1
Math 649 Midfinal
Answer as many questions as you can! Make sure you state clearly any theorems from class that
you use.
Part I. Definitions.
1. Give two different characterizations of the Jacobson radical J(R) of a left Artinian ring R.
Solution.
(1) It is the smallest two-sided ideal Jof Rsuch that R/J is a semisimple left R-module.
(2) It is the largest nilpotent left ideal of R.
Part II. True or False. Justify your answers briefly.
1. For any commutative ring R, the R-algebras R[x]⊗RR[x] and R[x, y] are isomorphic.
True. An isomorphism maps x⊗1 to xand 1 ⊗xto y.
2. If Ris a ring having no non-trivial two-sided ideals, then Ris a division algebra.
False. For example M2(F) for Fa field. Its a simple ring (e.g. because it is Morita equivalent
to the field F), but there are non-zero 2 ×2 nilpotent matrices which are not invertible.
3. If Ris a principal ideal domain, then Ris Noetherian.
True. Take an ascending chain I1⊆I2⊆· · · of ideals in R. We need to show it stabilizes.
Let I=!n≥1In. It is also an ideal of R. Since Ris a PID, I= (a) for some a∈R. But
then a∈Infor some n, and hence In=Ialready. Hence In=In+1 =· · · and the chain has
stabilized.
4. If Ris a commutative Noetherian ring, then every R/J(R)-module is semisimple.
False. Take for instance R=Z. It is a PID so its Noetherian. Its Jacobson radical is the
intersection of all maximal ideals (p) for pprime. That is an ideal of the form (n) where n
is divisible by all primes. Hence it is (0). But not every Z-module is semisimple (e.g. Z/(4)
is not).
5. If Fis a field of characteristic p > 0 and Gis a finite abelian p-group, then there is only one
irreducible F G-module up to isomorphism.
True. Note Gis a product of cyclic groups of order pifor various i. So it suffices to prove
the result just for Gcyclic of order pi. But then F G ∼
=F[x]/(xpi−1) = F[x]/(x−1)pi.
This clearly has just one irreducible module by the structure theorem for modules over PIDs
(there is just one prime in the ring F[x]/(x−1)pi), namely, (x−1)).
Part III. Longer problems.
1. Let Gbe a finite group and Fbe an algebraically closed field of characteristic p≥0.
(i) Prove that up to isomorphism, there are only finitely many irreducible F G-modules
L1, . . . , Lr.
(ii) Let ni= dimFLi,i= 1, . . . , r. Prove that "r
i=1 n2
i≤|G|, with equality if and only if
p= 0 or p!|G|.
(iii) Is it true that the inequality "r
i=1 n2
i≤|G|holds even if Fis not algebraically closed?
