Spectrum of Ring is Nonempty

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

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

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

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

Let $I, J, K \in \Sigma$.

Then it is trivial from the definitions that:
 * $I \subseteq I$
 * If $I \subseteq J$ then $J \subseteq I$
 * $I \subseteq J$ and $J \subseteq K$ then $I \subseteq K$.

Therefore $\subseteq$ is an order on $\Sigma$.

Let $\left\{ {I_\alpha}\right\}$ be a chain in $\Sigma$.

It is to be demonstrated that $J = \bigcup_\alpha I_\alpha$ is a proper ideal of $A$.

Aiming for a contradiction, suppose $J = A$.

Then $1 \in J$.

So for some $\alpha$, $1 \in I_\alpha$.

This implies that $I_\alpha = A$.

This contradicts the definition of $\Sigma$.

Therefore $J \ne A$.

Now let $a \in A$.

Let $x, y \in J$ be arbitrary.

Then there exist $\alpha$, $\beta$ such that $x \in I_\alpha$, $y \in I_\beta$.

By the definition of a chain, either $I_\alpha \subseteq I_\beta$, or $I_\beta \subseteq I_\alpha$.

Let $K$ be the larger of these two sets.

Thus $x, y \in K$.

Then $K$ is an ideal.

So:
 * $x y, x \pm y \in K$

and:
 * $a x \in K$

Since $K \subseteq J$ it follows that $J$ is an ideal of $A$.

So 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, $\operatorname{Max}\:\operatorname{Spec}(A) \ne \varnothing$.

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

Also see

 * Maximal Spectrum of Ring is Nonempty