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A function that is both one-to-one and onto, that is if f(x) = f(y) then x = y, and for every y of the domain there is an x of the range so that f(x) = y. A function f : AmapsB is said to be injective (or one to one) if whenever f(x) = f(y), we have x = y. A function f : AmapsB is said to be surjective (or onto) if for every y element of B there exists x element of A such that f(x) = y. The set of integers. The set of rational numbers. The set of real numbers. The set residue classes mod n. Let R and S be rings and let f :RmapsS be a surjective isomorphism. Then f -1:SmapsR is a surjective isomorphism. Let f :RmapsS and g:SmapsT be ring homomorphisms. Then gf :RmapT is a ring homomorphism. Let R be a ring. An ideal of R is a subring I of R such that if a element of I, r element of R, then a · r element of I and r · a element of I.

Isomorphisms and Homomorphisms

In this section we will imitate our approach in the theory of groups, whereby we introduced isomorphisms and homomorphisms between groups.

Definition 1: Let R and S be rings. A ring homomorphism from R to S is a function f :RmapsS such that for a,b element of R, we have

(a) f(a + b) = f(a) + f(b),

(b) f(a · b) = f(a) · f(b).

Note that in condition (a) the sum on the left refers to addition in R, while the sum on the left refers to addition in S. Similar comments apply to the multiplication in (b).

Condition (a) implies that a ring homomorphism is a homomorphism from the additive group of R to the additive group of S. Therefore, we may apply our results on group homomorphisms to get that

(1)
f(-a) = -f(a)    (a element of R),
(2)
f(0) = 0.

[Note that there is a slight ambiguity in (2). The zero on the left refers to the zero element of R, while the zero on the right refers to the zero element of S.]

Definition 2: Let R and S be rings. A ring homomorphism f : RmapsS is said to be isomorphic if f is injective. We say that R is isomorphic to S if there exists a surjective isomorphism from R to S, and write Risomorphic toS.

Example 1: Let R be any commutative ring, X indeterminate over R, a element of R. If f = a0 + a1X + ... + anXn element of R[X], let us define the value of f at a, by

f(a) = a0 + a1a + ... + anan.

Let phia:R[X]mapsR be the mapping defined by

phia(f ) = f(a)   (f element of R[X]).

Then phia is a homomorphism. Indeed, if

g = b0 + b1X + ... + bnXn + ... element of R[X]

then

phia(f + g) = phia((a0+b0)+(a1+b1)X+...+(an+bn)Xn+...)
= (a0+b0)+(a1+b1)a+...+(an+bn)an+...
= (a0 + a1a + ... + anan + ...)
     +(b0 + b1a + ... + bnan + ...)
= phia(f) + phia(g).

Similarly, since

f · g = c0 + c1X + ... + cnXn + ...,

where

ci = a0bi + a1bi-1 + ... +aib0,

we see that

phia(f · g) = c0 + c1a + c2a2 + ... + cnan + ...

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

=phia(f ) · phia(g).

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

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

f(x + y) = x + y = x + y = f(x) + f(y),
f(x · y) = x · y = x · y = f(x) · f(y),

Therefore f :RmapsS is a homomorphism.

Example 3: Let R = S = Zsquare root of 2 and define f :R mapsS by

f(a + square root of 2) = a - bsquare root of 2
where (a + bsquare root of 2 element of Zsquare root of 2)

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(square root of 2), S = Q[X]/(X2 - 2)Q[X], and let f :SmapsR be defined by

f(a0 + a1X + I) = a0 + a1square root of 2
(a0,a1 element of Q, I = (X2 - 2)Q[X]).

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

Q(square root of 2)isomorphic toQ[X]/(X2 - 2)Q[X].

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

f :Rmaps toS,
f(x) = 0     (x element ofR)

is called the zero homomorphism.

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

phiI:RmapsR/I,
phiI(x) = x + I   (x element of R).

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

phiI(x + y) = (x + y) + I
= (x + I) + (y + I)
= phiI(x) + phiI(y),
phiI(x · y) = x · y + I
= (x + I) · (y + I)
= phiI(x) · phiI(y).

Therefore, phiI is a ring homomorphism, called the canonical ring homomorphism from R to R/I. Note that if we consider phiI only as a homomorphism from the additive group of R to the additive group of R/I, then phiI 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 :RmapsS be a surjective isomorphism. Then f -1:SmapsR is a surjective isomorphism.

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

f -1(x + y) = f -1(x) + f -1(y),
(3)
f -1(x · y) = f -1(x) · f -1(y).

Since f is a homomorphism,

f(f -1(x) + f -1(y)) = f(f -1(x)) + f(f -1(y))
(4)
= x + y,
f(f -1(x) · f -1(y)) = f(f -1(x)) · f(f -1(y))
(5)
= x · y,

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 :RmapsS and g:SmapsT be ring homomorphisms. Then gf :RmapT is a ring homomorphism.

Proof: Let a,b element of 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) Risomorphic toR.

(2) If Risomorphic toS, then Sisomorphic toR.

(3) If Risomorphic toS and Sisomorphic toT, then Risomorphic toT.

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

(2) If f :RmapsS is a surjective isomorphism, then f -1:SmapsR is a surjective isomorphism by Lemma 3, so that Sisomorphic toR.

(3) If f :RmapsS, g:SmapsT are surjective isomorphisms, then gf:RmapsT is bijective and a homomorphism by Lemma 4. Therefore, gf is a surjective isomorphism and Risomorphic toT.

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 :RmapsS 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 element of f(U), and let x = f(r), y = f(t), where r,t element of U. Then

x - y = f(r) = f(t) = f(r - t),
xy = f(r)f(t) = f(rt).

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

(2) Let x,y element of S, and let x = f(r), y = f(t), where r,t element of 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 element of 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 :RmapsS as

{r element of R | f(r) = 0}.

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 = phia, where phia:R[X]mapsR is evaluation at a, then ker(f ) = {f element of R[X] | f(a) = 0}.

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

3'. If f :Z[square root of 2]mapsZ[square root of 2], f(a + bsquare root of 2) = a - bsquare root of 2, then ker(f ) = {0}.

4'. If f :Q[square root of 2]mapsQ[X]/(X2 - 2)Q[X], f(a0 + a1square root of 2) = a0 + a1X + I, then ker(f ) = {0}.

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

6'. If R is any ring, I an ideal in R, phiI:RmapsR/I, phiI = the canonical homomorphism, then ker(phiI) = 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 :mapsS 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 element of ker(f ), r element of R. Then

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

= 0 · f(r)   (since x element of ker(f ))

= 0.

Therefore, x · r element of ker(f ). Similarly, r · x element of 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 :RmapsS 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 )mapsS such that the diagram

ring isomorphism diagram

is commutative - that is, fphi = f. Here phi denotes the canonical homomorphism phi:RmapsR/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 :RmapsS be a homomorphism, phi:RmapsR/ker(f ) the canonical homomorphism associated with ker(f ). Then there exists a unique ring isomorphism f :R/ker(f )mapsS such that the diagram

ring isomorphism diagram

is commutative. In particular, f(R)isomorphic toR/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.