 
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 nonzero ideals of R is
nonzero, i. e. R has a unique minimal (nonzero) ideal which is contained in each nonzero ideal.
c) (Birkho#) Every ring is a subdirect product of subdirectly irreducible rings.
d) Z n 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 nonzero ideals is 0, then there is subdirect embedding of R into
# R/I with I running through
the nonzero 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 nonzero 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#=r#R Ir = 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
