What exactly is a maximal ideal?

Question

I am confused about the definition of maximal ideal. Suppose that there is ring $R$. Now if we select the whole $R$ to be an ideal, then wouldn't this be maximal ideal? Or is the definition of maximal ideal stating that whole $R$ cannot be maximal ideal?

Secondly, can anyone show why in the ring of integers $p\mathbb{Z}$ is maximal ideal whenever $p$ is prime?

Answer 1

An ideal $I$ is called maximal if it is proper and there are no other ideals other than the ring itself that properly contain $I$. In other words, you cannot have another ideal $J$ such that $I \subsetneq J \subsetneq R$. Another way to say this is that if we have $I \subseteq J$, then $I = J$ or $J = R$. To emphasize: any maximal ideal $I$ must be properly contained in $R$.

Approaching directly from the definitions, recall that $\mathbb{Z}$ is a principal ideal domain; so if $p\mathbb{Z}$ is not maximal, then $p\mathbb{Z} \subset m\mathbb{Z}$ for some integer $m \neq \pm 1$. Is this possible?

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Alternatively, we can use this important theorem:

$I \subset R$ is a maximal ideal if and only if the quotient ring $R/I$ is a field.

Because $\mathbb{Z}/p\mathbb{Z}$ is a field, this allows us to conclude that $p\mathbb{Z}$ is maximal in $\mathbb{Z}$.

Answer 2

The definition requires $I$ to be a proper ideal. Why? Because if it didn't, "maximal ideal" would be a really boring concept.

Here are two ways to show $p\mathbb{Z}$ is maximal in $\mathbb{Z}$. First is direct. Say there exists something larger, $p\mathbb{Z} \subset I$. Pick $a \in I \setminus p\mathbb{Z}$. Because $p$ is prime, and $a$ is not a multiple of $p$, they are coprime. By Bezout's lemma, we have some $x, y \in \mathbb{Z}$ such that $ax + py = 1$; this implies $1 \in I$. That forces $I = \mathbb{Z}$.

The other requires a bit more machinery, but it gives a good reason for why we should care about maximal ideals. It turns out that $I$ is a maximal ideal in $R$ iff $R/I$ is a field. Since $\mathbb{Z}/p\mathbb{Z}$ is a field, $p\mathbb{Z}$ is maximal in $\mathbb{Z}$.

Answer 3

We call an ideal M of a ring R to be a maximal ideal, if we cannot squeeze any other ideal between M and R. Suppose if we could do so, then either that ideal becomes M or R. Mathematically, M is a maximal ideal of R if $M\subset K \subset R$, then either $M=K $ $or $ $K=R$. Next consider $\mathbb{Z}_{p}$, the ring of integers modulo p, where p is a prime. We have a result, which says that $M$ $is$ $a$ $maximal$ $ideal$ $of$ $R$ $\iff$ $R/M$ $is$ $a$ $field$. Since $\mathbb{Z}/\mathbb{Z}_{p}$ is afield, so $\mathbb{Z}_{p}$ is a maximal ideal in $\mathbb{Z}$ whenever p is prime.