# Maximal ≠ Maximum!

/Suffixes are important!

Did you know that the words

*" maximal" and "maximum" generally do NOT mean the same thing*

in mathematics? It wasn't until I had to think about Zorn's Lemma in the context of maximal ideals that I actually thought about this, but more on that in a moment. Let's start by comparing the definitions:

Do you see the difference? An element is a maximum if it is larger than *every single* element in the set, whereas an element is *maximal* if it is not smaller than any other element in the set (where "smaller" is determined by the partial order $\leq$). Yes, it's true that the* maximum also satisfies this property, i.e. every maxim*um* element is also maxim*al*. But the converse is not true: if an element is maximal, it may not be the maximum! *Why?* The key is that these definitions are made on a *partially* ordered set. Basically, partially ordered just means it makes sense to use the words "bigger" or "smaller" - we have a way to compare elements. In a totally ordered set ALL elements are comparable with each other. But in a partially ordered set SOME, but not necessarily all, elements can be compared. This means it's possible to have an element that is maximal yet fails to be the maximum because it cannot be compared with some elements. It's not too hard to see that when a set is totally ordered, "maximal = maximum."**

How about an example? Here's one I like from this scholarly site which also gives an example of a miminal/minimum element (whose definitions are dual to those above).

## Example

Consider the set

- $\{d,o\}$ is
*minimal*because $\{d,o\}\not\supseteq x$ for every $x\in X$.- i.e. there isn't a single element in $X$ that is "smaller" than $\{d,o\}$

- $\{g,o,a,d\}$ is
*maximal*because $\{g,o,a,d\}\not\subseteq x$ for every $x\in X$- i.e. there isn't a single element in $X$ that is "larger" than $\{g,o,a,d\}$

- $\{o,a,f\}$ is both minimal and maximal because
- $\{o,a,f\}\not\supseteq x$ for every $x\in X$
- $\{o,a,f\}\not\subseteq x$ for every $x\in X$

- $\{d,o,g\}$ is neither minimal nor maximal because
- there is an $x\in X$ such that $x\subseteq \{d,o,g\}$, namely $x=\{d,o\}$
- there is an $x\in X$ such that $\{d,o,g\}\subseteq x$, namely $x=\{g,o,a,d\}$

- $X$ has NEITHER a maximum or a minimum because
- there is no $M\in X$ such that $x\subseteq M$ for
*every*$x\in X$ - there is no $m\in X$ such that $m\subseteq x$ for
*every*$x\in X$

- there is no $M\in X$ such that $x\subseteq M$ for

Let's now relate our discussion above to ring theory. One defines an ideal $M$ in a ring $R$ to be a **maximal ideal** if $M\neq R$ and the only ideal that contains $M$ is either $M$ or $R$ itself, i.e. if $I\trianglelefteq R$ is an ideal such that $M\subseteq I \subseteq R$, then we must have either $I=M$ or $I=R$.

Not surprisingly, this coincides with the definition of maximality above. We simply let $X$ be the set of all proper ideals in the ring $R$ endowed with the partial order of inclusion $\subseteq$. The only difference is that in this context, because we're in a ring, we have the second option $I=R$.

I think a good way to see maximal ideals in action is in the proof of this result:

As a final remark, the notions of "a maximal element" and "an upper bound" come together in Zorn's Lemma which is needed to prove that every proper ideal in a ring is contained in a maximal ideal. I should mention that an **upper bound** $B$ on a partially ordered set (a.k.a. a "poset") has the *same* definition as the maximum EXCEPT that $B$ is not required to be *inside* the set. More precisely, we define an upper bound on a *subset* $Y$ of $X$ to be an element $B\in X$ such that $y\leq B$ for every $y\in Y$.

So here's the deal with Zorn's Lemma: It's not too hard to prove that every *finite* poset has a maximal element. But what if we don't know if the given poset is finite? Or what happens if it's infinite? How can we tell if it has a maximal element? Zorn's Lemma answers that question:

As I mentioned above, it's this result which is needed to prove that every proper ideal is contained in a maximal ideal***. If you'd like to see the proof, I've typed it up in a separate PDF here. It actually implies a weaker statement, called Krull's Theorem (1929), which says that every non-zero ring with unity contains a maximal ideal.

*Footnotes*

*One can easily show that if a set has a maximum it must be unique, hence THE maximum.

** Here's the proof: Let $(X,\leq)$ be a totally ordered set and let $m\in X$ be a maximal element. It suffices to show $m$ is the maximum. Since $X$ has a total order, either $m\leq x$ or $x\leq m$ for every $x\in X$. If the latter, then $m$ is the maximum. If the former, then $m=x$ by definition of maximal. In either case, we have $x\leq m$ for all $x\in X$. Hence $m$ is the maximum.

*** Note this is NOT the same as saying that every maximal ideal contains all the proper ideals in a ring! Remember, maximal $\neq$ maximum!!