Since (R,+) is an abelian group, (-1)a + 1a = ((-1) + 1)a=0a=0.

(-1)a + 1a -1a = 0 -1a = -a (Adding -1a to both sides and use the fact that 1 is a unity in R).

Thus, (-1)a = -a.

I leave it to you to check is a commutative ring with unity. Then, check belongs to R.2) Show that the set of all real numbers of the form a+bsqrt(2), where a,b in Q, forms a field under ordinary addition and multiplication.

In , no nilpotent element can be a unit (excluding a trivial ring {0}). All non-zero nilpotent elements are zero divisors.3)

a) Show that every element of Zn (integers mod n) is either a unit or a zero-divisor. (Use the fact that an element of this ring is a unit iff (a,n) = 1.)

b) Which elements of Zn are nilpotent?

Thus, if you remove unit elements (satisfying (a,n)=1) from , then the remaining elements are nilpotents.

Suppose both x and y are unity elements in the ring R. Then, x = xy = y. Thus, x = y.4) Show that if R is a ring with unity, then the multiplicative identity element in R is unique.

Since r is idempotent in R, ,5) Let R be a ring with unity. R is calledBooleanif every element of R is idempotent. Show that if R is Boolean then:

a) 2r = 0 for every r in R (in other words, r = -r) (Hint: Consider (r+r)^2)

b) R is commutative.

.

We have . Since r is idempotent, r + r = 0. Thus, 2r = 0.

Let x, y be elements in a boolean ring R.

Since R is a boolean ring, we have xy = -yx. Let x = y. Then x = -x.

That means every element in R is its own additive inverse.

Thus, xy = -yx = yx. We conclude that R is commutative.