A definition of a maximal ideal in a ring $R$ is as follows:

$I$ is a maximal ideal if for any ideal $J$ with $I \subseteq J$, either $J = I$ or $J = R$.

Let's say $I \subseteq J = R$, thus by this definition, $I$ is a maximal ideal. However $J$ is also an ideal, with size larger than that of $I$ (since $J=R$). Thus using this definition, both $J$ and $I$ are ideals, where $I$ is maximal, while being a subset of $J$. How can this be?

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    $\begingroup$ A maximal ideal has to be proper by definition. $\endgroup$ – Sha Vuklia Apr 7 '18 at 9:42
  • $\begingroup$ Ah ok, without that assumption, the maximal ideal of any ring R is R itself, right? $\endgroup$ – Dylan Zammit Apr 7 '18 at 9:44
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    $\begingroup$ Yes, and hence it's not a very interesting notion, while the notion of maximal proper ideal is very interesting $\endgroup$ – Maxime Ramzi Apr 7 '18 at 10:13

$I$ is a maximal ideal if it is a proper ideal (so with $I \neq R$) such that whenever $J$ is any ideal with $I \subseteq J$ we have $J = I$ or $J=R$.

As $R$ is always an ideal, without the properness condition, there could only be one, namely $R$, in any ring $R$. This trivialises the notion, the interesting case is when $I$ is proper and we cannot enlarge it and have a larger proper ideal. And this is what the above amended definition comes down to.


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