Subdirect Products, Little Ideal
Little Ideal
A little ideal is a nonzero ideal that is contained in every nonzero ideal of R.
If R has a little ideal, the intersection of any set of nonzero ideals is nonempty,
and R cannot have a nontrivial subdirect product.
Conversely, if all nonzero ideals intersect in 0,
let each act as a kernel Ki,
and build the nontrivial subdirect product representation Si = R/Ki.
Thus R is subdirectly irreducible iff it has a little ideal.
If R is a division ring,
or even a simple ring,
R is subdirectly irreducible.
The direct product of two or more rings is never subdirectly irreducible.
Intersect the ideals R1*0 and 0*R2 and get 0.
Dedekind Domain
A dedekind domain does not have a little ideal
(the square of such an ideal is smaller),
hence it is not subdirectly irreducible.
For a nontrivial representation, map R onto R/Mj for M a maximal
ideal and arbitrarily large exponents j.
The intersection of this descending chain of kernels is 0,
so R embeds.
Alternatively, map R onto R/M for infinitely many maximal ideals M.
Nothing can be in all these ideals at once.
Mod out by Mj, and R/Mj becomes subdirectly irreducible.
The little ideal is the image of Mj-1.
Semisimple
If R is semisimple, it is subdirectly irreducible iff it is the
direct product of only one simple ring.
We already said direct products are out,
and a simple ring has its one and only ideal as a little ideal.
Group Ring
If R is not subdirectly irreducible then neither is the
group ring RG.
Let Ki be a collection of kernels of R that intersect in 0.
Let these act as coefficients on the elements of G,
whence KiG become ideals in RG that intersect in 0.
Prime Ring
Let R be prime with nonzero
socle H.
We will show H is a little ideal, whence R is subdirectly irreducible.
Let I be any ideal and let M be a minimal left ideal.
Since I*M is a left ideal inside M,
and since I*M is nonzero (R is prime),
then I*M = M.
Of course I*M lies in I since I is an ideal, hence I contains M.
This holds for every M, hence I contains H.
The converse does not hold.
Let R be simple, which
Implies
R is prime.
Assume R is not left artinian.
Being simple, R is subdirectly irreducible.
Its socle H is either 0 or R.
Suppose it is R.
Then a finite sum of elements, from a finite set of minimal left ideals,
equals 1.
These are simple left ideals, and simple modules are pairwise disjoint.
Thus R is the finite sum of simple left ideals,
R is a semisimple ring,
and R is left artinian.
This is a contradiction, hence H = 0.
A prime ring can have a 0 socle and still be subdirectly irreducible.
K[x]
If K is a division ring that is nonalgebraic over its center,
K[x] is left primitive.
Since it is a left pid, any ideal can be squared to get something smaller.
There is no little ideal,
and the ring is not subdirectly irreducible.
Reduced and Commutative
If R is reduced and commutative,
it is subdirectly irreducible iff it is a field.
Let x lie in the little ideal.
Since x2 is not zero, it generates the same little ideal, hence x2*y = x for some y.
Multiply through by y, and (xy)2 = xy.
Remember that x2y = x, and x is nonzero, hence xy is a nonzero idempotent.
There can be no nontrivial idempotents, else R is a direct product of rings, with no little ideal.
Thus xy = 1, x is a unit, and the little ideal is all of R.
There can be no smaller ideals, other than 0, hence R is a field.