It is easy to verify that a ring has no zero divisors iff the left and right cancellation laws hold in .
The set of units forms a group under multiplication.
Proof. Use that for . Use induction. ◻
Proof. (ii). If
is the least positive integer such that
,
.
(iii). If
,
, then
implies that
or
, a contradiction. ◻
Proof. Let and define multiplication in by is a ring with identity and characteristic zero and the map given by is a ring monomorphism. If , use a similar proof with and multiplication defined by Then . ◻
If is any ring, the center of is the set . is easily a subring of but may not be an ideal. A left ideal of that is not or is called a proper left ideal. Observe that if has an identity and is an ideal of , iff . A nonzero ideal of is proper iff contains no units of . A division ring has no proper left or right ideals since every nonzero element of is a unit. The ring of matrices over a division ring has proper left and right ideals, but no proper ideals.
The elements of are called generators of the ideal . If then the ideal is denoted and said to be finitely generated. An ideal generated by a single element is called a principal ideal. A principal ideal ring is a ring in which every ideal is principal. A principal ideal domain is an integral domain and a principal ideal ring.
Let be nonempty subsets of a ring . Denote by . If and are nonempty subsets of let denote . More generally, let denote the set of all finite sums of the form . In the special case when all are the same set , denote it by .
Let be a ring and an ideal of . Since the additive group of is abelian, is a normal subgroup. is a well-defined quotient group.
Isomorphism theorems also exist for rings.
Proof. is an additive subgroup of . If whence . Thus is an ideal. is an epimorphism of groups with kernel . . is also an epimorphism of rings. ◻
Proof. Let . Then and . Thus has the same effect on every element of . The map defined by is well-defined. Since is a homomorphism, is easily shown to be a homomorphism of rings. is unique since it is completely determined by . Clearly and iff . . is an epimorphism iff is an epimorphism. is a monomorphism iff . ◻
Proof. is a homomorphism of rings and . There is a unique homomorphism of rings such that . , . iff . iff iff . Note that and implies . ◻
(i) is the second isomorphism theorem and (ii) is the third isomorphism theorem.
Proof. Suppose and are ideals such that and . whence or . Thus so and is prime. Conversely, if is a prime ideal, is commutative, and , then . Note that whence . Either or , whence or . ◻
Proof. If is prime, since , . has no zero divisors since implies implies or implies or . Therefore is an integral domain. If is an integral domain, then whence . Thus . Also, implies implies or . ◻
Proof. Suppose but . and properly contains . By maximality, . Since is commutative and , . This contradicts that . Thus or , whence is prime. ◻
In particular, whenever has an identity.
Proof. (i). If
is maximal, then
is prime. Whence
is an integral domain. We must show if
,
has a multiplicative inverse in
.
is properly contained in
.Since
is maximal,
.
for some
,
.
.
Thus
is a multiplicative inverse of
in
.
(ii). If
is a division ring, then
whence
and
. If
is an ideal such that
, let
.
has a multiplicative inverse say
.
.
. But
and
implies that
. Thus
. Therefore
is maximal. ◻
Proof. is a field iff is maximal. is maximal iff has no proper ideals. ◻
Let be an ideal and . is said to be congruent to modulo denoted iff .
Proof. Since and , Since , . Thus . Assume that . Then and hence . Thus and the induction step is proved. . Similarly, . Thus such that . Note that and for . Let . Verify that . Finally, if is such that for each , then for each . Whence . ◻
Proof. Let be the canonical epimorphism. The induces a homomorphism with . . Thus induces a monomorphism . If the hypotheses of the Chinese remainder theorem are satisfied, for , there exists such that for all . Thus . Whence is an isomorphism. ◻
Proof. (i). If
is prime,
. If
is a nonzero prime ideal,
.
(ii). If
is irreducible, then
is a proper ideal of
. If
, then
. Since
is irreducible,
or
is a unit. Hence
is maximal. Conversely, if
is maximal in
, then
is a nonzero nonunit in
. If
, then
whence
or
. If
, then
is a unit. If
, then
hence
. Thus
is a unit. Therefore
is irreducible.
(iii). If
,
. Say
. Then
and
. Thus
is a unit.
(iv). If
is irreducible, then
is maximal, hence prime, thus
is prime.
(v). If
is irreducible,
is an associate of
,
where
is a unit. If
, then
whence
is a unit or
is a unit. If
is a unit, so is
hence
is irreducible.
(vi). If
is irreducible and
, then
whence
or
. Thus
is an associate of
or a unit. ◻
Proof. Let . is an ideal. Let . . Thus . ◻
Proof. Let be PID and be the set of all nonzero nonunit elements of which cannot be factored as a finite product of irreducible elements. Suppose is not empty and . Then is a proper ideal and is contained in a maximal ideal . is irreducible. . Therefore, it is possible to choose for each an irreducible divisor of . Since is an integral domain, uniquely determines a nonzero such that . We claim . If were a unit, would be irreducible hence is not a unit. If were not in , then has a factorization as a product of irreducibles, whence also does. Thus . We claim . Since implies for some whence . Contradicting that is irreducible and hence a nonunit. The function given by is well defined. By the recursion theorem, there is a function such that , . Denote . There is an ascending chain of ideals contradicting the previous lemma. Thus must be empty. Finally, if then divides some . Since is not a unit, is associate to . We can cancel and (with a factor of a unit), and proceed by induction to canceling the associates. If , this would imply that some of the or are units, a contradiction. ◻
Proof. If is a nonzero ideal in , choose such that is the least integer in the set . If , then with or and . so that , whence . . is a principal ideal ring. Since itself is an ideal, for some . . If . Thus . Whence is a multiplicative identity for . ◻
Greatest common divisors need not exist. When it exists, it may not be unique. However, two greatest common divisors are associates by (ii). Furthermore, any associate of a greatest common divisor is a greatest common divisor. If has an identity and have as a greatest common divisor, then are said to be relatively prime.
Proof. (i). Routinely follows. (ii) follows from (i). (iii). Each has a factorization with distinct irreducible elements and each . where is a greatest common divisor. ◻