3) A ring R such that is called a Boolean ring. Prove that every Boolean ring R is commutative and .
Proof- Let . Since R has an underlying additive group structure we know that . But also, . Thus we have . Then we have Thus so that R is commutative. Lastly, .
5) If A is the abelian group , then End A is a noncommutative ring.
Proof- Recall that End A is defined to be the set of all group homomorphisms from A to A with addition being pointwise and multiplication being composition. Let and . Then we have which shows that addition is well defined. Composition is similarly well defined. Addition is clearly associatve and commutative, with additive identity the 0 map and additive inverse of f being . Composition is also clearly associative and distributive over addition so that End A is a ring. Now we show that End A is noncommutative. Consider the function defined by and the function defined by It is easily checked that they are both homomorphisms. We have that while so that End A is not commutative.
6) A finite ring with more than one element and no zero divisors is a division ring.
Proof – Let be nonzero. Since we have a finite ring, there exists natural numbers such that . First we show that there is a multiplicative identity element. Since we are in a ring we have additive inverses so we can say . Then we can rewrite Lastly, because we have no zero divisors one of the two factors must be 0, but a was assumed to be nonzero so we have which gives us the right identity. We can similarly get the left identity so that R has a multiplicative identity, call it 1. Now we must show that is a a unit, i.e. that it has an inverse. Again we can write . We can write this as so that the right inverse of a is . Similarly it is the left inverse of a so that we have a is invertible and thus R is a division ring.
7) Let R be a ring with more than one element such that for each nonzero there is a unique such that . Prove:
a)R has no zero divisors
c) R has an identity
d) R is a division ring
Proof- a) Assume that is nonzero and that and assume . Then we can say . Let such that . Then we have . However we know b is the unique element with this property so that . Thus R has no zero divisors.
b). This gives since a was nonzero.
c) . This gives the element ba as the left identity. Then . This gives that the element ba is the right identity, and thus is the identity, call it 1.
d) gives that b is a left inverse of a and gives that b is also a right inverse. Thus a is a unit.
11) Let R be a commutative ring with identity of prime characteristic p. If then for all integers .
Proof- Since R is commutative with identity we have that . Then since for all and R has characteristic p, we know that . Therefore . Thus . Also . When p = 2 we have that , otherwise . This completes the proof.
12) An element of a ring is nilpotent if for some n. Prove that in a commutative ring a+b is nilpotent if a and b are.
Proof- Let a and b be two nilpotent elements such that for some r,s. Take and consider . We have two cases. If then . If then . Thus so that a + b is nilpotent.
13) In a ring R, the following are equivalent.
(a) R has no nonzero nilpotent elements
(b) If and then .
Proof- Assume that R has no nonzero nilpotent elements. If and then a is nilpotent, but by assumption there are no nonzero nilpotent elements so that a must be 0.
Assume that if and then . Want to show that there are no nonzero nilpotent elements. We do this by showing that using induction. If n is one then it is trivially true. Assume that , for some . Then if by assumption and by inductive hypothesis. Thus any element which is nilpotent is zero, so the claim is true.
14) Let R be a commutative ring with identity and prime characteristic p. The map given by is a homomorphism of rings called the Frobenius homomorphism.
Proof- Call the map f. Let . Then to check that f is a ring homomorphism we note that by exercise 11. Also, .
15)(a) Give an example of a nonzero homomorphism of rings with identity such that .
(b) If is an epimorphism then .
(c) If is a homomorphism of rings with identity and u is a unit in R such that is a unit in S, then and . Give an example to show that need not be a unit in S even though u is a unit in R.
Proof- a) Consider by . Then for we have and but note that in .
b) Let . Since f is surjective we have that for some . Then and similarly . Since the identity element is unique and is acting as the identity we must have that .
c) First note that . Then since is given to be a unit we can multiply both sides by the inverse so that . Then using this fact we have and similarly . Then since inverses are unique and is acting as the inverse of we must have that . Lastly, use the same map as in part a) for the counterexample and note that 2 is a unit , but is not a unite in .
16) Let be a homomorphism of rings such that for some nonzero . If R has identity and S has no zero divisors, then S is a ring with identity .
Proof- Let be nonzero such that . Then we have . Then since S has no zero divisors and it was assumed that we must have that . Similarly we can get that which means that must be the multiplicative identity element in S, .
18) Let be the field of rational numbers and R any ring. If are ring homomorphisms such that , then f = g.
Proof – Let x be an arbitrary rational number so that we can write where m and n are integers and n is nonzero. Then note that . Then also Using this finally we have that . Thus f and g agree on the rationals.