Ideals
An ideal of is defined to be a additive subgroup of that is closed under multiplication with elements of ; that is, . The quotient ring inherits a uniquely defined multiplication from , making it into a subring, the quotient ring or the residueclass ring. is a ring homomorphism, and it maps every to the coset .
There is a onetoone orderpreserving correspondence between ideals of which contain , and ideals of , given by . Notice that the kernel of the homomorphism is , image is a subring , and induces the homomorphism .
A zerodivisor of an element is, if it exists, an element . A ring with no zerodivisors is called an integral domain. A unit of an element is an element ; then . The units in form a multiplicative abelian group, and a ring in which every element has an inverse is called a field; obviously every field is an integral domain. The multiples form the principal ideal, and is denoted by .
is a prime ideal of if and for . is a maximal ideal of if , and there exists no ideal strictly between and , that is no . In , prime ideals consist of , maximals, and irreducibles.
Three elementary results (proved in ringshw):

is maximal is a field.

is prime is an integral domain.

Every maximal ideal is a prime ideal.
If is a ring homomorphism, and is a prime ideal of , then is a prime ideal of , for is isomorphic to a subring of , and hence has no zerodivisors. But if is a maximal ideal in , need not be a maximal ideal in ; all we can say is that it is prime.
Every ring has a maximal ideal, and this can be proved using Zorn's lemma. If is an ideal of , then there is some maximal ideal containing . It follows that every nonunit of is contained in some maximal ideal. A ring with exactly one maximal ideal is called a local ring, and the field is called the residue field.
Let be a ring, then:

Let be an ideal such that every is a unit in . Then, is a local ring, and is a maximal ideal.

Let be a maximal ideal such that every is a unit in . Then, is a local ring.
To prove (i), notice that every ideal consists of nonunits and is hence contained within .
To prove (ii), let . Since is maximal, the ideal generated by and is , hence there exists such that or ; this is contained in and is hence unit.
What follows is a bunch of examples:

Let , the ring of polynomials. If is an irreducible polynomial, then, by unique factorization, is a prime ideal. Similarly, if is a ring of polynomials with constant term zero, then every ideal in is a maximal ideal; it is the kernel of the homomorphism , which maps every to .

Let with ideals of the form . Then is a prime ideal is prime number or . All prime ideals are maximal: is a field.

A principal ideal domain is an integral domain where all ideals are principal. In such a ring, every nonzero prime ideal is a maximal ideal.

Principal ideal domain unique factorization domain commutative rings.
Nilradicals
The set of all nilpotent elements of is an ideal, and does not have any nilpotent elements. To prove this, let be nilpotent elements of order . By the binomial theorem, has terms of the form ; but we cannot have both and , so each of these terms vanish, and . Let be a representation of . Then , and . Ideal is called the nilradical of .
The nilradical of is the intersection of all prime ideals of . To prove this, let be the intersection of prime ideals. If is nilpotent, and is a prime ideal, for some , and hence (because is prime) and . Conversely, suppose that is not nilpotent. Let be the set of ideals such that . is not zero, and by Zorn's lemma, there is some maximal element in . Let be this maximal element: it remains to be shown that is a prime ideal. Let , so contain , and therefore do not belong to . If:
then, , hence the ideal , and . Hence, we have a prime ideal , and .
The Jacobson radical of is defined to be the intersection of all maximal ideals of .
is a unit in for all . To prove this, let is first assume that is not a unit. But belongs to some maximal ideal ; and , so , and hence , which is absurd. To prove the converse, let , for some maximal ideal . Then, and generate , and , for some . Hence, is not unit.
Operations on ideals
The sum of two ideals , denoted , is the smallest set, , containing both ideals. The product, denoted is the ideal generated by . In , let ; then the sum is the l.c.m of , and is the h.c.f of . In this case, are coprime; more generally, .
In , we have the distributive law as in the case of . However, unlike the case,
Ideals are said to be coprime if .
Define the following homomorphism on direct products of ideals:
by the rule . Then:

If are coprime whenever , then .

is surjective are coprime whenever .

is injective .
In the ring :

Let be prime ideals, and any ideal such that . Then, for some .

Let be ideals, and a prime ideal containing . Then, . In particular, if , then for some .
To prove (i), use induction of the form:
Certainly, this is true for . It holds for , if it holds for . Let such that . If , we are done. Otherwise, for all . Consider an element such that . Hence, we have .
To prove (ii), let , and . Then, . But , since is prime. Hence, . Finally, if , then , and hence , for some .
If are ideals on ring , the ideal quotient is defined as:
which is an ideal in itself. In particular is called the annihilator of , also written as .
As an example, in , if , then , where , with denoting the h.c.f of .
What follows is a bunch of elementary results about ideal quotients:

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If is any ideal of , then the radical of is defined as:
If is a ring homomorphism, then , hence the radical of an ideal is an ideal.
What follows is a set of results about radicals:

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For prime ideal , .
The radical of ideal is the intersection of prime ideals containing . In general, we can define a radical over a set , but this is not an ideal in general, and .
Set of zerodivisors of is equal to , and hence, .
Let be ideals on ring , such that are coprime. Then, are coprime. This can be proved by observing .
Extensions and contractions
Let be a ring homomorphism. Then for ideal , is not necessarily an ideal in . We define extension to be the set , generated by . Alternatively, if is an ideal of , then is necessarily an ideal in , and this is termed contraction . We can hence factorize as follows:
Here, is surjective and is injective. For , the situation is very simple: there is a onetoone correspondence between ideals of and ideals of that contain , and prime ideals correspond to prime ideals. For , however, the situation is complicated.
As an example, consider . Primes in may not stay prime in . In fact, is a principal ideal domain, and the situation is as follows:

, the square of a prime ideal in .

If , then is the product of two distinct prime ideals. Example: .

If , then is prime in .
Let , and as before. Then:

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If is the set of contracted ideals in , and is the set of extended ideals in , then , and is a bijective map that maps onto with inverse .
Let us prove (iii), since (i) and (ii) are trivial. If , then . Conversely, if , then is the contraction of . A similar argument holds for .
Modules
Modules let us do linear algebra on general rings, even noncommutative ones. For instances of noncommutative rings, there's no need to look further than matrices over , the field of real numbers. Another instance of noncommutative algebra can be found in Quaternions, that have different properties when cycling clockwise and counterclockwise. The ideal and quotient ring are both examples of modules, and can hence, to a certain extent, be treated on an equal footing.
A module over ring is formally defined as the multiplication map satisfying the following axioms:




where and . What follows is a bunch of examples:

If is a field in , then module vector space.

If , then module abelian group.

If where is a field, then the module is a vector space with linear transformation.

If , where is a group, then module representation of .
Let be modules, and be an module homomorphism, or linear. Then,
for all . If is a field, then an module homomorphism is the same thing as linear transformation over vector spaces.
The composition of any two module homomorphisms is again an module homomorphism. The set of all module homomorphisms from to can be turned into an module with the axioms:
This module is denoted by . Homomorphisms and induce the homomorphisms:
where:
For any module , there is the natural isomorphism : any module homomorphism is determined uniquely determined by , which can be any element of .
Submodules
Submodule of is a subgroup of closed under multiplication by elements of . The abelian group inherits the module structure from , given by . The module is termed quotient of by . There is a natural homomorphism between and , and this is an module homomorphism. There is a onetoone orderpreserving bijection from submodules of containing and submodules of , as in the case of ideals.
If is an module homomorphism, then its kernel and image are also submodules defined by:
and
The cokernel of is a quotient module of , defined by:
If , then gives rise to the homomorphism , where . Taking , we then have the following isomorphism of modules:
Operations on submodules
Let be a module, and let be the family of submodules of . Then, the direct sum is defined as , the smallest submodule of containing all . is again a submodule.
What follows is a result on modules:

If are modules, then .

For submodules , .
To prove (i), define homomorphism by . Then, is a welldefined module homomorphism with kernel .
To prove (ii), consider the composition . It is surjective with kernel .
We cannot define a product of submodules. However, we can define the product , where is an ideal. If are two modules, we can define as the set of all such that ; it is an ideal. In particular is termed annihilator, and written as . An module is faithful if ; if , then the module is faithful as an module.
Two elementary results on annihilators:

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If , then the are termed the set of generators of the module . This means that every element of can be expressed as finite linear combinations (not necessarily unique) of s with coefficients in . is said to be finitely generated if this is the case.
Direct sum and product
If are modules, then their direct sum is defined as the pairs for . Sum and scalar multiplication are defined in the usual way:
More generally, we can define the direct sum as to have elements in the family . For a finite index set , direct sum and direct product, , are the same thing, but not otherwise.
Suppose ring is a direct product , then the set of elements of of the form where , form an ideal . This is not a subring in general, because it does not contain identity. By considering ring as a module, we have its decomposition into the direct sum of ideals. Conversely, given a ideal decomposition of a ring:
we have:
where . Each is a ring isomorphic to .
Finitely generated modules
A free module, , is one which is isomorphic to an module of the form , where each , as an module. A finitely generated module is therefore isomorphic , with summands, written as .
is a finitely generated module is isomorphic to the quotient of for some . To prove this, let generate , and be defined by . Then, is an module homomorphism onto and . Conversely, if , with being in the place, then s generate , and hence generate .
Let be a finitely generated module, be an ideal of , and be an module endomorphism such that . Then,
where . To prove this, let be the generators of . Then, each , so that . That is:
where is the Kronecker delta. Multiplying on the left by the adjoint of the matrix , we see that annihilates each ; expanding out this determinant gives us the required result.
Nakayama's lemma, corollary: Let be a finitely generated module, and be an ideal in . If , then and . This can be proved by taking identity and in the previous result.
Nakayama's lemma: Let be a finitely generated module, and be an ideal contained in the Jacobson radical of . Then . To prove this, notice that we have for some , by the corollary. But is a unit in , and hence .
Let be a finitely generated module, a submodule of , and an ideal . Then, . This can be proved by observing that .
Let be a local ring, its maximal ideal, its residue field, and a finitely generated module. Then, is annihilated by , hence is an module i.e. a vector space that is finitedimensional. Let s be the elements of whose image is in . Then, s generate . To prove this, let be the submodule of generated by , so it remains to be shown that . The composite maps onto , and hence , and .
Homomorphism theorems of modules
Proofs for these correspond exactly to those in vector spaces.

, and are submodules.

Let be a submodule; then a surjective quotient homomorphism , ker . Elements of can be constructed either as equivalence classes , or as cosets in .

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Exact sequences
If , , are Modules, and , are homomorphisms, then:
is exact if is injective, is surjective, and induces the isomorphism onto .
Alternatively, the condition for a short exact sequence can be written as: , and , such that maps and maps .
These two results are easily proved:

The sequence is exact for any , the sequence is exact.

The sequence is exact for any , the sequence is exact.
Define the following commutative diagram of modules, and module homomorphisms:
Then, there exists an exact sequence:
where are restrictions of , and are induced by .
The boundary homomorphism is defined as follows: if , then for some , and , hence so that for some . Then is defined to be the image of in .
Let be a class of modules, and a function with values in abelian group . is additive if, for each s.e.s with terms in , we have . As an example, let be a vector space , and be the class of all vector spaces . Then is additive.
Let be an exact sequence with modules and kernels of all homomorphisms in class . Then, for any additive function , we have:
This is proved by splitting up the long exact sequence into short exact sequences.
The Koszul complex of pair is an example of an s.e.s:
Tensor product
Let be three modules. The mapping is said to be bilinear if, for each , the mapping of onto , and the mapping of onto , are both linear.
We construct an module , called the tensor product, if the bilinear mapping is in onetoone correspondence with linear mappings , for all modules . More precisely, there exists a pair consisting of module and bilinear mapping , such that , where is a unique linear mapping. In other words, every mapping factors uniquely through .
Note that tensor product notation is ambiguous; it may be zero in , submodules of , while being zero in modules . For example, consider submodule of while . Then, , but is nonzero as an element of .
Let , and in . Then, there exist finitely generated submodules of such that in . To prove this, consider ; then is a finite sum of generators of . Let be the submodule of generated by the and all the elements of that occur as the first coordinates in these generators of . Define similarly. Then, in , as required.
Let be modules. Then, there exist unique isomorphisms:

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Let be a rings, an module, a module, and an module (i.e. it is simultaneously an module and module, and the two structures are compatible in that for all ). Then is naturally a module, is an module, and we have:
Let be two module homomorphisms. Define by . Then, is bilinear and induces the homomorphism:
such that:
Let be two module homomorphisms. Then, and agree on all elements of the form in . Since the elements generate , it follows that:
Restriction and extension of scalars
Let be a homomorphism of rings, and let be a module. Then has an module structure defined as follows: if , is defined to be . This module is derived from by restriction of scalars. In particular, defines an module structure on .
Suppose is a finitely generated module, and is a finitely generated module, then is finitely generated as an module. To prove this, let generate over , generate as an module. Then the products generate over .
Let be an module. Since an can be regarded as an module, we can form the module . In fact, carries a module structure since , for all . is said to be obtained from by extension of scalars.
If is a finitely generated module, is finitely generated as a module. To prove this, observe that, if generate over , then generates over .
Exactness of the tensor product
Let be an bilinear mapping. Then, the mapping given by is linear, and hence gives rise to the mapping , which is also linear, because is linear in the variable . Conversely, the homomorphism is a bilinear map given by . Hence the set of bilinear mappings is in natural onetoone correspondence with . On the other hand, is in onetoone correspondence with , by definition of tensor product. Hence:
Consider the following exact sequence of module homomorphisms:
Let be any module. Then, the following sequence is also exact:
To prove this, let denote the first sequence, the second sequence, and any module. Since the first sequence is exact, is exact is exact is exact.
Let . Then, for all . is termed left adjoint of , and is termed right adjoint of . From category theory, any functor which is left adjoint is right exact, and right adjoint is left exact.
If tensoring with translates all exact sequences into exact sequences, then is said to be a flat module. This is not true in general, and can be illustrated with an example: consider the exact sequence , where . Tensoring with yields , which is not exact because, for any :
So is the zero mapping, while is nonzero.
The following statements about an module are equivalent:

is flat.

The exact sequence , when tensored with , is exact.

If is injective, so is .

If is injective, and are finitely generated, then is injective.
If is a ring homomorphism, and is a flat module, then is a flat module.
Algebras
Let be a ring homomorphism, . Define the operation . This definition makes the ring into an module. Thus, has module structure as well as ring structure. Ring equipped with is said to be an algebra.
What follows are two examples:

If , a vector space, then is injective, and therefore can be identified with its image in . Thus a algebra is simply a ring containing as a subring.

Let be any ring. Since has identity, there is a unique homomorphism , given by . Thus every ring is automatically a algebra.
Let and be two ring homomorphisms. Then, is an algebra homomorphism, which is both a ring homomorphism and an module homomorphism. We also have .
A ring homomorphism is of finite, and is a finite algebra if is finitely generated as an module. is of finite type, and is a finitely generated algebra, if there is a finite set of elements in such that every element of can be written as a polynomial in with coefficients in ; equivalently, there is an algebra homomorphism from the polynomial ring onto .
Ring is said to be finitely generated, if it is finitely generated as a algebra. This means that there is a finite set of elements in , such that every element in can be written as a polynomial in s with rational coefficients.
Tensor product of algebras
Let be two algebras, and be two homomorphisms. Since are two modules, we may form the tensor product , which is an module. Now, consider defined by:
This is linear, and therefore induces the homomorphism:
And hence by module homomorphism,
This corresponds to the bilinear mapping:
which is such that:
We have therefore defined multiplication on the tensor product ; for elements of the form , this is given by:
Furthermore is an algebra, and the mapping is a ring homomorphism . We hence end up with the following commutative diagram: