Rings and Fields

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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 : RS 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, yR.

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 0R to 0S.
    Even if the rings R and S have multiplicative identities a ring homomorphism will not necessarily map 1R to 1S.

  3. It is easy to check that the composition of ring homomorphisms is a ring homomorphism.

Definition

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

If f : RS 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.

Examples

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

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

  3. The map from Z 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 xx mod n 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 R given by (e½)f = f (1/2) for f belongs C[0, 1] is a ring homomorphism.

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

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

  7. The map from C 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, bZ3} and the multiplication rule x2 = -1.
    Let S be the field with 9 elements {a + by | a, bZ3 } and the multiplication rule y2 = y + 1.
    Then the map defined by 1 ↦ 1 and xy + 1 defines a ring isomorphism.

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


The connection of this with the last section is given by:

Definition

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

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

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.

Examples

  1. The kernel of the above map from Z to Zn is the ideal nZ.

  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 belongs ker(f) and we have r - s = 0.
    (<=) If aker(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 R[x] to R is the ideal < x > of polynomials with zero constant terms.

  4. The kernel of the "evaluation at 0 taken modulo 2" map from Z[x] to Z2 is the ideal < 2, x > of polynomials with even constant terms.

Remark

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: RS is a subring of S.

Proof

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

Example

The image of e√2 from Z[x] to R is the subring {a + b √2 | a, bZ } we met earlier.


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JOC/EFR 2004