2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

$A$ 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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.