If spectrum of a commutative ring is empty

Question

I have found this proposition: "If the spectrum of a commutative ring A is empty then A is the zero ring". By absurdum, if A is not the zero ring, there exists in A an element $x\ne0$. By Zorn's lemma, there exists a maximal ideal M such that $x\notin M$. If A has not a unit, M is not necessarily a prime ideal... So I have no idea to complete the proof. May you help me with some hint? Thank You.

Answer 1

$A$ has a unit in this context. This is necessary both to claim the existence of a maximal ideal using Zorn's lemma and to claim that a maximal ideal is prime.

The example Wikipedia provides, is the ring whose underlying additive group is $\mathbf Q$ with the usual addition and whose multiplication is $a \cdot b = 0$ for all $a, b$. A subset $A \subseteq \mathbf Q$ is an ideal if and only if it is an additive subgroup of $\mathbf{Q}$.

It is clear that no proper ideal can be prime since if $x \notin A$ then $x^2 = 0 \in A$.

Also, if $A$ is proper then $(\mathbf{Q}/A,+)$ is a non-zero divisible group, which means it has a non-zero proper subgroup, which by correspondence means there is a larger proper ideal than $A$. So no proper ideals of $\mathbf{Q}$ are maximal.

Answer 2

I very much doubt you are working with rings lacking identity.

Indeed, the statement is false for rings without identity: $2\mathbb Z/4\mathbb Z$ has no prime ideals.