Ring Homomorphisms and Ideals

Centered at the study of algebra is the study of homomorphisms between algebraic structures. For the case of ring, a ring homomorphisms is connected by a special kind of subring called the ideal, which is the kernel of the homomorphism and has absorption property.

In this text, we will first compare and migrate the isomorphism theorems into ring theory, and then categorize the ideals into principal, maximal and prime ideals. This should establish a firm understanding of ring homomorphisms and ideals, and will be helpful in the future study.

Ring Homomorphisms

Just like how we define the group homomorphisms, we define the ring homomorphisms as maps such that .

A ring homomorphism is a group homomorphism in the first place, mapping to . And by , multiplication is closed in , inheriting 's properties and .

This does not appeal to us since any subgroup of abelian group is normal in it and can be used as the kernel of group homomorphism. However, when we put into multiplication:

So it's required for the kernel of a ring homomorphism to absorb elements in the ring under multiplication.

On the other hand, consider a special kind of subring called ideal, which is a subring fulfiling , and absorbing 's elements under multiplication. Obviously by , we can define quotient group and canonical homomorphism that . Noted that , which means , and it won't be hard to verify is also a ring (and is called the quotient ring specifically, analogous to quotient group in group theory):

is abelian group.
is semigroup:
1. so .
2. .
Multiplication is distributive over addition:
1. .
2. .

Now we know ideal has normal subgroup like properties in ring theory: the canonical homomorphism (under addition) of is also a ring homomorphism, and each kernel of ring homomorphism must essentially be an ideal of . In this way we establish the first isomorphism theorem in ring theory.

We conventionally denote such relationship between ideal and ring as . There should be no ambiguity since is trivial and not worth mentioning in ring theory context.

It's possible for some subring to absorb 's element on only one side, and we call such that the left ideal, and such that the right ideal. And when we refer to ideal barely, it must be both left ideal and right ideal, or we can call it two-sided ideal to emphasize.

Where does the unity go

For a ring with unity , under the ring homomorphism , let's consider how will the unity be mapped.

Assume , then there will be:

Which means the whole ring is pinched to its kernel and quotient ring is pinched to zero ring, if unity is in the kernel. So for any proper ideal , there must be .

A proper ideal cannot contain a unit either, since for a unit , there's:

So we can conclude that it's impossible for a proper ideal to contain unity or any unit. We can furtherly conclude that fields do not have non-trivial ideals.

On the other hand, assume , there's:

So is also a ring with unity if . And if is also a ring with unity , we have by the uniqueness of unity.

Analogy of congruence relationship

We have already been familiar with congruences of integers that : by group theory, is a subgroup of , and the congruence means lies inside the coset ; while by ring theory, obviously is an ideal of since , and the addition and multiplication in and matches by 's defining a ring homomorphism between them.

The study of ring homomorphisms and ideals reveals the fundamentals of congruence relationship, while extending and generalizing it on general ring , that as long as there's ring homomorphism defined by ideal , we may write:

Of course we are not limited to introducing just another congruence defined by ideals. By studying ring homomorphisms and ideals, we compare an ideal to integers, inspecting under which circumstances an ideal behaves like an integer, and under which condition the ideal is in a completely different nature. Only then will we know what disciplines the behaviour of integers and more general rings.

Migration of the isomorphism theorems

While we would like to elevate the isomorphism theorems to be held in ring theory, we must be careful about the following points for well-formedness:

The symbol represents the existence of ring isomomorphism between and , not just trivial group isomorphism defined on and .
The symbol represents the quotient ring over ideal .

Recall how we established the isomorphism theorems of groups: first we proved the first isomorphism theorem of groups; then for certain type of quotient groups and normal subgroups, we came up with the potential group homomorphisms between them and identified their kernels, proving their isomorphic relationship by the first isomorphism theorem.

Then back to our circumstance of proving the case of isomorphism theorem of rings: for each isomorphism theorem, we will identify the group homomorphism used to prove it, alongside with its kernel; then we will prove the kernel is an ideal, so that the group homomorphism given is actually also a ring homomorphism, and the isomorphic relationship holds between rings; we will also ensure the well-formedness of quotient rings and ideals.

Let's take a look at the second isomorphism theorem, which states for subring and ideal . In group theory, this is done by showing there's a group homomorphism defined by , with . If we want to prove it to be held in ring theory, we must ensure that and .

The fact that can be shown by:

And since , we have .

The fact that can be shown by:

So in this way, we've proved the second isomorphism theorem also holds in ring theory.

Specially, when , we have due to:

Meantime we have due to:

These facts can be useful in the following text.

Let's have a look at the correspondence theorem, that the map defined by is bijective for ideal and subgroup : it's true that when viewing from the quotient group , the map defined by is bijective; and since , are all well-formed quotient rings, the bijective map here is not a ring homomorphism so no extra check is needed, and the correspondence theorem holds also for ring theory.

Finally, let's have a look at the third isomorphism theroem, which states for ideals that . In group theory, this is done by showing that there's a group homomorphism defined by , with kernel . The quotient rings are well-formed quotient rings, but we need to ensure that there's here.

By the correspondence theorem, we know corresponds to a subring of , so . And can be proved by:

By now, we've proved the third isomorphism theorem also holds in ring theory.

Migration of the direct products

The (external) direct product of ring is defined as:

One can easily check its operations fulfil the ring axioms so it's also a ring. And is a ring with unity iff is a ring with unity and is a ring with unity .

Consider the sum of ideals fulfiling , according to the second isomorphism theorem, we have and . And consider the map defined by , obviously it's a ring homomorphism. It has kernel , otherwise assume , there's , which means , and is a contradiction to . So we have .

Principal ideals

The idea of principal ideals is analogous to cyclic subgroups of groups, that we wish to construct an ideal generated by : by starting from , for every , we put into , while ensuring it's closed under addition and multiplication.

Without further information given, to fulfil our requirement, the element in the principal ideal generated by is in the form of:

The fact that can be verified by:

So the general form of principal ideal is ugly and giving us no information, but it can be simplified drastically by adding more constrain to it.

Let's try adding the existence of unity to the ring . In this way, we have , so they can all be merged into the first term and simplified into .

On the other hand, let's try adding commutativity to the ring , then we have , so they can be simplified into .

Finally, when both unity and commutativity are added to the ring , so that it's a commutative ring with unity, the elements in principal ideal generated by a can be simplified into , which means in a commutative ring with unity, all elements are in the form of product with . We'll see how this speciality affects other ideals of a ring in the remaining text.

All ideals of ring of integers are principal

It's worth mentioning that all ideals of rings of integers are principal, or formally . The whole ring and the zero ring are principal ideals generated by and trivially.

First, for two principal ideals , there's . due to the Bezout's theorem's constraining what linear combination of can cover, sufficiently and essentially.

Then, for any non-zero ideal of , select any , and its principal ideal must be ideal of , due to the closed property. Consider the emptiness of , if it's empty, , otherwise it's possible to select , and there's since and must be in in order for to be true. Then we check the emptiness of , select if it's not empty, and go on such process of checking and selecting on and so on. Such process won't last indefinitely since the descending chain of greatest common divisors has finite length and is capped at , at which point there's .

We can formalize such a process as the following algorithm with non-zero ideal as input:

Randomly select any .
If , then halt with .
Otherwise select , and update , and go to step 2.

When the input is zero ring, the algorithm halts at step 1 with . When the input is non-zero ideal, since is finite, monotonously decreasing and capped at , the algorithm must halt at step 2 with some such that . In this way we've proved that any non-zero ideal of must be principal.

Maximal ideals

A maximal ideal is a proper ideal such that .

Maximal ideals of commutative ring with unity

In the commutative ring with unity , is a maximal ideal iff is a field.

First, we would like to show is field when is maximal ideal. If is not a field, there should exists some non-unit . Consider the principal ideal generated by , we have , so there's . Since , there's . Noted that:

Then is a unit instead and is a contradiction to 's not being a field.

Conversely, if is a field, , so . For any ideal such that , it must include some element from . When we iterate through and encounter some such that , first by 's being ideal we have , then by 's closed property we have , so and is the maximal ideal.

In the ring of integers , if then there's obviously , and the ideal is maximal iff is prime.

Maximal ideals of non-commutative ring with unity

While one might suspect such property is still applicable to non-commutative ring with unity by merely replacing 's being field with 's being division ring, ~~that is a maximal ideal iff is a division ring~~. This will not happen generally and we are about to see why.

Let's start by inspecting a counterexample: denote the ring of all 2-by-2 real-valued matrices as , which is a non-commutative ring with unity and zero . For any proper ideal , there's , otherwise is a unit and .

Assume , without losing generality, let's pick up a matrix , and manipulate with the following steps:

With , we have , .
With , we have .
With , we have , .
With the closed property of , we have .

So if contains any , it will eventually contain and thus , which is a contradiction to 's being proper ideal, so it's only possible for to be . It won't be hard to verify that is maximal ideal. However, there's and is not a division ring.

When putting the cases of commutative ring with unity and non-commutative ring with unity together, we will see it's the complexity of the principal ideal prevent the quotient ring over maximal ideal in the non-commutative ring with unity from being a division ring: if would like to be a division ring, then for any , there's some such that or , this will require either or to be true. When is commutative, the simplicity of principal ideal generated by ensures the existence of such , but when is not commutative, the principal ideal generated by can be as complex as , and when it's possibly the case that , while . Like in our counterexample, for the principal ideal generated by , we have , and the unity can be represented as . However there's no such that (otherwise there will be ), so there's no inverse corresponding to and the quotient ring over the maximal ideal is thereby uninvertible.

Prime ideals

A prime ideal of a commutative ring is a proper ideal such that for any , either or . The adjective "prime" symbolizes 's behaviour of acting as if it were the prime in Euclid's lemma.

There's definition of prime ideals in non-commutative rings but the discussion requires excessive knowledge of ring theory, so we will omit them for now.

In the commutative ring , the proper ideal is prime ideal iff is an integral domain.

First, we would like to show is an integral domain. Since is also a commutative ring, if it's not an integral domain, it must be possible to pick up some non-trivial zero divisors such that . However this means . by is prime there's there's or , and there's either or , which is a contradiction and are both non-trivial zero divisors. So is an integral domain.

Then, assume is an integral domain, , but there's either or , which means either or .

In commutative ring , every maximal ideal is prime, since field is also an integral domain. But prime ideal is not necessarily maximal, as integral domain is not necessarily a field. Like in the ring of integers, the ideal , where is prime, is maximal and prime. The zero ideal is prime but not maximal, given that there's no non-trivial zero divisor in , but any ideal is a super ideal of .

On the other hand, we define the multiplicatively closed set of ring as and . An proper ideal of commutative ring is prime iff its complement is a multiplicatively closed: when is prime, if there were any such that , then by is prime there's either or , which is a contradiction; when is ideal and is multiplicatively closed, if there's any case that but and , or equivalently and , then is not multiplicatively closed, which is a contradiction.

Chinese remainder theorem for rings

Finally, let's see if we are able extend the Chinese remainder theorem from integers, or just ring , to general rings.

According to our text of Chinese remainder theorem for integers, in order to find the solution to the linear congruences , we reconstruct them into two linear Diophantine equations, that . And since it can be furtherly transformed into , the system of linear congruences is solvable regardless of the choice of , only if there's . And if it's so, by using the extended Euclidean algorithm, we are guaranteed to find the Bezout's identity , and the solution is in the form of .

So the Chinese remainder theorem for integers specifies the existence of solution to system of linear congruences in the first place. As a starting point, for the ring and its ideals , let's think of under what circumstance will the system of congruences be unsolvable. Noted that the map defined by is a ring homomorphism, and its codomain enumerating all possibly solvable system of congruences. So when is not surjective, each one of corresponds to an unsolvable system of congruences.

Then let's see how the theorem guaranteed such a map for to be surjective: for each chosen, the will able to represent it for some , backed up by the existence of Bezout's identity , with which one can multiply the identity by to come up with a solution. Such capaiblity can also be intepreted in ring theory, that all elements representable by are in the form of . If it's capable of handling each incoming , then by , there's . When combined with , there's . Motivated by such instance in , when it comes to the case of general rings, let's try , and we have , which can be transformed into . Let , then by and , it's the solution and is surjective.

Noted that when ring has unity , there exists "Bezout's identity" due to . One can conveniently utilize it to obtain a solution in the form of , by and . But even without the Bezout's identity, is strong enough to come up with solution for each incoming . So the existence of Bezout's identity is not mandatory, we are still able to establish 's surjectivity even in the ring without unity.

On the other hand, if would like to be surjective, there's , since inside the intersection lies the solution to the system of congruences, corresponding to each . And if it's true, we have , and , so . When combined with , there's .

So in conclusion, the ring homomorphism defined by is surjective iff , and are said to be comaximal ideals in such case.

It won't be hard to verify that , since , so . And by first isomorphism theorem, there's:

In this way, we've extended and established the Chinese remainder theorem from the ring of integers to general rings.

Conclusion

In this text, we studied the ring homomorphism, and derived the first isomorphism theorem for rings that a ring homomorphism can be seen as modular arithmetics on an ideal, which is the kernel of the homomorphism and has absorption property. Based on the first isomorphism theorem, we also migrate the other isomorphism theorems into ring theory, using quotient rings and ideals in place of quotient group and normal subgroups. We've also briefly discussed about some common operations between ideals during the migration.

Then we studied the principal, maximal and prime ideals: the principal ideal is the ideal generated by an element ; the maximal ideal is the ideal such that only the whole ring can be its super ideal; the prime ideal discussed on our current stage, is the defined for commutative ring, that for any , there's either or .

Specially in the ring of integers , all ideals are principal. For an arbitary ideal , we must be capable of selecting some , and there's . Remove from , if it's empty, there's , otherwise select another , and there's . Such process forms a monotonically descending chain , which is of finite length and corresponds to an ascending chain of principle ideals , so must be among them and principal.

Specially in the commutative rings with unity , the principal ideal generated by has the form . This brings special property to maximal ideals since for any non-zero element and the maximal ideal it generates, we have . So the unity is in the form of , and given that:

All non-zero elements in are units and is a field. We've proved that is integral domain iff is prime, and as the field is also an integral domain, every maximal ideal is also prime. But there's prime ideal that is not maximal, like .

Finally, we've extended the Chinese remainder theorem to the ring theory. For ring , The existence of unsolvable system of congruences over ideals is due to that ring homomorphism defined by is not surjective. It's surjectivity can be proved to be equivalent to . And when is surjective, given that , by the first isomorphism theorem there's . This is a generalized form of Chinese remainder theorem in ring theory.

By now, we've got used to the ring homomorphisms and ideals, and ready to dive into further topics in ring theory.

June 18, 2023

algebra ring-theory