Are primary ideals always contained in unique maximal ideal? Just wondering, is this a standard fact? I notice a couple Banach algebra texts define primary ideals in this way. Another question: does this property, i.e. being contained in a unique maximal ideal, characterize proper primary ideals in a commutative ring with identity?
 A: Any prime ideal is primary. $(0)$ is prime in any integral domain. There are integral domains with more than one maximal ideal.
For a more interesting answer, consider $k[x,y]$ and the ideal $(x)$. As before, $(x)$ prime implies $(x)$ is primary. Now $(x)$ is contained in many maximal ideals, for example $(x, y-a)$ where $a \in k$.
A: On the geometric side of things, the points of an irreducible affine variety are in one-to-one correspondence with the maximal ideals containing the prime ideal defining the variety. Thus any irreducible affine variety with more than one point produces a counter-example.
A: In a commutative ring $A$ an ideal $\mathfrak q$ is defined to be primary if in the quotient ring $\,A/\mathfrak q$ zero-divisors are nilpotent.
If $A$ is noetherian, then the radical of $\mathfrak q$ is a prime ideal $\mathfrak p$, which consists of all zero-divisors modulo $\mathfrak q$.
Example: let $K$ be a field; in the ring $K[X,Y]$, the ideal $(X^3)$ is $(X)$-primary. It is contained in any maximal that contains $X$, e. g. $(X, Y)$, $(X, Y-1)$ and more generally $(X, p(Y))$, where $p(Y)$ is any irreducible polynomial in $K[Y]$.
Similarly, $(X^3,Y^2)$ is $(X,Y)$-primary.
