The characteristic of a ring is the smallest positive integer such that ; if no such exists, then the characteristic is zero. For instance, is a ring of characteristic for each positive integer and is a ring of characteristic 0.
 Prove that the map defined by , for , and for is a ring homomorphism whose kernel is , where is the characteristic of . (This explains the use of the terminology “characteristic zero” instead of the archaic phrase “characteristic infinity” if no finite sum of 1 is zero.)
 Determine the characteristics of , , and .
 Prove that if is a prime and is a commutative ring of characteristic then for all .
 We begin by showing that for nonnegative by induction. For the base case , we have . For the inductive step, suppose that for some , for all , . If , then . Thus the conclusion holds for all positive . Suppose now that . If , then . If , then , and we have . Thus for all integers and , .
To show that , we again proceed for nonnegative by induction. For the base case, note that . For the inductive step, suppose the result holds for all for some . Then . Thus by induction the result holds for all nonnegative . Now suppose . Then . Thus is a ring homomorphism.
Now we show that , where . Suppose ; then . Suppose now that does not divide . Then we have for some with by the division algorithm. Note moreover that by definition, and that is minimal with this property. Now , a contradiction. Thus divides , and . Suppose . Then for some . Note that by definition, so that . Thus .

 Consider . We claim that , the inclusion map. To see this, note that , that for , , and that for , .
Since is injective, . Thus .
 Consider . We claim that , the inclusion map. The proof of this proceeds exactly as in the previous example.
 Consider . Note that . Thus . Now let and suppose does not divide . By the division algorithm, we have where . Then , a contradiction since is minimal. Thus divides , and . Thus .
 Consider . We claim that , the inclusion map. To see this, note that , that for , , and that for , .
 Let be a commutative ring with characteristic , and let . Note that divides but not or when . Thus divides for . Using the Binomial Theorem, we have .