I’ve shown the case for $\mathbb F_3[x]$ but this is a lot simpler since this a field and hence a UFD and follows the same proof for prime numbers. I’ve tried using that maximal ideals are prime and that an ideal is maximal if and only if $R/I$ is a field but struggling to construct any maximal ideals from this
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Hint: combine that with: $,(3,f),$ is prime in $,R \iff f,$ is prime in $\Bbb F_3[x],,$ since $,R/(3,f) \cong ,\Bbb F_3[x]/(f),$ by quotient reciprocity. Or, you can do the same thing $\bmod 2.\ \ $ – Bill Dubuque Nov 04 '20 at 22:13
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Well you are in luck because $(\mathbb Z/6\mathbb Z)[x]\cong F_2[x]\times F_3[x]$, and using what you just said about $F_3[x]$ having infinitely maximal ideals, you should see why this has infinitely many maximal ideals (hence infinitely many prime ideals.)
rschwieb
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Which - of course - is essentially the same as my prior hint (which I stated that simpler way in case the OP doesn't know enough yet to prove the above isomorphism). – Bill Dubuque Nov 04 '20 at 22:29
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2@BillDubuque Am I supposed to give my evaluation too? IMO the isomorphism hint is far simpler than your version. I sort of tuned out at “quotient reciprocity” since I had never heard of it. By that is neither here nor there, people will like one version or the other. Backing up a moment, would you clarify what the purpose of your comment comparing the two is? – rschwieb Nov 05 '20 at 00:56
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"tuning out" when one encounters terms one never heard of is not the best way to learn math. In any case, surely you know this is a dupe, so why answer it? – Bill Dubuque Nov 05 '20 at 01:39