Example of a finite commutative ring $R$ whose maximal ideal $I$ is not prime?
Is it possible? I am not able to find such example.
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You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Dec 31 '19 at 14:31
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Right after I answered I noticed a very good duplicate in the items to the right. Among the solutions is the one I offered ($2\mathbb Z/4\mathbb Z$), plus more. – rschwieb Dec 31 '19 at 14:37
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@hardmath I came up with a compromise, considering the asker's difficulty seeing the connection. – rschwieb Dec 31 '19 at 15:45
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@Kangmin If you had read the solutions at the duplicates, you would have noticed that this solution already does what you want. So indeed, your question is already answered elsewhere. Therefore a duplicate link is useful. – rschwieb Dec 31 '19 at 15:47
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This is only possible if you ring lacks an identity.
Perhaps you didn't notice, but this is trivially derived from this solution, where $R=2\mathbb Z$ and $M=4\mathbb Z$ is given as an example, but obviously in the quotient $R/M$ has order $2$ and its zero ideal is maximal and nonprime.
Similarly this question can be adapted by saying $R=6\mathbb Z/12\mathbb Z$ has a maximal nonprime zero ideal (it is isomorphic to the ring above, as a matter of fact.)
And in fact this solution is exactly what you're asking for. So remember when you are reading duplicates, check their solutions too, because sometimes they do more than what the question asks of them.
rschwieb
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No, it's not possible. For any commutative ring with $1$, any maximal ideal is prime.
Hint. use characterization of prime/maximal ideals in terms of quotient rings
GreginGre
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