Rings and Fields

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Factor rings and the isomorphism theorems

We parallel the development of factor groups in Group theory.


If I is an ideal of a ring R and aR 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.


  1. This is the same as the (additive) group theoretic coset. As in the groups case the cosets partition the ring into disjoint subsets.

  2. The element a is called a coset representative of the coset. Note that each coset will have lots of coset representatives. Elements a1 and a2 represent the same coset of I if a1 - a2I.

  3. The ideal I is itself a coset : 0 + I.


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.


We need to check that the operations are "well-defined". That is if a1 and a2 are representatives of the same coset and b1 and b2 represent the same coset then a1 + b1 and a2 + b2 represent the same coset and so do a1b1 and a2b2.

We have a1 - a2 I and b1 - b2 I and so adding these shows that (a1 + b1) - (a2 + b2) ∈ I and so these do represent the same coset.

Similarly, for the product, observe that a1b1 - a2b2 = (a1 - a2)b1 + a2(b1 - b2) 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.


  1. We call this ring a factor ring rather than a quotient ring to avoid confusion with something called a "ring of quotients".

  2. 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".

  3. Here is a "picture":


  1. Let I = nZZ.
    Then Z/nZ has n elements: I, 1 + I, 2 + I, ... , (n - 1) + I and the factor ring is Zn.

  2. Let I be the ideal of R[x] generated by the polynomial x2 + 1.

    Then every coset has a representative of the form (a + bx) + I since if we choose a coset representative (say) x3 then we have x3 = -x + (x2 + 1)x - x + I and so -x will do for a representative instead. You should see that by thinking of x2 + 1 as 0 one can use the "rule" x2 = -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 x2 = -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]/ < x2+ 1 > C .

  3. Using a similar argument we can see that the finite field GF(9) constructed earlier is in fact the factor ring Z3[x]/< x2 + 1 > since the "multiplication rule" x2 = -1 we used is equivalent to x2 + 1 = 0.

  4. The ring Z5[x]/< x2 + 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 x2 + 1 factorises in Z5[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 :RS is a ring homomorphism the factor ring R/ker(f) is isomorphic to im(f).


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.


  1. This shows that one can always think of factor rings as "homomorphic images" rather than as something involving operations of subsets (cosets).

  2. 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 IJ are also ideals and the factor rings (I + J)/J and I/(IJ) are isomorphic.


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 ii + J.

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

If iker(θ) then i + J = J and so iJ and hence is in IJ.

The result now follows from the First Isomorphism Theorem.


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.


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


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

Define a map θ : R/IR/J by a + Ia + 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 IJ) will only happen if aJ. Thus ker(θ) = J/IR/I and the result follows from the the First Isomorphism Theorem.

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JOC/EFR 2004