Number of Associated Prime Ideals vs. Number of Maximal Ideals This is a very naive (and presumably basic) question, but suppose you have an ideal $I \subset R$ in a commutative Noetherian ring with unity.  Does the number of maximal ideals containing $I$ have any relation to the number of associated primes of $I$?  What if $R$ is Dedekind, or Artinian?
Thanks for any elucidation!
 A: Typically, $I$ is contained in infinitely many maximal ideals. For example, the ideal $(x) \subseteq k[x,y]$ is contained in all $(x,y-\alpha)$ for $\alpha \in k$. This corresponds to the geometric fact that there are infinitely many points on the $y$-axis.
If you replace maximal by minimal containing $I$, then there is always the obvious inequality: every prime ideal that is minimal over $I$ is an associated prime of $I$, but the converse need not hold.
Maybe you already know this, but an example of an associated prime that is not minimal is given by the prime ideal $(x,y)$ in $R = k[x,y]/(x^2,xy)$: it is the annihilator of $x$, but the only minimal prime of $R$ is $(x)$. Thus $(x,y)$ is an embedded prime.
(The name embedded prime comes from the fact that $V(x,y) \subsetneq V(x)$, i.e. the origin is an embedded component contained in the unique irreducible component.)
However, if $R$ is Dedekind, then either $I = 0$ which has no associated primes except $0$ itself, or $I \neq 0$, in which case every prime ideal containing $I$ is minimal over $I$ (since $R/I$ is Artinian). This is more a coincidence than a general philosophy.
Similarly, in an Artinian ring, there are no embedded primes, since there are no inclusions between the prime ideals.
