ Commented Sep 26, 2016 at 7:35

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