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I come up with a problem that I cannot figure out. Let $B$ be a UFD and $A=B[y]$ the polynomial ring. Given $f\in A$ containing a term $by^i$ with $i>0$ and $b$ not divisible by some prime element $p\in B$. Prove that $(f)$ is not maximal.

I tried to construct a larger ideal containing $(f)$ by taking $(f,py)$ or $(f,p)$, but I cannot prove that $(f,py)\neq (1)$ or $(f,p)\neq 1$.

user26857
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LonelyQuantum
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    The question is poorly stated and consequently dismissed by a trivial counterexample. The real question is this: https://math.stackexchange.com/questions/1591558, and in a comment under my answer I pointed out how to prove it for UFDs having infinitely many non-associated primes. – user26857 Jun 01 '23 at 19:19

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It looks like the statement you are trying is not true. Here are my arguments.

Any field has just two ideals, the ones generated by 0 and 1 respectively; hence it is PID, and consequently a UFD.

Now take $B$ to be the field of real numbers, a UFD. Here the polynomial $f(y) = y^2 +1$ satisfies your hypothesis as it has the term $y^2$, and the coefficient being 1 is not divisible by any prime.

The ideal generated by $f$ is maximal as the quotient by that ideal is the FIELD of complex numbers.

Are we missing something? Perhaps you have misstated the hypothesis

P Vanchinathan
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  • I think the assumption is $\exists p\in B$ prime such that $p\nmid b$, which is different from $\forall p\in B$ prime, $p\nmid b$. – Acrobatic Jun 01 '23 at 05:32
  • @Acrobatic: yes your point is valid, and I had missed it. But still the "counter-example" I have given is valid even when a stronger hypothesis is assumed. – P Vanchinathan Jun 01 '23 at 05:37
  • Thank you. I think all maximal ideals $(x-\alpha)\in\mathbb C[x]$ also serve as counterexamples. The problem is wrong. – LonelyQuantum Jun 01 '23 at 06:17
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    A field does not contain any prime, even though it is a UFD, because prime elements are not $0$ or unit by definition. – Acrobatic Jun 01 '23 at 06:31