Infinite PID with finitely many units must have infinitely many maximal ideals

abstract-algebramaximal-and-prime-ideals

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!

Best Answer

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.

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