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Suppose that $R$ is an infinite PID with finitely many units. Show that $R$ has infinitely many maximal ideals.

My thought is that $x$ is irreducible implies $(x)$ (here $(x)$ is the ideal generated by $x$) is maximal - is this right? If yes, then perhaps to try to make a Euclidean proof that there are infinitely many irreducible elements would suffice. However, I cannot properly use that $R$ is infinite.

Any help appreciated!

DesmondMiles
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  • True for any infinite ring with fewer units than elements - see this theorem. – Bill Dubuque Feb 18 '20 at 22:15
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    Let $m_1,\ldots,m_n\subset R$ be distinct maximal ideals. If the set $1+\prod m_i$ contains only units, then each of the $m_i$ is finite and hence there exists a non-unit in $R-\bigcup m_i$, which then generates a maximal ideal distinct from the $m_i$. If the set $1+\prod m_i$ contains a non-unit then this element is contained in a maximal ideal distinct from the $m_i$. – Servaes Feb 18 '20 at 22:27

1 Answers1

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Yes, mimic Euclid's proof in $\Bbb Z\!:\,$ if $\,p_i\,$ are primes then $\,1+p_1\cdots p_{n} R\,$ is infinite so it contains a nonzero nonunit, with prime factor $\,p\,$ being coprime (so comaximal) to all $\,p_i$.

Remark $ $ More generally Euclid's idea extends to rings with fewer units than elements.

Bill Dubuque
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