Let $R$ be a commutative ring without identity.
My question: is it true or false that every maximal ideal of $R$ is primary?
(An ideal I of R is said primary if is proper and
$\forall a,b\in R, ab\in I \Rightarrow a\in I \vee b\in\sqrt{I}$,
where $\sqrt{I} = \{r\in R| \exists n\in\mathbb{N}^+ : r^n \in I \} $. )
In a unitary commutative ring this is trivially true, because in such a ring is true:
$I$ maximal $\Rightarrow$ $I$ prime,
and in every ring holds:
$I$ prime $\Rightarrow$ $I$ primary.
I'm not sure about what happens in the non-unitary case. I've some results about the primary ideals, but mainly in unitary rings.
Definition: A simple ring (not necessarily commutative or with 1) is a non-zero ring $A$ that $A^2\neq0$ and it has no two-sided ideal besides the zero ideal and itself.
Suppose $R$ is a commutative ring and $I\subset R$ is a maximal ideal such that it is not primary. We want to get a contradiction. Then there are $a,b\in R$ such that $ab\in I$ but $a\not\in I$ and $\{b, b^2,\dots\}\cap I=\emptyset$ (obviusly, in this case, $a\neq0,b\neq0$ and $R^2\neq0$). Consider $R/I$; it is a simple (commutative) ring. It is not so difficult to prove that $R/I$ is a field (in fact, Theorem: Any commutative simple ring is a field). Since $a\not\in I$ and $b\not\in I$, we have $\overline{a}\neq0$ and $\overline{b}\neq0$ in $R/I$. Hence $0\neq\overline{a}\overline{b}=\overline{ab}$, i.e., $ab\not\in I$. This contradiction anwers your question.