# Non-unital ring of rational numbers with even numerator and odd denominator has no maximal ideals

I've been told that the non-unital ring $$\{\frac{2n}{2m+1}: n, m \in \mathbb{Z}\} \subseteq \mathbb{Q}$$ has no maximal ideals. I've been trying to crack this one to no avail.

This example was presented (with no proof) to stress the importance of the requirement of having an identity element in order to ensure the existence of maximal ideals.

EDIT: This assertion seems to be false.

• Did it really say "has no maximal ideals" or did it say "has an ideal not contained in a maximal ideal," which would be another question entirely? – rschwieb Jan 27 at 16:27
• It said "has no maximal ideals". Maybe it was a mistake as it was just a passing remark. – user430825 Jan 27 at 17:22
• An example of a rng with no maximal ideals is $\mathbb{Q}$ with zero multiplication; the ideals correspond to subgroups, and $\mathbb{Q}$ has no maximal subgroups. – Arturo Magidin Jan 27 at 17:45
• @user430825 Maybe you're actually supposed to be using a more strict definition of "maximal ideal"? I know Jacobson additionally talked about using maximal modular ideals. The ideal in Eric's example would not be a modular ideal. – rschwieb Jan 27 at 19:35
• Are you then trying to construct an example as in the last paragraph of that paper? It says to start with a commutative ring $R$ with $1$ and divisible underlying abelian group, localize at a maximal ideal $M$, and then taking the ring $MR_M$. So it seems to me that you are trying to start with the local commutative ring obtained by localizing $\mathbb{Z}$ at $(2)$, and then taking $(2)\mathbb{Z}_{(2)}$. But the problem is that $\mathbb{Z}_{(2)}$ does not have divisible underlying abelian group ($2$ is not $4$ times anything), so the construction does not match that of that paragraph. – Arturo Magidin Jan 27 at 20:01

This is false. Let $$R=\{\frac{2n}{2m+1}:n,m\in\mathbb{Z}\}$$ and let $$I=2R=\{\frac{4n}{2m+1}:n,m\in\mathbb{Z}\}$$, which is a proper ideal. Note that $$rs\in I$$ for all $$r,s\in R$$, so the quotient $$R/I$$ is a rng in which all products are $$0$$. This means ideals in $$R/I$$ are the same as additive subgroups. As an abelian group, $$R/I$$ is a vector space over $$\mathbb{Z}/(2)$$ since every element is annihilated by $$2$$. So, we can pick a codimension $$1$$ vector subspace $$J\subset R/I$$ and $$J$$ will be a maximal proper subgroup and hence a maximal ideal. The inverse image of $$J$$ in $$R$$ is then a maximal ideal in $$R$$.
One can create more rings satisfying the hypothesis of the corollary by starting with any commutative ring with identify and divisible additive group, taking its localization $$S_M$$ at a maximal ideal $$M$$, and letting $$R=MS_M$$.
The corollary in question says, inter alia, that if $$S$$ is a commutative ring with identity and a unique maximal ideal $$R$$, and its additive group is divisible, then $$R$$ has no maximal ideals. Alternatively, that if $$R^2+pR=R$$ for every integer prime $$p$$, then $$R$$ has no maximal ideals.
It looks to me like perhaps this is an attempt at doing that construction by taking the ring of rationals with odd denominator; that is, the localization of $$\mathbb{Z}$$ at $$(2)$$, as the ring, and then taking $$(2)\mathbb{Z}_{(2)}$$. The problem is that this ring does not satisfy the hypotheses of the corollary: the underlying additive group of $$\mathbb{Z}_{(2)}$$ is not divisible, and $$(2)^2+2(2)\neq(2)$$ in that ring. So this is not an instance of the given construction. You can also not start with $$\mathbb{Q}$$ itself, because then the maximal ideal is $$(0)$$ and you just get the trivial ring.