Irreducible Components of the Prime Spectrum of a Quotient Ring and Primary Decomposition
In order to find an ideal which doesn't have a primary decomposition, the following construction is useful. Let $R$ be a commutative ring and $M$ an $R$-module. On the set $A=R\times M$ one defines the following two algebraic operations:
$$(a,x)+(b,y)=(a+b,x+y)$$
$$(a,x)(b,y)=(ab,ay+bx).$$
With these two operations $A$ becomes a commutative ring with $(1,0)$ as unit element. ($A$ is called the idealization of the $R$-module $M$ or the trivial extension of $R$ by $M$). Let's list some important properties of this ring:
$\{0\}\times M$ is an ideal of $A$ isomorphic to $M$ (as $R$-modules) and there is a ono-to-one correspondence between the ideals of $R$ and the ideals of $A$ containing $\{0\}\times M$.
$A$ is a Noetherian ring if and only if $R$ is Noetherian and $M$ is finitely generated.
All prime (maximal) ideals of $A$ have the form $P\times M$, where $P$ is a prime (maximal) ideal of $R$.
If $R$ is an integral domain and $M$ is divisible, then all the ideals of $A$ have the form $I\times M$ with $I$ ideal of $R$, or $\{0\}\times N$ with $N$ submodule of $M$.
Now I suggest to consider $R=\mathbb{Z}_{(2)}$ (the localization of $\mathbb{Z}$ at the prime ideal $2\mathbb{Z}$), $M=\mathbb{Q}$, $A=R\times M$ (as before) and $\mathfrak{a}=\{0\}\times H$ with $H$ a proper $\mathbb{Z}_{(2)}$-submodule of $\mathbb{Q}$. There are only two prime ideals of $A$ containing $\mathfrak{a}$ (and one of them is minimal over $\mathfrak{a}$), and $\mathfrak{a}$ has no finite primary decomposition because the primary ideals of $A$ have the following form: $\{(0,0)\}$, $\{0\}\times\mathbb{Q}$ and $2^n\mathbb{Z}_{(2)}\times\mathbb{Q}$, $n\in\mathbb{N}^*$.