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I know a prime ideal is a maximal ideal because a maximal ideal occurs in an field and an integral domain is a field but the converse may not be true. I am struggling to connect the ideas for $Q[x]$ and this specific problem.

DMcMor
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wolf
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  • You should prove that prime ideal : $R/I$ is an integral domain, maximal ideal : $R/I$ is a field. – reuns Apr 15 '17 at 04:36

1 Answers1

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The statement in your question is incorrect: every maximal ideal is a prime ideal, but the converse is not necessarily true.

However, because $\mathbb{Q}[x]$ is a principal ideal domain, every non-zero prime ideal in $\mathbb{Q}[x]$ is maximal. In particular, $(x^2+1)$ is a maximal ideal because $x^2+1$ is an irreducible polynomial over $\mathbb{Q}$.

It might help to contrast this with the case of $\mathbb{Z}[x]$: the ideal $(x^2+1)$ is still prime in $\mathbb{Z}[x]$, but it is not maximal because it is contained in the proper ideal $(3,x^2+1)$.

carmichael561
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