Ring homomorphisms and isomorphisms

Just as in Group theory we look at maps which "preserve the operation", in Ring theory we look at maps which preserve both operations.

Definition

A map f : R S between rings is called a ring homomorphism if
f(x + y) = f(x) + f(y) and f(xy) + f(x)f(y) for all x, y R.

Remarks

1. The operations on the left are in the ring R; those on the right are in S.

2. Since such a homomorphism is a group homomorphism from (R, +) to (S, +) it maps 0_R to 0_S.
   Even if the rings R and S have multiplicative identities a ring homomorphism will not necessarily map 1_R to 1_S.
   
3. It is easy to check that the composition of ring homomorphisms is a ring homomorphism.
   

A ring homomorphism which is a bijection (one-one and onto) is called a ring isomorphism.

If f : R S is such an isomorphism, we call the rings R and S isomorphic and write R S.

Remarks

1. Isomorphic rings have all their ring-theoretic properties identical. One such ring can be regarded as "the same" as the other.

2. The inverse map of the bijection f is also a ring homomorphism.
   

1. The map from to _n given by x x mod n is a ring homomorphism. It is not (of course) a ring isomorphism.

2. The map from to given by x 2x is a group homomorphism on the additive groups but is not a ring homomorphism.

3. The map from to the ring of 2 2 real matrices given by x is a ring homomorphism which does not map the multiplicative identity to the multiplicative identity.
   Of course the map x is a homomorphism which does map the identity to the identity.

4. The "evaluation at ¹/₂ map" from the ring of continuous real-valued functions on the interval [0, 1] to given by (e_1/2)f = f(¹/₂) for f C[0, 1] is a ring homomorphism.

5. The "evaluation at 1 map" from the ring [x] to given by (e₁)p = p(1) for p [x] is a ring homomorphism.
   That is: e₁(a₀ + a₁x + a₂x² + ... + a_nxⁿ) = a₀ + a₁ + a₂ + ... + a_n.

6. The "evaluation at 2 map" from [x] to given by p p(2) is a ring homomorphism.
   This will turn out to be a surprisingly important example.

7. The map from to ring of 2 2 real matrices given by a + bi is a ring isomorphism.

   Proof: Exercise.

8. Let R be the field with 9 elements {a + bx | a, b ₃ } and the multiplication rule x² = -1.
   Let S be the field with 9 elements {a + by | a, b ₃ } and the multiplication rule y² = y + 1.
   Then the map defined by 1 1 and x y + 1 defines a ring isomorphism.

   Proof: We'll see a neat way of proving this later.
   

Definition

The kernel of a (ring) homomorphism is the set of elements mapped to 0.

That is, if f: R S is a ring homomorphism, ker(f) = f^-1(0) = {r R | f(r) = 0_S }.

Theorem

The kernel of a ring homomorphism is an ideal.

Proof
An easy verification

Remarks

1. Note the similarity with the corresponding result for groups: the kernel of a group homomorphism is a normal subgroup.
   
2. If the ring R is not commutative, the kernel is a two-sided ideal.
   

1. The kernel of the above map from to _n is the ideal n.

2. Just as in the group-theory case, the kernel of a homomorphism is {0} if and only if the homomorphism is one-one.

   Proof
   () If f(r) = f(s) then f(r - s) = 0 and so r - s ker(f) and we have r - s = 0.
   () If a ker(f) and a 0 then a, 0 0 and so the map is not one-one.
   

3. The kernel of the "evaluation at 0" map from [x] to is the ideal < x > of polynomials with zero constant terms.

4. The kernel of the "evaluation at 0 taken modulo 2" map from [x] to ₂ is the ideal < 2, x > of polynomials with even constant terms.
   

We will see later that every ideal is the kernel of a ring homomorphism. This is similar to the group theory result that every normal subgroup is the kernel of a group homomorphism.

The last result in this section also parallels the corresponding example in Group theory,

Theorem

The image of a ring homomorphism f: R S is a subring of S.

Proof
The image is the set im(f) = {s S | s = f(r) for some r R }. It is easy to do the verification.

Example

The image of e₂ from [x] to is the subring {a + b2 | a, b } we met earlier.