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If R and S are both K-algebras, then it follows that λm = (1R λ)m = m(λ1S ) = mλ for all λ ∈ K. 1 Prove that Z(K n×n ) = {λ1n | λ ∈ K}. 2 Show that a homomorphism of R-algebras is an isomorphism if and only if it is bijective. 3 Show that the ring L of lower triangular matrices in K n×n is a K-algebra; determine Z(L), the inclusion K → Z(L) and the dimension of L over K. 4 Show that, if A is an R-algebra, then the set {r1A | r ∈ R} is contained in Z(A). 5 Explain why complex conjugation C → C, a + bi → a − bi is an isomorphism of R-algebras, but not even a homomorphism of C-algebras.

The radical rad A of a finite-dimensional algebra A is a nilpotent ideal. Proof. Note that we have tow possible interpretation of rad A, namely the Jacobson radical and the radical of A as left A-module. However, it follows directly from the definition that they coincide. Let I = rad A. By definition we have I m = rad I m−1 , that is, I m equals the radical of I m−1 , viewed as left A-module. 20, I m is a proper submodule of I m−1 and therefore a proper subspace. Since A is finite-dimensional, we must have I m = 0 for m ≥ dimK A.

Since EndA (M ) is a vector space of finite-dimension, the latter can only happen if ϕ itself is an isomorphism, hence invertible. This shows that EndA (M ) is local. Assume now that M is not indecomposable. Then there exist M1 , M2 = 0 such that M = M1 ⊕ M2 . Denote by π1 : M → M1 the canonical projection and by ι1 : M1 → M the canonical inclusion. Then ϕ = ι1 π1 ∈ EndA (M ) is idempotent, neither invertible nor nilpotent. Hence EndA (M ) is not local. The proof for right A-modules is completely similar.