Rings and Algebras Problem set #1: Solutions Sept. 15, 2011. 1. Prove that the following are equivalent for a ring R Summary: Rings and Algebras Problem set #1: Solutions Sept. 15, 2011. 1. Prove that the following are equivalent for a ring R: a) R has an identity element; b) whenever R # S for a ring S (possibly without an identity), then R is a ring direct summand of S (i. e. there is an ideal T # S such that R # T = {0} and R + T = S). Solution. a) # b). Let e be the identity element of R, and consider the set T = {s - se | s # S}. A straightforward calculation shows that T is a left ideal in S. Moreover, if r # R # T , then r = s - se for some s # S, and then r = re = (s - se)e = se - se = 0. Hence R # T = {0}. Finally, we can observe that s = se + (s - se) for each s # S, and here se # R and s - se # T . This means that S is the direct sum as a left S­module of R and T , hence every element in S can be (uniquely) written in the form r + t with r # R and t # T . We will now show that T is also a right ideal in S. Let s # S be arbitrary and compute (s - se)(r + t) with r # R and t # T . We have: (s - se)(r + t) = (sr - ser) + (s - se)t. Since e is the identity element in R and r # R, we get that sr - ser = sr - sr = 0. On the other hand (s - se)t # T since T is a left ideal. Thus T is a two sided ideal in S, hence S = R # T , a ring direct sum. --- b) # a) We know that any ring R can be embedded as an ideal into a ring S with identity. If R is also a direct summand of S, then the projection of 1 S in R is going to be an identity element in R. 2. Give an example of a non­commutative ring R for which the multiplicative group of invertible elements is commutative. Solution. Take R = K#x, y#, the free algebra over a (commutative field) K. Here the invertible elements are only the non­zero elements of K and the multiplication in K is commutative. -- Second solution: Take R = T 2 (Z 2 ), the set of upper triangular matrices over Z 2 . The multiplication is non­comutative (for example Collections: Mathematics