Summary: Rings and Algebras Problem set #5: Solutions Oct. 13, 2011.
1. Show the following statements:
a) A ring R is semiprimitive if and only if R is a subdirect product of primitive rings.
b) A ring R is subdirectly irreducible if and only if the intersection of all the non-zero ideals of R is
non-zero, i. e. R has a unique minimal (non-zero) ideal which is contained in each non-zero ideal.
c) (Birkhoff) Every ring is a subdirect product of subdirectly irreducible rings.
d) Zn is subdirectly irreducible if and only if n = p
for some prime p.
Solution. a) A ring R is semiprimitive if and only if J(R) = 0. Since the Jacobson radical is the intersection of the
primitive ideals in R, this means that there is an embedding of R into R/P with P running through the set of primitive
ideals. Clearly, R projects surjectively on each of the primitive rings R/P, thus we have a subdirect decomposition of R.
b) If the intersection of non-zero ideals is 0, then there is subdirect embedding of R into R/I with I running through
the non-zero ideals of R, and none of the projections is an isomorphism. Thus R is not subdirectly irreducible. On the
other hand, if there is an ideal I which is contained in each non-zero ideal of R then in any subdirect decomposition of R
at least one of the projections must be an isomorphism, hence R is subdirectly irreducible. c) For any 0 = r R let us
take an ideal Ir maximal with respect to the property that r Ir. Such a maximal ideal exists by Zorn's lemma. Clearly
0=rRIr = 0 so R is the subdirect product of the rings R/Ir. Now we just have to observe that R/Ir is subdirectly
irreducible since any ideal I properly containing Ir must contain the ideal generated by r, hence there is a unique minimal
ideal in R/Ir. d) The socle of Zp is simple, hence it is subdirectly irreducible. On the other hand if n is not a prime
power, then the socle contains disjoint subgroups hence it is not subdirectly irreducible.