Maximal Spectrum of Ring is Nonempty

Theorem
Let $A$ be a non-trivial, commutative ring with unity.

Let $\operatorname{Max}\:\operatorname{Spec} \left({A}\right)$ be the maximal spectrum of $A$.

Then:
 * $\operatorname{Max}\:\operatorname{Spec} \left({A}\right) \ne \varnothing$

Proof
Let $(\Sigma, \subseteq)$ be the set of all ideals $I \ne A$ under inclusion.

From Spectrum of Ring Nonempty, we have shown that every non-empty chain in $\Sigma$ has an upper bound in $\Sigma$.

Therefore by Zorn's Lemma, $\Sigma$ has a maximal element.

That is, an ideal $\mathfrak m \ne A$ such that if $I \in \Sigma$ with $\mathfrak m \subseteq I$, then $\mathfrak m = I$.

This is precisely the definition of a maximal ideal.

Therefore, $\operatorname{Max}\:\operatorname{Spec} \left({A}\right) \ne \varnothing$.

Also because a Maximal Ideal is Prime, $\operatorname{Spec} \left({A}\right) \ne \varnothing$.