Exercise set 2

Throughout F denotes a field and A a k-algebra. A ring is said to be semisimple if it a semisimple as a (left) module over itself.

(1)

(difficulty level: 2) Show that any ring A with 1≠0 admits simple modules.

(2)

(difficulty level: 2) Show that any module over a semisimple ring is semisimple.

(3)

(Easy if you follow the hint) (Example of a non-semisimple finite dimensional complex algebra) Let B be the set of 2 × 2 upper triangular matrices with complex entries. With respect to standard matrix addition and multiplication, B is a ℂ-algebra. Let M be the set of 2 × 1 matrices with complex entries. Then with respect to multiplication from the left, M is a B module. Show that M is not semisimple. (Hint: Consider the submodule N consisting of 2 × 1 matrices with the (2,1) entry being 0. Show that N has no complement. In fact, show that N is the only non-trivial proper submodule of M.) Conclude that B is not semisimple.

(4)

(difficulty level: 3) True or false?: If N is a semisimple submodule of a module M such that M∕N is also semisimple, then M is semisimple.

(5)

(difficulty level 3 without hint) A finite dimensional algebra over a field admits only finitely many isomorphism classes of simple modules. Solution: For any ring A, any simple module is a quotient of AA: indeed, for any m≠0 in M, the homomorphism AA → M by aam is onto. Now let A be an algebra of finite dimension over a field F. Fix a sequence AA = N₀ ⊋ N₁ ⊋ … ⊋ N_k-1 ⊋ N_k = 0 of submodules of AA with all quotients N_j∕N_j+1 being simple. Such a sequence exists: indeed the length k of such a strictly decreasing sequence is bounded by dim _FA. Given a simple A-module M, let φ : A ↠ M be an A-module epimorphism, and let r be the least integer such that φ|_Nr = 0. Then r ≥ 1 and M ≃ N_r-1∕N_r. Thus any simple module is isomorphic to one of the quotients N_j∕N_j+1.

Additional problems (optional)

(1)

(Difficulty level 4 without hint) (Equivalence in general of the conditions in the definition of a semisimple module) Let M be a module such that any submodule of it admits a complement. Then M is a direct sum of (some of its) simple submodules. Solution: Fix a maximal collection ℭ of simple submodules of M whose sum is their direct sum: such a collection exists by Zorn. Let N be the sum of submodules in ℭ, and suppose that N ⊊ M. Choose y ∈ M \ N. Choose, by Zorn, a maximal proper submodule P containing N of N + Ay. Let S be a complement to P in N + Ay (it exists because the hypothesis on M passes to submodules). Being isomorphic to (N + Ay)∕P, it is simple. And its existence violates the maximality of ℭ.

(2)

(Difficulty level 3 without hint) Show that a ring that admits a finitely generated faithful1 semisimple module is itself semisimple. Find a ring that is not semisimple but admits a faithful (necessarily non-finitely generated) semisimple module. (HInt: A commutative ring admits a faithful semisimple module iff its Jacobson radical is trivial.)

(3)

(Difficulty level 3 without hint) (A version of Nakayama’s lemma) Let M be a finitely generated A-module. Given a submodule N, there exists a maximal (proper) submodule of M containing N. (Hint: Zorn.) If M is finitely generated and non-zero, then there exists a primitive ideal 𝔞 such that 𝔞M ⊊ M.2 (Hint: Let N be a maximal proper submodule of M and take 𝔞 to be the annihilator of M∕N.) Deduce the following: if A is a commutative local ring with maximal ideal 𝔪 and M a non-zero finitely generated A-module, then 𝔪M ⊊ M.