I've : $I=<p,x>$ is not a principal ideal in $Z[x]$ where p is prime. My question : Is $I=<p,x>$ a principal ideal in $Z[x]$ where p is not a prime? More particularly, is the ideal generated by ${4,x}$ a principal ideal in $Z[x]$ ?
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If you have already that $I=(p,x)$ is not maximal, then where did you use there that $p$ is prime? Can you replace $2$ by $4$ in the standard proof? – Dietrich Burde Jan 12 '19 at 16:37
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You may want to check this answer of mine. – Andreas Caranti Jan 12 '19 at 17:38
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If $\langle4,x\rangle$ was a principal ideal, then you would have $\langle4,x\rangle=\langle p(x)\rangle$, for some $p(x)\in\mathbb{Z}[x]$. Is this true? What can you tell about a polynomial $p(x)\in\mathbb{Z}[x]$ if you know that $p(x)\mid4$ and that $p(x)\mid x$?
José Carlos Santos
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