Does maximal ideal exist in a commutative ring without identity?

Question

I was reading this proof https://equatorialmaths.wordpress.com/2009/03/11/existence-of-maximal-ideals/ where they use Zorn's lemma on the partially ordered set of proper ideals, and here we need identity to prove that the union of a chain is a proper ideal. What will happen otherwise? I didn't get a similar question in stack if it is a duplicate one let me know. Thanks

As the duplicate says, "no not in general", and of course that doesn't mean a nonunital ring can't have maximal ideals. Another thing is that the ordinary definition for maximal ideals in rings with identity may not always be useful for rings without identity. — rschwieb, Jun 24 '19 at 15:36

score 4 · Accepted Answer · answered Jun 24 '19 at 13:13

4

Any abelian group $G$ can be made into a trivial commutative ring without identity by taking $x \times y = 0$ for all $x,y$, and then ideals are exactly subgroups of $G$. This then comes down to the fact that some abelian groups do not have maximal subgroups, for example ($\mathbb{Q},+,0)$ (see $(\mathbb{Q},+)$ has no maximal subgroups)

answered Jun 24 '19 at 13:13

Junkyards

1,132

So you want to say maximal ideals don't exist for such a ring where the group has no maximal subgroup? – Rick Jun 24 '19 at 15:19

Does maximal ideal exist in a commutative ring without identity?

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