 
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
