f(-a) = -f(a)    (a R),

f(0) = 0.

f(a) = a₀ + a₁a + ... + a_naⁿ.

_a(f ) = f(a)   (f R[X]).

g = b₀ + b₁X + ... + b_nXⁿ + ... R[X]

then

Similarly, since

where

we see that

_a(f · g) = c₀ + c₁a + c₂a² + ... + c_naⁿ + ...

= (a₀ + a₁a + ...)·(b₀ + b₁a + ...) (R is commutative)

=_a(f ) · _a(g).

Therefore, _a is a ring homomorphism. We usually refer to _a as "evaluation at a," or "specialization at a"

Example 2: Let R = Z, S = Z_n. For x Z, define f(x) = x, where x Z_n denotes the residue class modulo n which contains x. Then

Therefore f :RS is a homomorphism.

Example 3: Let R = S = Z and define f :R S by

Then f is a ring homomorphism. It is trivial to see that f is injective, so that f is an isomorphism.

Example 4: Let R = Q(), S = Q[X]/(X² - 2)Q[X], and let f :SR be defined by

Then the mapping f was discussed in the previous section. It is a consequence of that discussion that f is a surjective isomorphism. Therefore,

Example 5: Let R be any ring and let S be any ring. The mapping

is called the zero homomorphism.

Example 6 Let R be any ring, I an ideal of R. Let us define the mapping

Then by the definitions of addition and multiplication in R/I.

Therefore, _I is a ring homomorphism, called the canonical ring homomorphism from R to R/I. Note that if we consider _I only as a homomorphism from the additive group of R to the additive group of R/I, then _I is the canonical homomorphism which we considered in our development of the first isomorphism theorem for groups. This naturally prompts us to ask whether the latter theorem can be extended to ring homomorphisms. This will be taken up below.

Lemma 3: Let R and S be rings and let f :RS be a surjective isomorphism. Then f ^-1:SR is a surjective isomorphism.

Proof: First note that it makes sense to speak of f ^-1 since f is bijective. Let x,y S. Then we must show that

(3)

Since f is a homomorphism,

(4) (5)

By applying f ^-1 to both sides of (4) and (5), we get (3). Therefore, f ^-1 is a homomorphism. But it is clear that f ^-1 is bijective. Therefore, f ^-1 is a surjective isomorphism.

Lemma 4: Let f :RS and g:ST be ring homomorphisms. Then gf :RT is a ring homomorphism.

Proof: Let a,b R. Then gf(a + b) = g(f(a + b)) = g(f(a) + f(b)) = g(f(a)) + g(f(b)) = gf(a) + gf(b). Similarly, gf(ab) = gf(a) · gf(b).

Proposition 5: Ring isomorphism is an equivalence relation. That is, the following properties hold. Let R, S, and T be rings.

(1) RR.

(2) If RS, then SR.

(3) If RS and ST, then RT.

Proof: (1) If i:RR is the identity function, then i is a surjective isomorphism.

(2) If f :RS is a surjective isomorphism, then f ^-1:SR is a surjective isomorphism by Lemma 3, so that SR.

(3) If f :RS, g:ST are surjective isomorphisms, then gf:RT is bijective and a homomorphism by Lemma 4. Therefore, gf is a surjective isomorphism and RT.

Just as was the case in the theory of groups, it is easy to see that isomorphic rings have identical properties differing only in the naming of the elements. The surjective ring isomorphisms preserve completely the structure of a ring. Although homomorphisms do not preserve all of the properties of a ring, they nevertheless preserve a number of very significant properties. The next proposition gives us some examples of properties preserved by homomorphisms.

Proposition 6: Let R and S be rings, f :RS a homomorphism. Then:

(1) If U is a subring of R, then f(U) is a subring of S.

Moreover, if f is surjective, then the following are true

(2) If R is commutative, then S is commutative.

(3) If R has an identity 1, then S has an identity f(1).

Proof: (1) Let x,y f(U), and let x = f(r), y = f(t), where r,t U. Then

Therefore, x - y and xy belong to f(U) is a subring of S

(2) Let x,y S, and let x = f(r), y = f(t), where r,t U. Then, since R is commutative, we see that xy = f(r)f(t) = f(rt) = f(tr) = f(t)f(r) = yx. Therefore, S is commutative.

(3) Let x S, and let x = f(r). Then f(1)x = f(1)f(r) = f(1 · r) = f(r) = x. Similarly, xf(1) = x. Therefore, f(1) is an identity for S.

Note that without the surjectivity of f, assertions (2) and (3) are false.

In analogy with the situation in group theory, let us define the kernel of the ring homomorphism f :RS as

Let us denote the kernel of f by ker(f ). Let us compute the kernels of the homomorphisms considered in Examples 1-6:

1'. If f = _a, where _a:R[X]R is evaluation at a, then ker(f ) = {f R[X] | f(a) = 0}.

2'. If f :ZZ_n, f(x) = x, then ker(f ) = {n · r | r Z}.

3'. If f :Z[]Z[], f(a + b) = a - b, then ker(f ) = {0}.

4'. If f :Q[]Q[X]/(X² - 2)Q[X], f(a₀ + a₁) = a₀ + a₁X + I, then ker(f ) = {0}.

5'. If R is any ring and f :R{0}, f(x) = 0 for all x R, then ker(f ) = R.

6'. If R is any ring, I an ideal in R, _I:RR/I, _I = the canonical homomorphism, then ker(_I) = I.

In group theory, the kernel of a homomorphism is always a normal subgroup, whereas the kernel of an isomorphism consists of just the identity element. These facts generalize to ring homomorphisms:

Proposition 7: Let R and S be rings, f :S a ring homomorphism, then

(1) ker(f ) is an ideal of R.

(2) f is an isomorphism if and only if ker(f ) = {0}.

Proof: Since f is a homomorphism from the additive group of R to the additive group of S, our results on group homomorphisms imply (a) that ker(f ) is a subgroup of the additive group of R and (b) that f is injective if and only if ker(f ) = {0}. Let x ker(f ), r R. Then

f(x · r) = f(x) · f(r)   (since f is a ring homomorphism)

= 0 · f(r)   (since x ker(f ))

= 0.

Therefore, x · r ker(f ). Similarly, r · x ker(f ). Thus by (a) we conclude that part (1) holds. Furthermore, (b) implies part (2).

Let us now take up the ring analogue of the first isomorphism theorem of group theory. Let R and S be rings and let f :RS be a ring homomorphism. Then, in particular, R and S are groups under addition and f is a homomorphism of the additive group of R into the additive group of S. Moreover, the kernel of f as a homomorphism of rings is just the same as the kernel of f as a homomorphism of additive groups. Therefore, the first isomorphism theorem of group theory implies that there exists a unique group isomorphism f :R/ker(f )S such that the diagram

is commutative - that is, f = f. Here denotes the canonical homomorphism :RR/ker(f ). Moreover, the proof of the first isomorphism theorem yields an explicit formula for f, namely f(r + ker(f )) = f(r). However, from the way in which we defined multiplication in R/ker(f ), we see that f((r + ker(f ))(r' + ker(f ))) = f((rr' + ker(f )) = f(rr') = f(r)f(r') = f(r + ker(f ))f(r' + ker(f )). Therefore, the unique group isomorphism which is provided by the first isomorphism theorem for group theory is also a ring isomorphism. Thus, we may summarize our findings in the following theorem.

Theorem 8 (First Isomorphism Theorem of Ring Theory) Let R and S be rings, and let f :RS be a homomorphism, :RR/ker(f ) the canonical homomorphism associated with ker(f ). Then there exists a unique ring isomorphism f :R/ker(f )S such that the diagram

is commutative. In particular, f(R)R/ker(f ).

The interpretation of Theorem 8 is analogous to the interpretation of the corresponding theorem for groups. On one hand, the theorem asserts that up to isomorphism, all homomorphisms of rings can be obtained as canonical homomorphisms. On the other hand, the first isomorphism theorem gives us a way of guaranteeing the existence of isomorphism without having to actually write them down.