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I've been trying without success to figure out what are the rings $R$ such that whenever $M_n, n \in \omega$ is a countably infinite collection of pairwise distinct maximal ideals then $\bigcap_{n \in \omega}M_n=0$. If $R$ is a Dedekind domain then this obviously holds, and if $R$ has this property and has infinitely many maximal ideals then it has to have zero radical. Thanks for any input or hint.

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What if all ideals are equal ? –  Amr Dec 2 '12 at 11:11
    
Maximal ideals $M_n$ are supposed to be distinct, I should have probably written pairwise distinct. –  Fred.Fred Dec 2 '12 at 11:15
    
A related notion is a semiprimitive ring, which is a ring such that the Jacobson radical is zero. –  JSchlather Dec 2 '12 at 11:18
    
The set $\{0,1,2,\cdots\}$. –  Fred.Fred Dec 2 '12 at 12:05

1 Answer 1

The condition as you've written it basically just characterizes rings which admit a faithful semisimple module of "countable length," that is, it is of the form $\oplus_{i\in\omega}S_i$ with the $S_i$ all simple.

Since it has Jacobson radical zero, it is a subclass of the semiprimitive rings (they admit faithful semisimple modules, but depending on the ring they have to have more than $\omega$ summands.)

The best you can get structure-wise is that the ring is a subdirect product of primitive rings, which in turn are dense subrings of (potentially infinite dimensional) matrix rings over division rings. Perhaps you can go even further for your type of rings to say "a subdirect product of at most countably many primitive rings" and maybe something about the dimensions of the primitive rings being countable.

Here are some examples: On one hand, the ring could be very "wide". Take for example, $\prod_{i\in \kappa} \Bbb F$ where $\kappa$ is your favorite infinite cardinal. With $\kappa=\omega$, you have a ring with the condition you are interested in. With $\kappa>\omega$, you will not be able to find countably many maximal ideals which intersect to zero.

Another example is the endomorphism ring of $\oplus_{i\in \kappa}\Bbb F$ for some field $\Bbb F$. When $\kappa=\omega$ you get "sqaure" matrices with rows and columns indexed by $\omega$, each of whose columns have finitely many nonzero entries. You can check that if you form the collection $I_i$ of matrices whose $i$th row are all zero, $I_i$ is a maximal left ideal of the ring, and $\cap_{i\in \omega} I_i=\{0\}$. If $\kappa>\omega$, then it will be impossible for countably many maximal right ideals to intersect to zero (since then it would have to be isomorphic to the ring with $\kappa=\omega$, but that's impossible due to cardinality reasons.)

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