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

Question

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.

Thanks in advance.

EDIT: This assertion seems to be false.

Answer 1

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 $R$.

Answer 2

This looks like an attempt to use the description found in the final paragraph of this paper linked to in the comments. It reads:

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.