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We want to examine whether the ideal $I=(X^2+4,X)$ is a maximal ideal of $\Bbb Z[X]$, which as we know is not a PID.

This result tells us precisely which are the maximal ideals of $\Bbb Z[X]$. Thus it seems that this is not maximal, but I can't see why if so. It might be easy but I feel I stuck here.

Any help is appreciated, thanks.

user26857
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Chris
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    Consider: what could you add to the ideal without it becoming the unit ideal? For example, the link you mentioned references ideals of the form (p, f), where p is a prime number. So, consider the ideal generated by $I$ and $2$, and check if it's the unit ideal. – WanderingMathematician Jun 17 '21 at 16:49
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    Hint: $(X^2+4,X)=(4,X)$ – Pomponazzo Jun 17 '21 at 16:51
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    Use the third isomorphism theorem. – Bernard Jun 17 '21 at 16:53
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    $I$ is a maximal ideal of $A$ if and only if the quotient $A/I$ is a field. Can you compute the quotient ring? – WhatsUp Jun 17 '21 at 16:56
  • This was what I couldn't see in fact: $(X^2+4,X)=(4,X)$. Then, $$\Bbb Z[X]/(X^2+4,X) = \Bbb Z[X]/(4,X) \cong \Bbb Z_4[X]/(X) \cong \Bbb Z_4$$ and the since the latter is not a field, the first ring is not a field and thus our ideal is not maximal. Thank you all. – Chris Jun 17 '21 at 17:07
  • @Desperado Could I ask in general how can we "smell" equalities like this? – Chris Jun 17 '21 at 17:15
  • @Chris the idea is to play around with the generators, and ues the fact that any multiple of a generator can be removed by the other generators. – Pomponazzo Jun 17 '21 at 17:22

1 Answers1

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The first way is something more sophisticated. Recall that an ideal $I \subset R$ is a maximal ideal iff $R/I$ is a field. In this case, note that we have $$\frac{\Bbb Z[X]}{(X^2 + 4, X)} \cong \frac{\Bbb Z}{(4)}.$$ The latter is not a field and thus, $(X^2 + 4, X)$ is not maximal.

The other way is to explicitly produce a larger proper ideal. To see this, one can note that $$(X^2 + 4, X) \subsetneq (X^2 + 4, X, 2) \subsetneq \Bbb Z[X].$$ (Why are the inclusions proper?)

Aryaman Maithani
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  • Thank you for your answer. The first way is a bit nicer indeed and it's in fact a short version of what I wrote in the comments above. As for the second, this was precisely the point I stuck constructing an increasing chain $(X^2 + 4, X) \subsetneq (X^2 + 4, X, 2) \subsetneq (1)$! – Chris Jun 17 '21 at 17:12
  • @Chris: Glad to have helped. If you're comfortable with the first way, then that is good because that is, in a way, more "methodical". Indeed, the point is that you just have to compute a quotient and then check if it's a field. In the second way, it seems that one must "cook up" a bigger ideal from nowhere. However, if you wish to explicitly give an example, the first way does helps you with it. Indeed, if $R/I$ is not a field, then it has a non-zero proper ideal. By the ideal correspondence, it is of the form $J/I$ for some ideal $I \subsetneq J \subsetneq R$. This again shows $I$ isn't max. – Aryaman Maithani Jun 17 '21 at 17:17