Does Krull's theorem hold for left weakly reductive rings? Let $(R, +, \cdot)$ be a ring (non-unital, non-commutative). We call it left weakly reductive, or lwr, if the following property holds: $xa = 0$ for all $x$ implies $a = 0$. This is equivalent to $xa = xb$ for all $x$ implying $a = b$.
An lwr ideal in a lwr ring is an ideal $J$ such that $R\cdot A \subseteq J \implies A\subseteq J$.
Does the following generalization of Krull's theorem hold for lwr rings?
Conjecture. Suppose that $R$ is a non-trivial lwr ring. Then there is a maximal (proper) lwr ideal in $R$.
Note that if the conjecture holds, $I\subset R$ is a proper lwr ideal, then we can find maximal lwr ideal $J\subset R/I$. Then, letting $\pi:R\to R/I$ be the canonical map, $\pi^{-1}(J)$ will be a lwr ideal, for $R\cdot a \subseteq \pi^{-1}(J)$ would imply $(R/I)\cdot \pi(a)\subseteq J$, from which $\pi(a)\in J$, hence $a\in \pi^{-1}(J)$. Then, since $\pi$ induces a lattice isomorphism between ideals of $R$ containing $I$ and ideals of $R/I$, $\pi^{-1}(J)$ would necessarily be a maximal lwr ideal. So that any proper lwr ideal in $R$ is contained in some maximal lwr ideal.
Also note that lwr ideals are precisely ideals $I$ for which $R/I$ is an lwr ring.
 A: No. Let $R$ be the (nonunital) subring of the Levi-Civita field over some field $F$ consisting of elements of positive valuation i.e. we permit sums of the form $a=\sum_{n=0}^\infty a_nT^{e_n}$, where $e_n$ is an increasing sequence of positive rationals tending to infinity. The valuation of a nonzero element like this is defined to be $v(a)=e_n$ for the least $n$ such that $a_n\neq 0$. Using the fact Levi-Civita field is a field we easily verify the following claim: if $a,b\in R$ and $v(a)>v(b)$, then $a=bc$ for some $c\in R$.
$R$ is an integral domain, so is lwr. I claim $R$ has no maximal lwr ideal. Indeed, suppose $I$ is any maximal lwr ideal in $R$. Let $e$ be the infimum of valuations of elements of $I$. If $e=0$, then for any $a\in R$ there is $b\in I$ such that $v(a)>v(b)$, and then $b$ divides $a$ by the claim above and hence $a\in I$, contradicting the fact $I$ is proper. Hence $e>0$. Pick some rational $0<d<e$ and let $J$ be the ideal of $R$ consisting of elements of valuation at least $d$. We then check $J\supsetneq I$ and $J$ is an lwr ideal, contradicting maximality of $I$.
