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We parallel the development of factor groups in Group theory.

**Definition**

If *I* is an ideal of a ring *R* and *a* ∈ *R* then a **coset** of *I* is a set of the form *a* + *I* = {*a* + *s* | *s* *I* }.

The set of all cosets is denoted by *R*/*I*.

**Remarks**

- This is the same as the (additive) group theoretic coset. As in the groups case the cosets partition the ring into disjoint subsets.
- The element
*a*is called a**coset representative**of the coset. Note that each coset will have lots of coset representatives. Elements*a*_{1}and*a*_{2}represent the same coset of*I*if*a*_{1}-*a*_{2}∈*I*. - The ideal
*I*is itself a coset : 0 +*I*.

**Theorem**

*If I is an ideal of a ring R, the set R/I is a ring under the operations*

(*a* + *I*) + (*b* + *I*) = (*a* + *b*) + *I* and (*a* + *I*) . (*b* + *I*) = (*ab*) + *I*.

**Proof**

We need to check that the operations are "well-defined". That is if *a*_{1} and *a*_{2} are representatives of the same coset and *b*_{1} and *b*_{2} represent the same coset then *a*_{1} + *b*_{1} and *a*_{2} + *b*_{2} represent the same coset and so do *a*_{1}*b*_{1} and *a*_{2}*b*_{2}.

We have *a*_{1} - *a*_{2} *I* and *b*_{1} - *b*_{2} *I* and so adding these shows that (*a*_{1} + *b*_{1}) - (*a*_{2} + *b*_{2}) ∈ *I* and so these do represent the same coset.

Similarly, for the product, observe that *a*_{1}*b*_{1} - *a*_{2}*b*_{2} = (*a*_{1} - *a*_{2})*b*_{1} + *a*_{2}(*b*_{1} - *b*_{2}) and the result follows from the properties of the ideal.

Once you know that the operations are well-defined the ring axioms follow easily.

Note that the zero of the factor ring is the coset 0 + *I* or the ideal *I* itself.

**Remarks**

- We call this ring a factor ring rather than a quotient ring to avoid confusion with something called a "ring of quotients".
- The way you should think of the factor ring
*R*/*I*is that it is the ring*R*in which "all the elements in the ideal*I*have been made into zero". This is sometimes called "killing of the elements of*I*". - Here is a "picture":

**Examples**

- Let
*I*=*n***Z**⊆**Z**.

Then**Z**/*n***Z**has*n*elements:*I*, 1 +*I*, 2 +*I*, ... , (*n*- 1) +*I*and the factor ring is**Z**_{n}. - Let
*I*be the ideal of**R**[*x*] generated by the polynomial*x*^{2}+ 1.Then every coset has a representative of the form (

*a*+*bx*) +*I*since if we choose a coset representative (say)*x*^{3}then we have*x*^{3}= -*x*+ (*x*^{2}+ 1)*x*-*x*+*I*and so -*x*will do for a representative instead. You should see that by thinking of*x*^{2}+ 1 as 0 one can use the "rule"*x*^{2}= -1 to get rid of all higher powers.So we get a ring in which we can add and multiply things like

*a*+*bx*using the rule*x*^{2}= -1 to do multiplication.If you replace the indeterminate

*x*by*i*you will see that this is the complex numbers.That is:

**R**[*x*]/ <*x*^{2}+ 1 >**C**. - Using a similar argument we can see that the finite field
*GF*(9) constructed earlier is in fact the factor ring**Z**_{3}[*x*]/<*x*^{2}+ 1 > since the "multiplication rule"*x*^{2}= -1 we used is equivalent to*x*^{2}+ 1 = 0. - The ring
**Z**_{5}[*x*]/<*x*^{2}+ 1 > has 25 elements since (as above) coset representatives can be chosen to be of the form*a*+*bx*with*a*,*b*∈ 5. However, this is not a field since the cosets (*x*+ 3) +*I*and (*x*+ 2) +*I*have a product (*x*+ 3)(*x*+ 2) +*I*= 0 +*I*. Hence the ring has zero-divisors and is not a field.Notice that the reason that this happened was that the polynomial x

^{2}+ 1 factorises in**Z**_{5}[*x*] into a product of factors of lower degree. (You could see it had linear factors since the polynomial has roots in 5.)

We saw in the last section that the kernel of a ring homomorphism is an ideal and the image is a subgroup. In view of the corresponding result for groups it will come as no surprise that we have

**The Isomorphism Theorem for Rings**

If *f* :*R*→ *S* is a ring homomorphism the factor ring *R*/*ker*(*f*) is isomorphic to *im*(*f*).

**Proof**

Define the map *θ*: *R*/*ker*(*f*)→ *im*(*f*) *by* *θ*(*a* + *ker*(*f*)) = *f*(*a*).

Then the proof is exactly as in the group theory case except you also need to check that this map respects the ring multiplication as well as addition.

**Remarks**

- This shows that one can always think of factor rings as "homomorphic images" rather than as something involving operations of subsets (cosets).
- From this theorem every ideal is the kernel of some homomorphism (the map from
*R*to*R*/*I*).

The above result is sometimes called the **First Isomorphism Theorem** for Rings. Here are some others.

**The Second Isomorphism Theorem for Rings**

Let *I* and *J* be ideals of a ring *R*. Then *I* + *J* and *I* ∩ *J* are also ideals and the factor rings (*I* + *J*)/*J* and *I*/(*I* ∩ *J*) are isomorphic.

**Proof**

Here is a picture showing the inclusions. The double lines represent the two factor rings.

To prove the result, Define *θ* : *I*→ (*I* + *J*)/*J* by *i* ↦ *i* + *J*.

Since every coset in (*I* + *J*)/*J* has a representative of this form *θ* is onto.

If *i* ∈ *ker*(*θ*) then *i* + *J* = *J* and so *i* ∈ *J* and hence is in *I* ⊆ *J*.

The result now follows from the First Isomorphism Theorem.

**Remark**

In fact one only needs *I* to be a subring of *R* for this theorem to hold. One does need *J* to be an ideal.

**Example**

Let *I* = < *m* > and *J* = < *n* > be ideals of **Z**. Then (See Exercises 4 Qu 1) *I* + *J* = < *gcd*(*m*, *n*) > and *I* ∩ *J* = < *lcm*(*m*, *n*) > .

Then from this theorem one deduces that < *gcd*(*m*, *n*) > / < *n* > < *n* > / < *lcm*(*m*, *n*) > and from this you may deduce that *n*/*gcd*(*m*, *n*) = *lcm*(*m*, *n*)/*m* or that *gcd*(*m*, *n*) × *lcm*(*m*, *n*) = *mn* (See Exercises 3 Qu 6).

**The Third Isomorphism Theorem for Rings**

*Let I and J be ideals of a ring R with I ⊆ J. Then J/I is an ideal of R/I and (R/I)/(J/I) R/J*

**Proof**

It is easy to verify that *J*/*I* is an ideal of *R*/*I*.

Define a map *θ* : *R*/*I*→ *R*/*J* by *a* + *I* ↦ *a* + *J* for any coset *a* + *I* of *I* in *R*. This is clearly onto.

An element *a* + *I* is in the kernel of *θ* if *a* + *I* = *J* which (since *I* ⊆ *J*) will only happen if *a* ∈ *J*. Thus *ker*(*θ*) = *J*/*I* ⊆ *R*/*I* and the result follows from the the First Isomorphism Theorem.

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