§3.3 Isomorphisms and Homomorphisms

Example 3.3.1: Isomorphic Fields

If you were unfamiliar with Roman numerals and came across a discussion of integer arithmetic written solely with Roman numerals, it might take you some time to realize that this arithmetic was essentially the same as the familiar arithmetic in \(\mathbb{Z}\) except for the labels on the elements. Here is a less trivial example.

Consider the subset \(S = \{0, 2, 4, 6, 8\}\) of \(\mathbb{Z}_{10}\). With the addition and multiplication of \(\mathbb{Z}_{10}\), \(S\) is actually a commutative ring, as can be seen from these tables:

A careful examination of the tables shows that \(S\) is a field with five elements and that the multiplicative identity of this field is the element 6.

We claim that \(S\) is “essentially the same” as the field \(\mathbb{Z}_5\), except for the labels on the elements. You can see this as follows. Write out addition and multiplication tables for \(\mathbb{Z}_5\). To avoid any possible confusion with elements of \(S\), denote the elements of \(\mathbb{Z}_5\) by \(\bar{0}, \bar{1}, \bar{2}, \bar{3}, \bar{4}\). Then relabel the entries in the \(\mathbb{Z}_5\) tables according to this scheme:

• Relabel \(\bar{0}\) as 0,
• Relabel \(\bar{1}\) as 6,
• Relabel \(\bar{2}\) as 2,
• Relabel \(\bar{3}\) as 8,
• Relabel \(\bar{4}\) as 4.

Look what happens to the addition and multiplication tables for \(\mathbb{Z}_5\):

By relabeling the elements of \(\mathbb{Z}_5\), you obtain the addition and multiplication tables for \(S\). Thus the operations in \(\mathbb{Z}_5\) and \(S\) work in exactly the same way—the only difference is the way the elements are labeled. As far as ring structure goes, \(S\) is just the ring \(\mathbb{Z}_5\) with new labels on the elements. In more technical terms, \(\mathbb{Z}_5\) and \(S\) are said to be isomorphic.

Isomorphism

In general, isomorphic rings are rings that have the same structure, in the sense that the addition and multiplication tables of one are the tables of the other with the elements suitably relabeled, as in Example 3.3.1. Although this intuitive idea is adequate for small finite systems, we need a rigorous mathematical definition of isomorphism that agrees with this intuitive idea and is readily applicable to large rings as well.

There are two aspects to the intuitive idea that rings \(R\) and \(S\) are isomorphic: relabeling the elements of \(R\) and comparing the resulting tables with those of \(S\) to verify that they are the same. Relabeling means that every element of \(R\) is paired with a unique element of \(S\) (its new label). In other words, there is a function \(f: R \to S\) that assigns to each \(r \in R\) its new label \(f(r) \in S\).

In the preceding example, we used the relabeling function \(f: \mathbb{Z}_5 \to S\), given by:

\(f(\bar{0}) = 0, \quad f(\bar{1}) = 6, \quad f(\bar{2}) = 2, \quad f(\bar{3}) = 8, \quad f(\bar{4}) = 4\).

Such a function must have these additional properties:

1. Distinct elements of \(R\) must get distinct new labels:
   If \(r \neq r'\) in \(R\), then \(f(r) \neq f(r')\) in \(S\).
2. Every element of \(S\) must be the label of some element in \(R\):
   For each \(s \in S\), there is an \(r \in R\) such that \(f(r) = s\).

Statements (i) and (ii) simply say that the function \(f\) must be both injective and surjective, that is, \(f\) must be a bijection.

In order for a bijection (relabeling scheme) \(f\) to be an isomorphism, applying \(f\) to the addition and multiplication tables of \(R\) must produce the addition and multiplication tables of \(S\). So if \(a + b = c\) in the \(R\)-table, we must have \(f(a) + f(b) = f(c)\) in the \(S\)-table.

However, since \(a + b = c\), we must also have \(f(a) + f(b) = f(c)\). Combining this with the fact that \(f(a) + f(b) = f(a + b)\), we see that:

\(f(a + b) = f(a) + f(b).\)

This is the condition that \(f\) must satisfy in order for \(f\) to change the addition tables of \(R\) into those of \(S\). The analogous condition on \(f\) for the multiplication tables is \(f(ab) = f(a)f(b)\).

We now can state a formal definition of isomorphism:

A ring \(R\) is isomorphic to a ring \(S\) (in symbols, \(R \cong S\)) if there is a function \(f: R \to S\) such that:

1. \(f\) is injective;
2. \(f\) is surjective;
3. \(f(a + b) = f(a) + f(b)\) and \(f(ab) = f(a)f(b)\) for all \(a, b \in R\).

In this case, the function \(f\) is called an isomorphism.

Example 3.3.2: Isomorphism Between \(K\) and \(\mathbb{C}\)

In Example 12 on page 50, we considered the field \(K\) of all \(2 \times 2\) matrices of the form:

\(\begin{pmatrix} a & b \\ -b & a \end{pmatrix},\)

where \(a\) and \(b\) are real numbers. We claim that \(K\) is isomorphic to the field \(\mathbb{C}\) of complex numbers. To prove this, define a function \(f: K \to \mathbb{C}\) by the rule:

\(f\left(\begin{pmatrix} a & b \\ -b & a \end{pmatrix}\right) = a + bi.\)

To show that \(f\) is injective, suppose:

\(f\left(\begin{pmatrix} a & b \\ -b & a \end{pmatrix}\right) = f\left(\begin{pmatrix} r & s \\ -s & r \end{pmatrix}\right),\)

so that \(f\) is injective. The function \(f\) is surjective because any complex number \(a + bi\) is the image under \(f\) of the matrix:

\(\begin{pmatrix} a & b \\ -b & a \end{pmatrix}\)

in \(K\). Finally, for any matrices \(A\) and \(B\) in \(K\), we must show that \(f(A + B) = f(A) + f(B)\) and \(f(AB) = f(A)f(B)\). We have:

\(f\left(\begin{pmatrix} a & b \\ -b & a \end{pmatrix} + \begin{pmatrix} c & d \\ -d & c \end{pmatrix}\right) = f\left(\begin{pmatrix} a + c & b + d \\ -(b + d) & a + c \end{pmatrix}\right)\) = \((a + c) + (b + d)i\) = \((a + bi) + (c + di)\) = \(f\left(\begin{pmatrix} a & b \\ -b & a \end{pmatrix}\right) + f\left(\begin{pmatrix} c & d \\ -d & c \end{pmatrix}\right).\)

and

\(f\left( \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \begin{pmatrix} c & d \\ -d & c \end{pmatrix} \right) = f\left( \begin{pmatrix} ac - bd & ad + bc \\ -ad - bc & ac - bd \end{pmatrix} \right)\) = \((ac - bd) + (ad + bc)i\) = \((a + bi)(c + di)\) = \(f\left( \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \right) f\left( \begin{pmatrix} c & d \\ -d & c \end{pmatrix} \right).\)

Therefore, \(f\) is an isomorphism.

It is quite possible to relabel the elements of a single ring in such a way that the ring is isomorphic to itself.

Example 3.3.3: Complex Conjugation Map

Let \(f: \mathbb{C} \to \mathbb{C}\) be the complex conjugation map given by \(f(a + bi) = a - bi\). The function \(f\) satisfies:

\(f((a + bi) + (c + di)) = f((a + c) + (b + d)i) = (a + c) - (b + d)i = (a - bi) + (c - di)\)

and

\(f((a + bi)(c + di)) = f((ac - bd) + (ad + bc)i) = (ac - bd) - (ad + bc)i = (a - bi)(c - di).\)

You can readily verify that \(f\) is both injective and surjective (Exercise 17). Therefore, \(f\) is an isomorphism.

Example 3.3.4: Identity Map on a Ring

If \(R\) is any ring and \(\nu_R: R \to R\) is the identity map given by \(\nu_R(r) = r\), then for any \(a, b \in R\):

\(\nu_R(a + b) = a + b = \nu_R(a) + \nu_R(b)\)

and

\(\nu_R(ab) = ab = \nu_R(a)\nu_R(b).\)

Since \(\nu_R\) is obviously bijective, it is an isomorphism.

Homomorphisms

Our intuitive notion of isomorphism is symmetric: "R is isomorphic to S" means the same thing as "S is isomorphic to R". The formal definition of isomorphism is not.

Many functions that are not injective or surjective satisfy condition (iii) of the definition of isomorphism. Such functions are given a special name.

Homomorphism: Let \(R\) and \(S\) be rings. A function \(f: R \to S\) is said to be a homomorphism if:

\(f(a + b) = f(a) + f(b) \quad \text{and} \quad f(ab) = f(a)f(b)\)

for all \(a, b \in R\).

Thus every isomorphism is a homomorphism, but as the following examples show, a homomorphism need not be an isomorphism because a homomorphism may fail to be injective or surjective.

Example 3.3.5: Zero Map

For any rings \(R\) and \(S\), the zero map \(z: R \to S\) given by \(z(r) = 0_S\) for every \(r \in R\) is a homomorphism because for any \(a, b \in R\):

\(z(a + b) = 0_S = 0_S + 0_S = z(a) + z(b)\)

and

\(z(ab) = 0_S = 0_S \cdot 0_S = z(a)z(b).\)

When both \(R\) and \(S\) contain nonzero elements, the zero map is neither injective nor surjective.

Example 3.3.6: Homomorphism from \(\mathbb{Z}\) to \(\mathbb{Z}_6\)

The function \(f: \mathbb{Z} \to \mathbb{Z}_6\) given by \(f(a) = [a]\) is a homomorphism because of the way that addition and subtraction are defined in \(\mathbb{Z}_6\). For any \(a, b \in \mathbb{Z}\):

\(f(a + b) = [a + b] = [a] + [b] = f(a) + f(b)\)

and

\(f(ab) = [ab] = [a][b] = f(a)f(b).\)

The homomorphism \(f\) is surjective, but not injective (Why?).

Example 3.3.7: Homomorphism from \(\mathbb{R}\) to Matrices

The map \(g: \mathbb{R} \to M(\mathbb{R})\) given by:

\(g(r) = \begin{pmatrix} 0 & -r \\ r & 0 \end{pmatrix}\)

is a homomorphism because for any \(r, s \in \mathbb{R}\):

\(g(r) + g(s) = \begin{pmatrix} 0 & -r \\ r & 0 \end{pmatrix} + \begin{pmatrix} 0 & -s \\ s & 0 \end{pmatrix} = \begin{pmatrix} 0 & -(r + s) \\ r + s & 0 \end{pmatrix} = g(r + s)\)

and

\(g(r)g(s) = \begin{pmatrix} 0 & -r \\ r & 0 \end{pmatrix} \begin{pmatrix} 0 & -s \\ s & 0 \end{pmatrix} = \begin{pmatrix} -rs & 0 \\ 0 & -rs \end{pmatrix} = g(rs).\)

The homomorphism \(g\) is injective but not surjective (Exercise 26).

Not all Functions are Homomorphisms

The properties \(f(a + b) = f(a) + f(b)\) and \(f(ab) = f(a)f(b)\) fail for many functions. For example, if \(f: \mathbb{R} \to \mathbb{R}\) is given by

\(f(x) = x + 2,\)

then

\(f(3 + 4) = f(7) = 9,\)

but

\(f(3) + f(4) = 5 + 6 = 11,\)

so that \(f(3 + 4) \neq f(3) + f(4)\). Similarly,

\(f(3 \cdot 4) = f(12) = 14,\)

but

\(f(3)f(4) = 5 \cdot 6 = 30.\)

Theorem 3.10: Properties of Homomorphisms

Let \(f: R \to S\) be a homomorphism of rings. Then:

1. \(f(0_R) = 0_S\).
2. \(f(-a) = -f(a)\) for every \(a \in R\).
3. \(f(a - b) = f(a) - f(b)\) for all \(a, b \in R\).

If \(R\) is a ring with identity and \(f\) is surjective, then:

4. \(S\) is a ring with identity \(f(1_R)\).
5. Whenever \(u\) is a unit in \(R\), then \(f(u)\) is a unit in \(S\) and \(f(u^{-1}) = (f(u))^{-1}\).

Proof of Theorem 3.10

1. \(f(0_R) + f(0_R) = f(0_R + 0_R)\) \hfill [\(f\) is a homomorphism.]

\(f(0_R) + f(0_R) = f(0_R)\) \hfill [\(0_R + 0_R = 0_R\) in \(R\).]

\(f(0_R) + f(0_R) = f(0_R) + 0_S\) \hfill [Subtract \(f(0_R)\) from both sides.]

\(f(0_R) = 0_S\)

2. First, note that:

\(f(a) + f(-a) = f(a + (-a))\) \hfill [\(f\) is a homomorphism.]

\(f(a) + f(-a) = f(0_R)\) \hfill [\(a + (-a) = 0_R\) in \(R\).]

\(f(a) + f(-a) = 0_S\)

Therefore, \(f(-a)\) is a solution of the equation \(f(a) + x = 0_S\). But the unique solution of this equation is \(-f(a)\) by Theorem 3.3. Hence, \(f(-a) = -f(a)\) by uniqueness.

3. \(f(a - b) = f(a + (-b))\) \hfill [Definition of subtraction.]

\(f(a - b) = f(a) + f(-b)\) \hfill [\(f\) is a homomorphism.]

\(f(a - b) = f(a) + (-f(b))\) \hfill [Part (2).]

\(f(a - b) = f(a) - f(b)\) \hfill [Definition of subtraction.]

4. We shall show that \(f(1_R) \in S\) is the identity element of \(S\). Let \(s\) be any element of \(S\). Since \(f\) is surjective, \(s = f(r)\) for some \(r \in R\). Hence:

\(s \cdot f(1_R) = f(r)f(1_R) = f(r \cdot 1_R) = f(r) = s\)

and, similarly, \(f(1_R) \cdot s = s\).

Therefore, \(S\) has \(f(1_R)\) as its identity element.

5. Since \(u\) is a unit in \(R\), there is an element \(v\) in \(R\) such that \(uv = 1_R\). Hence, by (4):

\(f(u)f(v) = f(uv) = f(1_R) = 1_S\).

Similarly, \(vf(u) = 1_R\) implies that \(f(v)f(u) = 1_S\). Therefore, \(f(u)\) is a unit in \(S\), with inverse \(f(v)\). In other words, \(f(u^{-1}) = f(v)\). Since \(v = u^{-1}\), we see that \(f(u^{-1}) = f(v) = f(u)^{-1}\).

Image of a Homomorphism

If \(f: R \to S\) is a function, then the image of \(f\) is this subset of \(S\):

\(Im f = \{s \in S \mid s = f(r) \text{ for some } r \in R\} = \{f(r) \mid r \in R\}.\)

If \(f\) is surjective, then \(Im f = S\) by the definition of surjective. In any case, we have:

Corollary 3.11

If \( f: R \to S \) is a homomorphism of rings, then the image of \( f \) is a subring of \( S \).

Proof of Corollary 3.11

Denote \( Im \, f \) by \( I \). \( I \) is nonempty because \( 0_S = f(0_R) \in I \) by part (1) of Theorem 3.10. The definition of homomorphism shows that \( I \) is closed under multiplication: If \( f(a), f(b) \in I \), then \( f(a)f(b) = f(ab) \in I \). Similarly, \( I \) is closed under subtraction because

\( f(a) - f(b) = f(a - b) \in I \)

by Theorem 3.10. Therefore, \( I \) is a subring of \( S \) by Theorem 3.6.

Existence of Isomorphisms

If you suspect that two rings are isomorphic, there are no hard and fast rules for finding a function that is an isomorphism between them. However, the properties of homomorphisms in Theorem 3.10 can sometimes be helpful.

Example 3.3.8: Isomorphism between \(\mathbb{Z}_{12}\) and \(\mathbb{Z}_3 \times \mathbb{Z}_4\)

If there is an isomorphism \(f\) from \(\mathbb{Z}_{12}\) to the ring \(\mathbb{Z}_3 \times \mathbb{Z}_4\), then \(f(1) = (1,1)\) by part (4) of Theorem 3.10. Since \(f\) is a homomorphism, it has to satisfy:

\(f(2) = f(1 + 1) = f(1) + f(1) = (1,1) + (1,1) = (2,2)\)

\(f(3) = f(2 + 1) = f(2) + f(1) = (2,2) + (1,1) = (0,3)\)

\(f(4) = f(3 + 1) = f(3) + f(1) = (0,3) + (1,1) = (1,0)\)

Continuing in this fashion shows that if \(f\) is an isomorphism, then it must be this bijective function:

\(f(1) = (1,1) \quad f(2) = (2,2) \quad f(3) = (0,3)\)

\(f(4) = (1,0) \quad f(5) = (2,1) \quad f(6) = (0,2)\)

\(f(7) = (1,3) \quad f(8) = (2,0) \quad f(9) = (0,1)\)

\(f(10) = (1,2) \quad f(11) = (2,3) \quad f(0) = (0,0)\)

All we have shown up to here is that this is the only possible homomorphism. To show that \(f\) actually is an isomorphism, it is necessary to verify that \(f\) satisfies the required properties for addition and multiplication. This can be done either by writing out the tables (tedious) or by observing that the rule of \(f\) can be described this way:

\(f([a]_{12}) = ([a]_3, [a]_4),\)

where \([a]_{12}\) denotes the congruence class of the integer \(a\) in \(\mathbb{Z}_{12}\), \([a]_3\) denotes the class of \(a\) in \(\mathbb{Z}_3\), and \([a]_4\) the class of \(a\) in \(\mathbb{Z}_4\). (Verify that this last statement is correct.) Then:

\(f([a]_{12} + [b]_{12}) = f([a + b]_{12}) \quad \text{[Definition of addition in \(\mathbb{Z}_{12}\)]}\)

= \(([a + b]_3, [a + b]_4) \quad \text{[Definition of \(f\)]}\)

= \(([a]_3 + [b]_3, [a]_4 + [b]_4) \quad \text{[Definition of addition in \(\mathbb{Z}_3\) and \(\mathbb{Z}_4\)]}\)

= \(([a]_3, [a]_4) + ([b]_3, [b]_4) \quad \text{[Definition of addition in \(\mathbb{Z}_3 \times \mathbb{Z}_4\)]}\)

= \(f([a]_{12}) + f([b]_{12}) \quad \text{[Definition of \(f\)]}\).

An identical argument using multiplication in place of addition shows that \(f([a]_{12}[b]_{12}) = f([a]_{12})f([b]_{12})\). Therefore, \(f\) is an isomorphism and \(\mathbb{Z}_{12} \cong \mathbb{Z}_3 \times \mathbb{Z}_4\).

Example 3.3.9: Proving Rings are Not Isomorphic

Up to now, we have concentrated on showing that various rings are isomorphic, but sometimes it is equally important to demonstrate that two rings are not isomorphic. To do this, you must show that there is no possible function from one to the other satisfying the three conditions of the definition.

\(\mathbb{Z}_6\) is not isomorphic to \(\mathbb{Z}_{12}\) or to \(\mathbb{Z}\) because it is not possible to have a surjective function from a six-element set to a larger set (or an injective one from a larger set to \(\mathbb{Z}_6\)).

Example 3.3.10: Proving Non-Isomorphism between \(\mathbb{Z}_4\) and \(\mathbb{Z}_2 \times \mathbb{Z}_2\)

The rings \(\mathbb{Z}_4\) and \(\mathbb{Z}_2 \times \mathbb{Z}_2\) are not isomorphic. To show this, suppose on the contrary that \(f: \mathbb{Z}_4 \to \(\mathbb{Z}_2 \times \mathbb{Z}_2\)\) is an isomorphism. Then \(f(0) = (0,0)\) and \(f(1) = (1,1)\) by Theorem 3.10. Consequently,

\(f(2) = f(1 + 1) = f(1) + f(1) = (1,1) + (1,1) = (0,0).\)

Since \(f\) is injective and \(f(0) = f(2)\), we have a contradiction. Therefore, no isomorphism is possible.

Example 3.3.11: Non-Isomorphism of \(\mathbb{Z}_8\) and \(\mathbb{Z}_4 \times \mathbb{Z}_2\)

In the ring \(\mathbb{Z}_8\), the elements 1, 3, 5, and 7 are units by Theorem 2.10. Since being a unit is preserved by isomorphism, any isomorphism from \(\mathbb{Z}_8\) to another ring with identity will map these four units to four units in the other ring. Consequently, \(\mathbb{Z}_8\) is not isomorphic to any ring with less than four units. In particular, \(\mathbb{Z}_8\) is not isomorphic to \(\mathbb{Z}_4 \times \mathbb{Z}_2\) because there are only two units in this latter ring, namely \((1,1)\) and \((3,1)\), as you can readily verify.

Example 3.3.12: Non-Isomorphism of \(\mathbb{Q}\), \(\mathbb{R}\), and \(\mathbb{C}\) with \(\mathbb{Z}\)

None of \(\mathbb{Q}\), \(\mathbb{R}\), or \(\mathbb{C}\) is isomorphic to \(\mathbb{Z}\) because every nonzero element in the fields \(\mathbb{Q}\), \(\mathbb{R}\), and \(\mathbb{C}\) is a unit, whereas \(\mathbb{Z}\) has only two units (1 and -1).

Example 3.3.13: Isomorphism Implies Commutativity

Suppose \(R\) is a commutative ring and \(f: R \to S\) is an isomorphism. Then for any \(a, b \in R\), we have \(ab = ba\) in \(R\). Therefore, in \(S\),

\(f(a) f(b) = f(ab) = f(ba) = f(b) f(a).\)

Hence, the image of \(R\) in \(S\) is commutative, and so \(S\) is a commutative ring.