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Gregory T. LeeAbstract AlgebraSpringer Undergraduate Mathematics Serieshttps://doi.org/10.1007/978-3-319-77649-1_9

9. Ideals, Factor Rings and Homomorphisms

Gregory T. Lee¹

(1)

Department of Mathematical Sciences, Lakehead University, Thunder Bay, ON, Canada

Gregory T. Lee

We saw in Chapter 4 that for some purposes, subgroups are not quite good enough; we needed to consider normal subgroups. There is a similar concept in ring theory. In this chapter, we introduce the notion of an ideal, which is a subring with one additional condition imposed. We can then define factor rings and discuss ring homomorphisms and isomorphisms. Along the way, we will mention several important sorts of ideals, including principal, maximal and prime ideals.

9.1 Ideals

When we discussed normal subgroups of a group, our main concern was to find a condition that we could impose in order to make the group operation on a factor group well-defined. Our motivation is the same here. Of course, a subring is necessarily an additive subgroup of a ring, and as the additive group is abelian, we do not have to worry about normality. However, we need an additional condition to make the multiplication operation work properly.

Definition 9.1.

Let R be a ring. Then a subring I of R is said to be an ideal if $ir, ri\in I$ for all $i\in I$ and $r\in R$ . We call this the absorption property .

Note that closure under multiplication is not enough. We need to be able to multiply an element of the ideal by any element of the ring and stay within the ideal. Combining the definition with Theorem 8.5, we immediately obtain the following.

Theorem 9.1.

Let R be a ring and I a subset of R. Then I is an ideal if and only if

1.

$0\in I$ ;
2.

$i-j\in I$ for all $i, j\in I$ ; and
3.

$ir, ri\in I$ for all $i\in I$ , $r\in R$ .

Example 9.1.

Let n be any integer. Then $n\mathbb Z$ is an ideal of $\mathbb Z$ . Indeed, we already know that it is a subring. But also, if $nk\in n\mathbb Z$ , then for any integer r, $r(nk)=n(rk)\in n\mathbb Z$ .

Example 9.2.

Let I be the set of all polynomials $f(x)\in \mathbb R[x]$ such that $$f(0)=0$$ . We claim that I is an ideal in $\mathbb R[x]$ . Certainly I contains the zero polynomial. Also, if $$f(0)=g(0)=0$$ , then $$(f-g)(0)=f(0)-g(0)=0$$ , hence $f(x)-g(x)\in I$ . Also, if and $h(x)\in \mathbb R[x]$ , then $$h(0)f(0)=h(0)0=0$$ . Hence, $h(x)f(x)\in I$ .

Example 9.3.

Let I be the set of all polynomials in $\mathbb Z[x]$ whose constant term is a multiple of 5. Then I is an ideal. See Exercise 9.2.

Example 9.4.

For any ring R, $\{0\}$ and R are ideals of R.

Example 9.5.

In $\mathbb Q$ , the ring of integers is a subring but not an ideal. Indeed, $3\in \mathbb Z$ , but $3(1/5)\notin \mathbb Z$ . Thus, $\mathbb Z$ does not have the absorption property.

Actually, this last example is not particularly surprising. Fields do not have interesting ideals, as the following results illustrate.

Theorem 9.2.

Let R be a ring with identity. If an ideal I of R contains a unit, then $$I=R$$ .

Proof.

Let $u\in I$ be a unit. Then by the absorption property, $1=uu^{-1}\in I$ . But then for any $a\in R$ , we also have $a=1a\in I$ . Thus, $$I=R$$ . $\square$

Corollary 9.1.

Let F be a field. Then the only ideals of F are $\{0\}$ and F.

Proof.

Let I be an ideal of F. If $I=\{0\}$ , there is nothing to do, so assume that $0\ne a\in F$ . Then a is a unit, and by the preceding theorem, $$I=F$$ . $\square$

There are a number of ways in which we can obtain new ideals from old ones. For instance, let I and J be ideals of a ring R. Then we write $I+J=\{i+j:i\in I, j\in J\}$ .

Example 9.6.

In $\mathbb Z$ , if we let $I=4\mathbb Z$ and $J=6\mathbb Z$ , then $I+J=2\mathbb Z$ . Indeed, if $m\in I+J$ , then $$m=4a+6b=2(2a+3b)$$ , for some integers a and b. In particular, $I+J\subseteq 2\mathbb Z$ . On the other hand, for any $c\in \mathbb Z$ , $2c=4(-c)+6c\in I+J$ , and hence $2\mathbb Z\subseteq I+J$ . We can say something more general here. See Exercise 9.4.

Theorem 9.3.

If I and J are ideals of a ring R, then so is $$I+J$$ .

Proof.

As I and J are subgroups of the abelian additive group R, and hence normal, we know from Theorem 4.5 that $$I+J$$ is an additive subgroup of R. It remains only to check the absorption property. Take $i\in I$ , $j\in J$ and $r\in R$ . Then $$r(i+j)=ri+rj$$ . Now, $ri\in I$ and $rj\in J$ . Thus, $r(i+j)\in I+J$ . Similarly, $(i+j)r=ir+jr\in I+J$ . $\square$

Also, we can define $IJ=\{i_1j_1+i_2j_2+\cdots +i_nj_n:i_k\in I, j_k\in J, n\in \mathbb N\}$ . (We cannot simply take terms of the form ij with $i\in I$ and $j\in J$ , as sums of terms of that form cannot necessarily be written in the same form.)

Theorem 9.4.

Let I and J be ideals in a ring R. Then IJ is also an ideal.

Proof.

Clearly $0=0\cdot 0\in IJ$ . If we have $i_k\in I$ , $j_k\in J$ , then

$\begin{aligned}&(i_1j_1+\cdots +i_mj_m) -(i_{m+1}j_{m+1}+\cdots +i_nj_n) \\=&\, i_1j_1+\cdots +i_mj_m+(-i_{m+1})j_{m+1}+\cdots +(-i_n)j_n \in IJ.\end{aligned}$

Also, for any $r\in R$ ,

$r(i_1j_1+\cdots +i_mj_m)=(ri_1)j_1+\cdots +(ri_m)j_m\in IJ$

since $ri_1,\ldots , ri_m\in I$ and, similarly,

$(i_1j_1+\cdots +i_mj_m)r=i_1(j_1r)+\cdots +i_m(j_mr)\in IJ.$

$\square$

Notice that I and J are both subsets of $$I+J$$ , as we can take $$i+0$$ and $$0+j$$ for any $i\in I$ , $j\in J$ . But by the absorption property, $IJ\subseteq I\cap J$ .

One type of ideal is particularly important.

Definition 9.2.

Let R be a commutative ring with identity and $a\in R$ . Then the principal ideal generated by a, denoted (a), is the set $\{ra:r\in R\}$ .

Example 9.7.

In $\mathbb Z$ , we have $(n)=n\mathbb Z$ for any $n\in \mathbb Z$ .

Example 9.8.

The ideal from Example 9.2 is (x). Indeed, if $f(x)\in \mathbb R[x]$ , then $$f(0)=0$$ if and only if the constant term is 0; that is, if and only if the polynomial is a multiple of x.

Theorem 9.5.

If R is a commutative ring with identity, and $a\in R$ , then (a) is an ideal of R; indeed, it is the intersection of all ideals of R containing a.

Proof.

We have $0=0a\in (a)$ . If $r, s\in R$ , then $ra-sa=(r-s)a\in (a)$ . Also, if $ra\in (a)$ and $s\in R$ , then $s(ra)=(ra)s=(sr)a\in (a)$ . Furthermore, if I is an ideal of R containing a, then by the absorption property, $ra\in I$ for all $r\in R$ . Thus, (a) is a subset of every ideal containing a. As (a) is an ideal containing a, our result is proved. $\square$

Notice that the preceding proof does not work if R is not a commutative ring with identity. Indeed, if R is not commutative, then $\{ra:r\in R\}$ need not be an ideal; if R does not have an identity, then it may not contain a. See Exercise 9.5.

In a similar fashion, if R is a commutative ring with identity, and $a_1,\ldots , a_n\in R$ , we can construct the ideal generated by $a_1,\ldots , a_n$ , namely, the set of all elements $r_1a_1+\cdots +r_na_n$ , with $r_i\in R$ .

Exercises

9.1.

List all of the elements in each of the following principal ideals of $\mathbb Z_4\oplus \mathbb Z_6$ .

1.

((0, 5))
2.

((2, 3))

9.2.

Let I be the set of all polynomials in $\mathbb Z[x]$ whose constant term is a multiple of 5. Show that I is an ideal of $\mathbb Z[x]$ .

9.3.

Show that the intersection of ideals I and J in a ring R is also an ideal. Extend this to the intersection of an arbitrary collection of ideals.

9.4.

Let m and n be positive integers. Show that $m\mathbb Z+n\mathbb Z=(m, n)\mathbb Z$ .

9.5.

Given a ring R and an element a, let $S=\{ra:r\in R\}$ . Show by example that

1.

S need not contain a, even if R is commutative; and
2.

S need not be an ideal, even if R has an identity.

9.6.

Let R be a commutative ring. If $I=\{a\in R:a^n=0\ \text {for some}\ n\in \mathbb N\}$ , show that I is an ideal of R.

9.7.

Find ideals I and J in a ring R such that $IJ\ne I\cap J$ .

9.8.

Let R be a commutative ring with identity having exactly two ideals. Show that R is a field.

9.9.

Consider the additive group G and subgroup H from Exercise 3.42. Define a multiplication operation on G via $(a_1,a_2,\ldots )(b_1,b_2,\ldots )=(a_1b_1,a_2b_2,\ldots )$ . Show that G is a ring and H is an ideal.

9.10.

In the preceding exercise, show that H is not a principal ideal.

9.2 Factor Rings

Let R be a ring and I an ideal. Then R is an abelian group under addition, and I is a subgroup. Thus, we can consider the left cosets $$a+I$$ , for all $a\in R$ . We use these to form a factor ring. Remember that $$a+I=b+I$$ if and only if $a-b\in I$ .

Definition 9.3.

Let R be a ring and I an ideal of R. Then the factor ring (or quotient ring ), R / I, is the set of all left cosets $\{a+I:a\in R\}$ together with the operations $$(a+I)+(b+I)=a+b+I$$ and $$(a+I)(b+I)=ab+I$$ , for all $a, b\in R$ .

Theorem 9.6.

For any ring R and ideal I, the factor ring R / I is a ring.

Proof.

Since R is an abelian group under addition, I is necessarily a normal subgroup. Thus, we know from Theorem 4.6 that R / I is a group under addition. By Theorem 4.7, it is abelian.

Let us show that the multiplication operation is well-defined. Suppose that $$a_1+I=a_2+I$$

and

. Then notice that

$$a_1b_1-a_2b_2=a_1b_1-a_1b_2+a_1b_2-a_2b_2 =a_1(b_1-b_2)+(a_1-a_2)b_2.$$

Now, $b_1-b_2\in I$ . Thus, by absorption, $a_1(b_1-b_2)\in I$ . Similarly, $a_1-a_2\in I$ , and hence $(a_1-a_2)b_2\in I$ . Thus, $a_1b_1-a_2b_2\in I$ , and therefore $$a_1b_1+I=a_2b_2+I$$

. That is, the multiplication operation is well-defined.

We must check the remaining properties of a ring. Take any $a,b, c\in R$ . Then since $ab+I\in R/I$ , we have closure under multiplication. Also,

$\begin{aligned}(a+I)((b+I)(c+I))&=(a+I)(bc+I) \\ {}&=a(bc)+I\\ {}&=(ab)c+I\\ {}&=(ab+I)(c+I)\\ {}&=((a+I)(b+I))(c+I),\end{aligned}$

and associativity holds. Similarly,

$\begin{aligned}(a+I)((b+I)+(c+I))&=(a+I)(b+c+I)\\ {}&=a(b+c)+I\\ {}&=ab+ac+I\\ {}&=(ab+I)+(ac+I) \\ {}&=(a+I)(b+I)+(a+I)(c+I).\end{aligned}$

The other distributive law is proved in the same fashion. $\square$

Let us discuss a few examples of factor rings.

Example 9.9.

Let $R=\mathbb Z$ and $I=(5)=5\mathbb Z$ . Then $R/I=\{0+I, 1+I, 2+I, 3+I, 4+I\}$ and, for instance, $$(2+I)+(4+I)=6+I=1+I$$ and $$(3+I)(4+I)=12+I=2+I$$ .

Example 9.10.

Let $R=M_2(\mathbb Z)$ and let I be the ideal consisting of all matrices whose entries are even. Then notice that for any $a_{ij}\in \mathbb Z$ , we have

$\begin{pmatrix}a_{11}&{}a_{12}\\ a_{21}&{}a_{22}\end{pmatrix}+I =\begin{pmatrix}b_{11}&{}b_{12}\\ b_{21}&{}b_{22}\end{pmatrix}+I,$

where $b_{ij}$ is 0 if $a_{ij}$ is even and 1 if $a_{ij}$ is odd. Thus, R / I consists of the sixteen different elements $\begin{pmatrix}b_{11}&{}b_{12}\\ b_{21}&{}b_{22}\end{pmatrix}+I$ , $b_{ij}\in \{0,1\}$ . We perform arithmetic in the following fashion:

$\left( \begin{pmatrix}1&{}0\\ 1&{}1\end{pmatrix}+I\right) + \left( \begin{pmatrix}1&{}1\\ 0&{}1\end{pmatrix}+I\right) =\begin{pmatrix}2&{}1\\ 1&{}2\end{pmatrix}+I=\begin{pmatrix}0&{}1\\ 1&{}0\end{pmatrix}+I$

and

$\left( \begin{pmatrix}1&{}0\\ 1&{}1\end{pmatrix}+I\right) \left( \begin{pmatrix}1&{}1\\ 0&{}1\end{pmatrix}+I\right) =\begin{pmatrix}1&{}1\\ 1&{}2\end{pmatrix}+I=\begin{pmatrix}1&{}1\\ 1&{}0\end{pmatrix}+I.$

Example 9.11.

Let $R=\mathbb R[x]$ and $$I=(x^2+3)$$

. Readers familiar with polynomial long division will know that if $f(x)\in R$ , then $$f(x)=(x^2+3)q(x)+r(x)$$

, where q and r are polynomials, with $$r(x)=a+bx$$

, for some $a, b\in \mathbb R$ . (Those who are unfamiliar with polynomial long division can peek ahead to Section 10.1, where it will be discussed in more generality.) Since $(x^2+3)q(x)\in I$ by absorption, we know that elements of R / I are of the form $$a+bx+I$$

, with $a, b\in \mathbb R$ . Addition behaves as expected; for instance, $$(2+3x+I)+(7-4x+I)=9-x+I$$

. To deal with multiplication, observe that $x^2-(-3)\in I$ ; thus, $$x^2+I=-3+I$$

. Therefore, we have calculations such as

$\begin{aligned}(5+4x+I)(-7+2x+I)&=-35-18x+8x^2+I\\ {}&=-35-18x+8(-3)+I \\ {}&=-59-18x+I.\end{aligned}$

Let us also record a few basic facts about factor rings.

Theorem 9.7.

Let R be a ring and I an ideal. Then

1.

if R is commutative, then so is R / I;
2.

if R has an identity, then so does R / I; and
3.

if u is a unit of R, then is a unit of R / I.

Proof.

(1) If $a, b\in R$ , then $$(a+I)(b+I)=ab+I=ba+I=(b+I)(a+I)$$ .

(2) If $a\in R$ , then $$(1+I)(a+I)=a+I=(a+I)(1+I)$$ . Hence, $$1+I$$ is the identity of R / I.

(3) Observe that $(u+I)(u^{-1}+I)=1+I=(u^{-1}+I)(u+I)$ ; thus, $(u+I)^{-1}=u^{-1}+I$ . $\square$

Theorem 9.8.

Let R be a ring with ideals I and J, such that $I\subseteq J$ . Then J / I is an ideal of R / I.

Proof.

We see that I is a subring of J, and since I enjoys the absorption property in R, it enjoys it in J as well. Thus, I is an ideal of J, and so J / I makes sense. Now, $0\in J$ , and therefore $0+I\in J/I$ . If $j_1,j_2\in J$ , then $(j_1+I)-(j_2+I)=(j_1-j_2)+I\in J/I$ . Also, if $j\in J$ and $r\in R$ , then $(r+I)(j+I)=rj+I\in J/I$ , since $rj\in J$ . Similarly, $(j+I)(r+I)=jr+I\in J/I$ . The proof is complete. $\square$

Exercises

9.11.

Let $\mathbb R=\mathbb Z$ and $$I=(5)$$ . Write the addition and multiplication tables for R / I.

9.12.

Let $\mathbb R=\mathbb Z_6\oplus \mathbb Z_4$ and $$I=((4,2))$$ . Write the addition and multiplication tables for R / I.

9.13.

Let $R=\mathbb R[x]$ and $$f(x)=x^3+6x^2+2$$ . If $$I=(f(x))$$ , calculate the product $$(4x^2+3x+1+I)(2x^2-x+2+I)$$ in R / I. Reduce the answer to the form $$ax^2+bx+c+I$$ , for some $a,b, c\in \mathbb R$ .

9.14.

Let R be a ring and I an ideal. Show that R / I is commutative if and only if $ab-ba\in I$ for all $a, b\in R$ .

9.15.

Let I and J be ideals in a ring R. Show that R / I and R / J are both commutative rings if and only if $R/(I\cap J)$ is commutative.

9.16.

Let R be a ring and I a proper ideal.

1.

If R is an integral domain, does it follow that R / I is an integral domain? Prove that it does, or find a counterexample.
2.

If R / I is an integral domain, does it follow that R is an integral domain? Prove that it does, or find a counterexample.

9.17.

If F is a field of order 81, what are the possible orders of F / I, where I is an ideal of F?

9.18.

Let $I_1\subseteq I_2\subseteq I_3\subseteq \cdots$ be ideals of R. Let $I=\bigcup _{n=1}^\infty I_n$ .

1.

Show that I is an ideal of R.
2.

Suppose that R / I is commutative. Show that for every $a, b\in R$ , there exists an $n\in \mathbb N$ such that $ab-ba\in I_n$ .

9.19.

Let R and I be as in Exercise 9.6. Define the analogous ideal for R / I, namely $\{a+I\in R/I:(a+I)^n=0+I\ \text {for some}\ n\in \mathbb N\}$ . Show that this ideal is $\{0+I\}$ .

9.20.

If I is an ideal of a ring R, show that the subrings of R / I are precisely of the form S / I, where S is a subring of R containing I. Further show that S / I is an ideal of R / I if and only if S is an ideal of R.

9.3 Ring Homomorphisms

We recall that a group homomorphism is a function from one group to another that respects the group operation. There is a similar concept for rings, but both of the ring operations must be respected.

Definition 9.4.

Let R and S be rings. Then a ring homomorphism (or, simply, homomorphism) from R to S is a function $\alpha :R\rightarrow S$ satisfying

$\alpha (r_1+r_2)=\alpha (r_1)+\alpha (r_2)$

and

$\alpha (r_1r_2)=\alpha (r_1)\alpha (r_2)$

for all $r_1,r_2\in R$ .

Thus, a ring homomorphism is a homomorphism of additive groups, with the additional property that it respects the multiplication operation. The kernel is the same as the kernel of the additive group homomorphism.

Definition 9.5.

Let $\alpha :R\rightarrow S$ be a ring homomorphism. Then the kernel of $\alpha$ is

$\ker (\alpha )=\{r\in R:\alpha (r)=0\}.$

Example 9.12.

Let $n\ge 2$ be a positive integer. Then $\alpha :\mathbb Z\rightarrow \mathbb Z_n$ given by $\alpha (a)=[a]$ (where we insert the square brackets for clarity) is a homomorphism. Indeed, by Example 4.10, it respects the addition operation. Also, for any $a, b\in \mathbb Z$ , $\alpha (ab)=[ab]=[a][b]=\alpha (a)\alpha (b)$ . Furthermore, by Example 4.10, $\ker (\alpha )=n\mathbb Z$ .

Example 9.13.

Define $\alpha :\mathbb C\rightarrow \mathbb C$ via $\alpha (a+bi)=a-bi$ , for all $a, b\in \mathbb R$ . Then notice that

$\begin{aligned}\alpha ((a+bi)+(c+di))&=\alpha ((a+c)+(b+d)i) \\ {}&=a+c-(b+d)i\\ {}&=(a-bi)+(c-di)\\ {}&=\alpha (a+bi)+\alpha (c+di),\end{aligned}$

for all $a,b,c, d\in \mathbb R$ . Similarly,

$\begin{aligned}\alpha ((a+bi)(c+di))&=\alpha ((ac-bd)+(ad+bc)i)\\ {}&=ac-bd-(ad+bc)i \\ {}&=(a-bi)(c-di)\\ {}&=\alpha (a+bi)\alpha (c+di).\end{aligned}$

Thus, $\alpha$ is a homomorphism. Also, if $\alpha (a+bi)=0$ , then $$a-bi=0$$

, and hence $$a=b=0$$

. Thus, $\ker (\alpha )=\{0\}$ .

Example 9.14.

If R and S are any rings, then $\alpha :R\rightarrow S$ given by $\alpha (r)=0$ for all $r\in R$ is a homomorphism. Indeed, by Example 4.13, it is an additive group homomorphism, and $\alpha (r_1r_2)=0=0\cdot 0=\alpha (r_1)\alpha (r_2)$ , for all $r_1,r_2\in R$ . The kernel of $\alpha$ is R.

Let us record a few basic properties of ring homomorphisms. If $\alpha :R\rightarrow S$ is a ring homomorphism, then as with group homomorphisms, we write $\alpha (M)=\{\alpha (m):m\in M\}$ and $\alpha ^{-1}(N)=\{r\in R:\alpha (r)\in N\}$ , for any subring M of R, and any subring N of S.

Theorem 9.9.

Let $\alpha :R\rightarrow S$ be a ring homomorphism. Then

1.

$\ker (\alpha )$ is an ideal of R;
2.

$\alpha$ is one-to-one if and only if $\ker (\alpha )=\{0\}$ ;
3.

$\alpha (0)=0$ ; and
4.

if R is a ring with identity, then so is $\alpha (R)$ , and $\alpha (1)$ is the identity of $\alpha (R)$ .

Proof.

(1) By Theorem 4.11, $\ker (\alpha )$ is an additive subgroup of R. It remains to check the absorption property. But if $k\in \ker (\alpha )$ and $r\in R$ , then $\alpha (rk)=\alpha (r)\alpha (k)=\alpha (r)0=0$ , and hence $rk\in \ker (\alpha )$ . Similarly, $kr\in \ker (\alpha )$ .

(2) See Theorem 4.11.

(3) This follows immediately from Theorem 4.10.

(4) If $r\in R$ , then $\alpha (1)\alpha (r)=\alpha (1r)=\alpha (r)$ = $\alpha (r1)=\alpha (r)\alpha (1)$ . $\square$

We must be a bit careful with the final part of the last theorem. It certainly does not follow that $\alpha (1)$ is the identity of S, as the following example illustrates.

Example 9.15.

Define $\alpha :\mathbb R\rightarrow M_2(\mathbb R)$ via

$\alpha (r)=\begin{pmatrix}r&{}0\\ 0&{}0\end{pmatrix}$

for all $r\in \mathbb R$ . It is easy to verify that $\alpha$ is a homomorphism. Now, $\alpha (1)$ is the identity of $\alpha (\mathbb R)$ , but not of $M_2(\mathbb R)$ .

Theorem 9.10.

Let $\alpha :R\rightarrow S$ be a ring homomorphism. Let M be a subring of R and N a subring of S. Then

1.

$\alpha (M)$ is a subring of S;
2.

if M is an ideal of R, then $\alpha (M)$ is an ideal of $\alpha (R)$ ;
3.

$\alpha ^{-1}(N)$ is a subring of R; and
4.

if N is an ideal of S, then $\alpha ^{-1}(N)$ is an ideal of R.

Proof.

(1) By Theorem 4.12, $\alpha (M)$ is an additive subgroup of S. If $m_1,m_2\in M$ , then $\alpha (m_1)\alpha (m_2)=\alpha (m_1m_2)\in \alpha (M)$ , since $m_1m_2\in M$ .

(2) By (1), it remains only to check the absorption property. If $m\in M$ , $r\in R$ , then $\alpha (r)\alpha (m)=\alpha (rm)\in \alpha (M)$ , since $rm\in M$ . Similarly, $\alpha (m)\alpha (r)\in \alpha (M)$ .

(3) By Theorem 4.12, $\alpha ^{-1}(N)$ is an additive subgroup of R. If $r_1,r_2\in \alpha ^{-1}(N)$ , then $\alpha (r_1r_2)=\alpha (r_1)\alpha (r_2)\in N$ , since $\alpha (r_1),\alpha (r_2)\in N$ . Thus, $r_1r_2\in \alpha ^{-1}(N)$ .

(4) In view of (3), we only need to check the absorption property. Take $a\in \alpha ^{-1}(N)$ and $r\in R$ . Then $\alpha (ra)=\alpha (r)\alpha (a)\in N$ , since $\alpha (a)\in N$ , and N is an ideal of S. Therefore, $ra\in \alpha ^{-1}(N)$ . Similarly, $ar\in \alpha ^{-1}(N)$ . $\square$

Once again, we note that the second part of the preceding theorem does not say that $\alpha (M)$ is necessarily an ideal of S.

One more homomorphism will prove useful later.

Theorem 9.11.

Let R be a ring with identity of characteristic n. Then there is a homomorphism $\alpha :\mathbb Z\rightarrow R$ with kernel (n).

Proof.

Define $\alpha :\mathbb Z\rightarrow R$ via $\alpha (k)=k1$ , for all $k\in \mathbb Z$ . Let us check that $\alpha$ is a homomorphism. If $j, k\in \mathbb Z$ , then $\alpha (j+k)=(j+k)1=j1+k1=\alpha (j)+\alpha (k)$ . (This is Theorem 3.6, using additive notation.) Also, $\alpha (jk)=(jk)1$ , whereas $\alpha (j)\alpha (k)=(j1)(k1)$ . Again by Theorem 3.6, $$(jk)1=j(k1)$$

. If

, then

$(j1)(k1)=(\underbrace{1+\cdots +1}_{j\ \text {times}})(k1) =\underbrace{k1+\cdots +k1}_{j\ \text {times}}=j(k1).$

, then

and

, and we can use the $$j>0$$

argument. If $$j=0$$

, then

. Thus, in any case, $\alpha (jk)=\alpha (j)\alpha (k)$ , and $\alpha$ is a homomorphism.

We note that $k\in \ker (\alpha )$ if and only if $$k1=0$$ . If $$n=0$$ , then by Theorem 8.13, 1 has infinite additive order. Therefore, the only solution is $$k=0$$ ; thus, $\ker (\alpha )=(0)$ . If $$n>0$$ , then by Theorem 8.13, n is the additive order of 1. Furthermore, by Corollary 3.2, if and only if the additive order of 1 divides k; that is, if and only if n divides k. In other words, the kernel is the set of multiples of n. The proof is complete. $\square$

Exercises

9.21.

Decide if each of the following functions is a ring homomorphism.

1.

$\alpha :\mathbb Z\rightarrow \mathbb R$ , $\alpha (a)=2a$ .
2.

$\alpha :\mathbb R[x]\rightarrow \mathbb R$ , $\alpha (f(x))=f(2)$ .

9.22.

Decide if each of the following functions is a ring homomorphism.

1.

$\alpha :M_2(\mathbb R)\rightarrow M_2(\mathbb R)$ , $\alpha (A)=\begin{pmatrix}1&{}1\\ 0&{}1\end{pmatrix}A\begin{pmatrix}1&{}-1\\ 0&{}1 \end{pmatrix}.$
2.

$\alpha :M_2(\mathbb R)\rightarrow \mathbb R$ , $\alpha (A)=\det (A)$ .

9.23.

Let $\alpha :R\rightarrow S$ and $\beta :S\rightarrow T$ be ring homomorphisms. Show that $\beta \alpha :R\rightarrow T$ is also a ring homomorphism.

9.24.

Let $\alpha :R\rightarrow S$ and $\beta :S\rightarrow T$ be ring homomorphisms. Show that $\ker (\alpha )\subseteq \ker (\beta \alpha )$ . If $\beta$ is one-to-one, show that $\ker (\beta \alpha )=\ker (\alpha )$ .

9.25.

Define $\alpha :\mathbb Z\oplus \mathbb Z\rightarrow \mathbb Z\oplus \mathbb Z$ via $\alpha ((a,b))=(a, 0)$ . Is this a ring homomorphism? If so, find $\ker (\alpha )$ and $\alpha ^{-1}(2\mathbb Z\oplus 3\mathbb Z)$ .

9.26.

Define $\alpha :\mathbb Z_8\rightarrow \mathbb Z_{16}$ via $\alpha ([a])=[a]$ , for all $a\in \{0,1,\ldots , 7\}$ , where the square brackets represent the congruence classes. Is this a ring homomorphism? If so, find $\ker (\alpha )$ and $\alpha ^{-1}([3])$ .

9.27.

Let R be a ring and I an ideal. Show that there exist a ring S and a homomorphism $\alpha :R\rightarrow S$ such that $\ker (\alpha )=I$ .

9.28.

Let R be a commutative ring with prime characteristic p. Show that $\alpha :R\rightarrow R$ given by $\alpha (a)=a^p$ is a ring homomorphism.

9.29.

Let F be a field of order 16 and K a field of order 4. Find all homomorphisms from F to K.

9.30.

Find all homomorphisms from $\mathbb Z$ to $\mathbb Q$ .

9.4 Isomorphisms and Automorphisms

As with groups, we use isomorphisms to establish if two rings have the same structure.

Definition 9.6.

Let R and S be rings. Then a ring isomorphism (or, simply, isomorphism) is a bijective homomorphism from R to S. When such an isomorphism exists, we say that R and S are isomorphic rings.

Example 9.16.

Consider the function $\alpha :\mathbb Z_{15}\rightarrow \mathbb Z_3\oplus \mathbb Z_5$ given by $\alpha (a)=(a, a)$ for all a. By Example 4.16, this is an isomorphism of additive groups. We claim that it is, in fact, a ring isomorphism. All that remains is to show that $\alpha$ respects multiplication. But if $a, b\in \mathbb Z_{15}$ , then $\alpha (ab)=(ab,ab)=(a,a)(b, b)=\alpha (a)\alpha (b)$ . Thus, $\mathbb Z_{15}$ and $\mathbb Z_3\oplus \mathbb Z_5$ are isomorphic rings.

We must not, however, make the mistake of thinking that rings that are isomorphic as additive groups are necessarily isomorphic as rings!

Example 9.17.

Let $R=\mathbb Z$ and $S=5\mathbb Z$ . As additive groups, these are isomorphic, since $5\mathbb Z$ is infinite cyclic, being generated by 5, and we can apply Theorem 4.14. However, as rings, there cannot even be an onto homomorphism from R to S. Why not? If there were, by Theorem 9.9, 1 would have to map to the identity of S, which is sadly lacking. Thus, R and S are not isomorphic rings.

Example 9.18.

Define $\alpha :\mathbb C\rightarrow M_2(\mathbb R)$ via

$\alpha (a+bi)=\begin{pmatrix}a&{}-b\\ b&{}a\end{pmatrix},$

for all $a, b\in \mathbb R$ . Let us check that $\alpha$ is a homomorphism. If $a,b,c, d\in \mathbb R$ , then

$\begin{aligned}\alpha ((a+bi)+(c+di))&=\alpha ((a+c)+(b+d)i) \\&=\begin{pmatrix}a+c&{}-(b+d)\\ b+d&{}a+c\end{pmatrix} \\&=\alpha (a+bi)+\alpha (c+di).\end{aligned}$

Also,

$\alpha ((a+bi)(c+di))=\alpha ((ac-bd)+(ad+bc)i) =\begin{pmatrix}ac-bd&{}-(ad+bc)\\ ad+bc&{}ac-bd\end{pmatrix}$

whereas

$\alpha (a+bi)\alpha (c+di) =\begin{pmatrix}a&{}-b\\ b&{}a\end{pmatrix} \begin{pmatrix}c&{}-d\\ d&{}c\end{pmatrix},$

and these are the same. Clearly, $\ker (\alpha )=\{0\}$ , so $\alpha$ is one-to-one. Now, $\alpha$ is not onto, but we see that $\mathbb C$ is isomorphic to the image $\alpha (\mathbb C)$ , namely

$\left\{ \begin{pmatrix}a&{}-b\\ b&{}a\end{pmatrix}:a, b\in \mathbb R\right\} .$

Let us discuss a few properties of isomorphisms.

Theorem 9.12.

On any collection of rings, isomorphism is an equivalence relation.

Proof.

Reflexivity: The function $\alpha :R\rightarrow R$ given by $\alpha (a)=a$ for all a is an isomorphism. Symmetry: Let $\alpha :R\rightarrow S$ be an isomorphism. By Theorem 4.13, the inverse $\alpha ^{-1}:S\rightarrow R$ is an isomorphism of additive groups. We only need to check that it respects multiplication. Take any $s_1,s_2\in S$ , and suppose that $\alpha ^{-1}(s_i)=r_i$ . Then $\alpha (r_1r_2)=\alpha (r_1)\alpha (r_2)=s_1s_2$ ; that is, $\alpha ^{-1}(s_1s_2)=r_1r_2=\alpha ^{-1}(s_1)\alpha ^{-1}(s_2)$ . Transitivity: Let $\alpha :R\rightarrow S$ and $\beta :S\rightarrow T$ be isomorphisms. By Theorem 4.13, $\beta \circ \alpha :R\rightarrow T$ is an isomorphism of additive groups. Again, we must check that it respects multiplication. Take $r_1,r_2\in R$ . Then

$(\beta \circ \alpha )(r_1r_2)=\beta (\alpha (r_1r_2))=\beta (\alpha (r_1)\alpha (r_2)) =\beta (\alpha (r_1))\beta (\alpha (r_2)).$

The proof is complete. $\square$

Theorem 9.13.

Let $\alpha :R\rightarrow S$ be a ring isomorphism. Then

1.

if R is commutative, then so is S;
2.

if R has an identity, then so does S;
3.

if R is an integral domain, then so is S; and
4.

if R is a field, then so is S.

Proof.

(1) Take $s_1,s_2\in S$ . Then $s_i=\alpha (r_i)$ , for some $r_i\in R$ . Thus,

$s_1s_2=\alpha (r_1)\alpha (r_2)=\alpha (r_1r_2)=\alpha (r_2r_1) =\alpha (r_2)\alpha (r_1)=s_2s_1.$

(2) Use Theorem 9.9 and the fact that $\alpha$ is onto.

(3) By (1) and (2), S is a commutative ring with identity. If $$1=0$$ in S, then $S=\{0\}$ . As $\alpha$ is bijective, $R=\{0\}$ , which is impossible. It remains only to check for zero divisors. Suppose that $$s_1$$ and $$s_2$$ are nonzero elements of S with $$s_1s_2=0$$ . Let us say $\alpha (r_i)=s_i$ . Then $0=s_1s_2=\alpha (r_1)\alpha (r_2) =\alpha (r_1r_2)$ . As $\alpha$ is one-to-one, $$r_1r_2=0$$ . But R is an integral domain, so either $$r_1=0$$ or $$r_2=0$$ , which means that $$s_1=0$$ or $$s_2=0$$ , giving us a contradiction.

(4) Once again, S is a commutative ring with identity and $1\ne 0$ . Suppose that $0\ne s\in S$ . Then $s=\alpha (r)$ , for some $r\in R$ . Now, $r\ne 0$ , so r has an inverse in R. Since $\alpha$ is onto, we know that $\alpha (1)=1$ . Thus,

$1=\alpha (1)=\alpha (rr^{-1})=\alpha (r)\alpha (r^{-1})=s\alpha (r^{-1}).$

That is, $\alpha (r^{-1})=s^{-1}$ , and every nonzero element of S has an inverse. $\square$

Example 9.19.

The rings $2\mathbb Z$ , $M_2(\mathbb R)$ , $\mathbb Z_6$ , $\mathbb Z$ and $\mathbb Q$ are all nonisomorphic. Indeed, $2\mathbb Z$ does not have an identity, so it cannot be isomorphic to any of the others, which do have an identity. Also, all of the rings are commutative except for $M_2(\mathbb R)$ , so it is ruled out. Next, $\mathbb Z_6$ is not an integral domain, but the remaining two are. (Also, it is finite and the others are infinite.) Finally, $\mathbb Q$ is a field but $\mathbb Z$ is not.

Given a ring R, we might well ask if it is a subring of a field. For example, $\mathbb Z$ is a subring of $\mathbb Q$ . But not every ring can be a subring of a field. Indeed, a noncommutative ring or a ring containing zero divisors cannot exist inside a field. So, it seems that integral domains are a good place to start. In fact, if R is an integral domain, then we can construct a field F containing an isomorphic copy of R. The method we will use may seem somewhat familiar; actually, it is exactly the way in which $\mathbb Q$ is constructed from $\mathbb Z$ . We will need something comparable to numerators and denominators. Also, the denominator must not be zero. Furthermore, we need some way of recognizing that $$10/25=8/20$$ , for instance. This gives us an idea of how to proceed.

Let R be an integral domain. Then let S be the Cartesian product $R\times (R\backslash \{0\})$ . Note that S is only a set, not a ring. (If $R=\mathbb Z$ , for instance, when we look at $(10,25)\in S$ , we are thinking of the fraction 10 / 25.) Let us define a relation $\sim$ on S via $(a,b)\sim (c, d)$ if and only if $$ad=bc$$ . (Continuing our parenthetical thought, $(10,25)\sim (8,20)$ .)

We claim that $\sim$ is an equivalence relation. Reflexivity: As R is commutative, we see that $(a,b)\sim (a, b)$ . Symmetry: Suppose that $(a,b)\sim (c, d)$ . Then $$ad=bc$$ and again, by commutativity, $(c,d)\sim (a, b)$ . Transitivity: Suppose that $(a,b)\sim (c, d)$ and $(c,d)\sim (e, f)$ . Then , and hence $$adf=bcf$$ . Also, $$cf=de$$ , and hence $$bcf=bde$$ . Thus, $$adf=bde$$ . Now, $d\ne 0$ and R is an integral domain. Thus, we have $$af=be$$ , and hence $(a,b)\sim (e, f)$ . For the sake of simplicity, write [a, b] for the equivalence class of (a, b). (In our example in $\mathbb Z$ , we have $$[10,25]=[8,20]$$ , and this is the set of all pairs (a, b), with $a, b\in \mathbb Z$ , $b\ne 0$ and $$10b=25a$$ .) Let us write F for the set of all equivalence classes of S.

The addition and multiplication operations on F work precisely as we would expect with fractions. Specifically,

and

We must verify that these operations are well-defined. Suppose that $$[a_1,b_1]=[a, b]$$

and

. Then

$$[a_1,b_1]+[c_1,d_1]=[a_1d_1+b_1c_1,b_1d_1]$$

. But

$\begin{aligned}(ad+bc)(b_1d_1)-(a_1d_1+b_1c_1)bd&=dd_1(ab_1-a_1b)+bb_1(cd_1-c_1d)\\ {}&=dd_1(0)+bb_1(0)\\ {}&=0;\end{aligned}$

thus,

. Similarly, $$[a_1,b_1][c_1,d_1]=[a_1c_1,b_1d_1]$$

, and

$\begin{aligned}acb_1d_1-a_1c_1bd&=(acb_1d_1-a_1cbd_1)+(a_1cbd_1-a_1c_1bd) \\ {}&=cd_1(ab_1-a_1b)+a_1b(cd_1-c_1d)\\ {}&=cd_1(0)+a_1b(0)\\ {}&=0;\end{aligned}$

thus,

. Also, we must note that if b and d are nonzero, then so is bd, since R is an integral domain.

Definition 9.7.

Let R be an integral domain. Then the field of fractions (or field of quotients) of R is the field F constructed above.

Of course, the fact that F is indeed a field needs proving!

Theorem 9.14.

Let R be an integral domain. Then the field of fractions, F, is a field.

Proof.

The proof of this theorem is not difficult. However, there are many steps to complete, as we must verify that F is an abelian group under addition, then that it has all of the remaining properties of a ring, and finally that it is a field. We will prove a few selected properties, and leave the rest to the reader¹ as Exercise 9.38.

Take any $a,b,c,d,e, f\in R$ with b, d and f all nonzero. Let us show that the addition operation on F is associative. We have

$\begin{aligned}([a,b]+[c,d])+[e,f]&=[ad+bc,bd]+[e,f]\\ {}&=[adf+bcf+bde,bdf] \\ {}&=[a,b]+[cf+de,df]\\ {}&=[a,b]+([c,d]+[e, f]),\end{aligned}$

as required.

Next, let us prove a distributive law. Observe that

$$[a,b]([c,d]+[e,f])=[a,b][cf+de,df]=[acf+ade, bdf]$$

whereas

$$[a,b][c,d]+[a,b][e,f]=[ac,bd]+[ae,bf]=[abcf+abde, b^2df].$$

But as

, these are equal.

Notice that [0, 1] is the additive identity of F and [1, 1] is the multiplicative identity. Let us show that every nonzero element of F has an inverse. Take a nonzero $[a, b]\in F$ . Note that $$[0,b]=[0,1]$$ , so $a\ne 0$ . But then $[b, a]\in F$ as well, and $$[a,b][b,a]=[ab, ab]=[1,1]$$ . Thus, $[b,a]=[a, b]^{-1}$ . $\square$

What is the connection between R and F? The idea is that F contains a copy of R and it is, in a sense, the smallest field that does.

Theorem 9.15.

Let R be an integral domain and F its field of fractions. Then F has a subring isomorphic to R. Furthermore, if K is any field having R as a subring, then K has a subfield isomorphic to F, and this subfield contains R.

Proof.

Define $\alpha :R\rightarrow F$ via $\alpha (r)=[r, 1]$ , for all $r\in R$ . We claim that $\alpha$ is a homomorphism. If $r_1,r_2\in R$ , then

$\alpha (r_1+r_2)=[r_1+r_2,1]=[r_1,1]+[r_2,1]=\alpha (r_1)+\alpha (r_2)$

and

$\alpha (r_1r_2)=[r_1r_2,1]=[r_1,1][r_2,1]=\alpha (r_1)\alpha (r_2),$

proving the claim. Furthermore, if $r\in \ker (\alpha )$ , then $$[r, 1]=[0,1]$$

, and therefore $$r=0$$

. Thus, $\alpha$ is one-to-one, and hence R is isomorphic to $\alpha (R)$ which, by Theorem 9.10, is a subring of F.

Also, define $\beta :F\rightarrow K$ via $\beta ([a, b])=ab^{-1}$ , for all $a, b\in R$ with $b\ne 0$ . (Since K is a field, b has an inverse in K. But we still need to check that $\beta$ is well-defined. Suppose that $$[a, b]=[a_1,b_1]$$

. Then

, and hence $ab^{-1}=a_1b_1^{-1}$ .) Now, let us show that $\beta$ is a homomorphism. But

$\begin{aligned}\beta ([a,b]+[c,d])&=\beta ([ad+bc, bd])\\ {}&=(ad+bc)(bd)^{-1}\\ {}&=ab^{-1}+cd^{-1} \\ {}&=\beta ([a,b])+\beta ([c, d]).\end{aligned}$

Similarly,

$\beta ([a,b][c,d])=\beta ([ac, bd])=ac(bd)^{-1}=ab^{-1}cd^{-1} =\beta ([a,b])\beta ([c, d]).$

Thus, $\beta$ is a homomorphism. Now, $\ker (\beta )$ is an ideal of F. By Corollary 9.1, $\ker (\beta )=\{0\}$ or F. But $\beta ([1,1])=1\ne 0$ , and therefore $\beta$ is one-to-one. Thus, K has a subfield $\beta (F)$ , which is isomorphic to F. Also, for any $r\in R$ , we have $r=\beta ([r, 1])\in \beta (F)$ . Thus, R is a subring of the isomorphic copy of the field of fractions of R contained in K. $\square$

Example 9.20.

As we mentioned above, the field of fractions of $\mathbb Z$ is $\mathbb Q$ .

Example 9.21.

Let $R=\{a+b\sqrt{2}:a, b\in \mathbb Z\}$ . Then R is an integral domain, and its field of fractions is isomorphic to $\{a+b\sqrt{2}:a, b\in \mathbb Q\}$ . See Exercise 9.36.

One particular type of isomorphism deserves special mention.

Definition 9.8.

Let R be a ring. Then an automorphism of R is an isomorphism from R to R.

Example 9.22.

For any ring R, the function $\alpha :R\rightarrow R$ given by $\alpha (a)=a$ for all a is an automorphism.

Example 9.23.

The function $\alpha :\mathbb C\rightarrow \mathbb C$ given by $\alpha (a+bi)=a-bi$ for all $a, b\in \mathbb R$ is an automorphism. Indeed, we saw in Example 9.13 that it is a homomorphism. It is immediately obvious that $\alpha$ is one-to-one, and if $a+bi\in \mathbb C$ , then $\alpha (a-bi)=a+bi$ , and therefore $\alpha$ is onto.

We have something similar to an inner automorphism of a group as well.

Theorem 9.16.

Let R be a ring with identity and u a unit of R. Then $\alpha :R\rightarrow R$ , given by $\alpha (a)=u^{-1}au$ for all $a\in R$ , is an automorphism of R.

Proof.

Take $a, b\in R$ . Then

$\alpha (a+b)=u^{-1}(a+b)u=u^{-1}au+u^{-1}bu =\alpha (a)+\alpha (b).$

Also,

$\alpha (ab)=u^{-1}abu=u^{-1}auu^{-1}bu=\alpha (a)\alpha (b).$

Thus, $\alpha$ is a homomorphism. If $\alpha (a)=0$ , then $u^{-1}au=0$ , and therefore $a=uu^{-1}auu^{-1}=u0u^{-1}=0$ . Therefore, $\alpha$ is one-to-one. Finally, take any $a\in R$ . Then $\alpha (uau^{-1})=u^{-1}uau^{-1}u=a$ . Thus, $\alpha$ is onto, and the proof is complete. $\square$

Exercises

9.31.

Explain why the following pairs of rings are not isomorphic.

1.

$\mathbb Z_4\oplus \mathbb Z_4$ and $\mathbb Z_4\oplus \mathbb Z_2\oplus \mathbb Z_2$
2.

$\mathbb Z[x]$ and $2\mathbb Z[x]$

9.32.

Explain why the following pairs of rings are not isomorphic.

1.

$\mathbb R$ and $M_2(\mathbb R)$
2.

$\mathbb R$ and $\mathbb R\oplus \mathbb R$

9.33.

Show that if two rings are isomorphic, then their centres are isomorphic.

9.34.

Let R and S be any rings. Show that $R\oplus S$ is isomorphic to $S\oplus R$ .

9.35.

Let F be a field. Show that F is isomorphic to its field of fractions by constructing an explicit isomorphism.

9.36.

Let $R=\{a+b\sqrt{2}:a, b\in \mathbb Z\}$ . Show that R is an integral domain, and that its field of fractions is isomorphic to $\{a+b\sqrt{2}:a, b\in \mathbb Q\}$ .

9.37.

Let R and S be integral domains. If the fields of fractions of R and S are isomorphic, does it follow that R and S are isomorphic? Prove that it does, or give an explicit counterexample.

9.38.

Complete the proof of Theorem 9.14.

9.39.

Let R be a ring. An involution on R is a function $\alpha :R\rightarrow R$ such that, for all $r_i\in R$ , we have $\alpha (r_1+r_2)=\alpha (r_1)+\alpha (r_2)$ , $\alpha (r_1r_2)=\alpha (r_2)\alpha (r_1)$ and $\alpha (\alpha (r_1))=r_1$ . Show that the following functions $\alpha$ are involutions on $M_2(\mathbb R)$ .

1.

$\alpha \left( \begin{pmatrix}a&{}b\\ c&{}d\end{pmatrix}\right) =\begin{pmatrix}a&{}c\\ b&{}d\end{pmatrix}$ (called the transpose involution)
2.

$\alpha \left( \begin{pmatrix}a&{}b\\ c&{}d\end{pmatrix}\right) =\begin{pmatrix}d&{}-b\\ -c&{}a\end{pmatrix}$ (called the symplectic involution)

9.40.

Let R be a ring. Using the definition of an involution from the preceding question,

1.

determine under what circumstances an involution on R is an automorphism; and
2.

show that the composition of two involutions on R is an automorphism.

9.5 Isomorphism Theorems for Rings

We recall that the three isomorphism theorems for groups were presented in Section 4.5. Let us now state the analogues for rings. The first is certainly the most important.

Theorem 9.17

(First Isomorphism Theorem for Rings). Let $\alpha :R\rightarrow S$ be a ring homomorphism. Then $R/\ker (\alpha )$ is isomorphic to $\alpha (R)$ .

Proof.

Let $K=\ker (\alpha )$ . Define $\beta :R/K\rightarrow \alpha (R)$ via $\beta (a+K)=\alpha (a)$ . From the proof of Theorem 4.18, we see that $\beta$ is an isomorphism of additive groups. Thus, it remains only to check that $\beta$ respects multiplication. Take any $a, b\in R$ . Then

$\beta ((a+K)(b+K))=\beta (ab+K)=\alpha (ab)=\alpha (a)\alpha (b)=\beta (a+K)\beta (b+K),$

as required. $\square$

Whenever we are asked to show that a ring modulo an ideal is isomorphic to some other ring, it is usually a good indication that we should employ the First Isomorphism Theorem.

Example 9.24.

We already know that for any $n\ge 2$ , the additive groups $\mathbb Z/n\mathbb Z$ and $\mathbb Z_n$ are isomorphic. Indeed, in Example 4.21, we showed that $\alpha :\mathbb Z\rightarrow \mathbb Z_n$ , given by $\alpha (a)=[a]$ , is an onto group homomorphism with kernel $n\mathbb Z$ . But, in fact, we also have $\alpha (ab)=[ab]=[a][b]=\alpha (a)\alpha (b)$ , for all $a, b\in \mathbb Z$ . Thus, $\alpha$ is actually an onto ring homomorphism, and we now know that $\mathbb Z/n\mathbb Z$ and $\mathbb Z_n$ are isomorphic rings.

Example 9.25.

Let us show that $\mathbb R[x]/(x)$ is isomorphic to $\mathbb R$ . To this end, let us define $\alpha :\mathbb R[x]\rightarrow \mathbb R$ via $\alpha (f(x))=f(0)$ . Now, if $f(x), g(x)\in \mathbb R[x]$ , then

$\alpha (f(x)+g(x))=f(0)+g(0)=\alpha (f(x))+\alpha (g(x)).$

Furthermore,

$\alpha (f(x)g(x))=f(0)g(0)=\alpha (f(x))\alpha (g(x)).$

Thus, $\alpha$ is a homomorphism. Also, if $r\in \mathbb R$ , then simply regarding r as a constant polynomial, we have $\alpha (r)=r$ ; hence, $\alpha$ is onto. The kernel of $\alpha$ is the set of all polynomials f(x) satisfying $$f(0)=0$$

. But f(0) is the constant term of the polynomial. Thus, $\ker (\alpha )$ is the set of all polynomials with zero constant term, that is, the set of all polynomials that are multiples of x. Now apply Theorem 9.17.

A couple of rather interesting consequences follow.

Corollary 9.2.

Let R be a ring with identity of characteristic n. If $$n=0$$ , then R has a subring isomorphic to $\mathbb Z$ . If $n\ge 2$ , then R has a subring isomorphic to $\mathbb Z_n$ .

Proof.

By Theorem 9.11, there is a homomorphism $\alpha :\mathbb Z\rightarrow R$ with kernel $n\mathbb Z$ . Now, Theorem 9.10 says that $\alpha (\mathbb Z)$ is a subring of R, and Theorem 9.17 tells us that this subring is isomorphic to $\mathbb Z/n\mathbb Z$ . If $$n=0$$ , then $n\mathbb Z=\{0\}$ , and there is nothing to do. Otherwise, we use Example 9.24. $\square$

Corollary 9.3.

Let F be a field. If F has characteristic 0, then F has a subfield isomorphic to $\mathbb Q$ . If F has prime characteristic p, then F has a subfield isomorphic to $\mathbb Z_p$ .

Proof.

If char $$F=p>0$$ , then we use the preceding corollary. If char $$F=0$$ , then we note that F has a subring isomorphic to $\mathbb Z$ . By Theorem 9.15, F also has a subfield isomorphic to the field of fractions of $\mathbb Z$ , namely, $\mathbb Q$ . $\square$

The subfield discussed in Corollary 9.3 (either $\mathbb Q$ or $\mathbb Z_p$ ) is the smallest subfield of F, and it is called the prime subfield .

Theorem 9.18

(Second Isomorphism Theorem for Rings). Let R be a ring with ideals I and J. Then $I/(I\cap J)$ is isomorphic to $$(I+J)/J$$ .

Proof.

Define $\alpha :I\rightarrow (I+J)/J$ via $\alpha (i)=i+J$ , for all $i\in I$ . Consulting the proof of Theorem 4.19, we see that $\alpha$ is an onto homomorphism of additive groups with kernel $I\cap J$ . In view of the First Isomorphism Theorem for Rings, it suffices to show that $\alpha$ respects multiplication. But for any $i_1,i_2\in I$ , we have $\alpha (i_1i_2)=i_1i_2+J=(i_1+J)(i_2+J)=\alpha (i_1)\alpha (i_2)$ . The proof is complete. $\square$

Example 9.26.

Let $R=\mathbb Z$ , $$I=(4)$$ and $$J=(6)$$ . Then the preceding theorem tells us that $(4)/((4)\cap (6))$ is isomorphic to $$((4)+(6))/(6)$$ . That is, (4) / (12) is isomorphic to (2) / (6).

Theorem 9.19

(Third Isomorphism Theorem for Rings). Let R be a ring, and let I and J be ideals of R with $I\subseteq J$ . Then (R / I) / (J / I) is isomorphic to R / J.

Proof.

Define $\alpha :R/I\rightarrow R/J$ via $\alpha (a+I)=a+J$ , for all $a\in R$ . The proof of Theorem 4.20 shows us that $\alpha$ is an onto additive group homomorphism with kernel J / I. It remains only to show that $\alpha$ respects multiplication, for then we can apply Theorem 9.17. Take $a, b\in R$ . Then

$\alpha ((a+I)(b+I))=\alpha (ab+I)=ab+J=(a+J)(b+J)=\alpha (a+I)\alpha (b+I).$

We are done. $\square$

Exercises

9.41.

Let R and S be rings. Show that $(R\oplus S)/(R\oplus \{0\})$ is isomorphic to S.

9.42.

Let m and n be positive integers, both greater than 1. Show that the rings $(\mathbb Z\oplus \mathbb Z)/((m)\oplus (n))$ and $\mathbb Z_m\oplus \mathbb Z_n$ are isomorphic.

9.43.

Let I be the set of all polynomials $f(x)\in \mathbb Z[x]$ such that the constant term of f(x) is a multiple of 5. Show that $\mathbb Z[x]/I$ is isomorphic to $\mathbb Z_5$ .

9.44.

Let I be the set of all matrices in $M_2(\mathbb Z)$ in which every entry is even. Show that $M_2(\mathbb Z)/I$ is isomorphic to $M_2(\mathbb Z_2)$ .

9.45.

Show that the rings $(3\mathbb Z/60\mathbb Z)/(12\mathbb Z/60\mathbb Z)$ and $3\mathbb Z/12\mathbb Z$ are isomorphic. Then show that both are isomorphic to $\mathbb Z_4$ .

9.46.

Let I and J be ideals in a ring R such that $$I+J=R$$ . Show that $R/(I\cap J)$ is isomorphic to $(R/I)\oplus (R/J)$ .

9.6 Prime and Maximal Ideals

We conclude this chapter by discussing two special sorts of ideals.

Definition 9.9.

Let R be a ring. An ideal M of R is said to be maximal if

1.

$M\ne R$ ; and
2.

if I is an ideal of R containing M, then or .

Example 9.27.

Let $R=\mathbb Z$ and let n be a nonnegative integer. Then we claim that (n) is a maximal ideal of R if and only if n is prime. Indeed, (0) is certainly not maximal, as $(0)\subsetneq (2)\subsetneq R$ . Also, (1) is not maximal, since $$(1)=R$$ . If n is composite, say $$n=kl$$ , with $$1<k, l<n$$ , then we note that $(n)\subsetneq (k)\subsetneq R$ , so (n) is not maximal. Finally, let n be prime. Suppose that I is an ideal of R with $(n)\subsetneq I\subsetneq R$ . Take $a\in I\backslash (n)$ . Since a is not divisible by n, and n is prime, we know that $$(a, n)=1$$ . Thus, by Corollary 2.1, we can find integers u and v such that $$au+nv=1$$ . But as $a, n\in I$ , this implies that $1\in I$ , hence $$I=R$$ , giving us a contradiction. (As we shall see shortly, there is another way to prove this.)

Example 9.28.

In any field, the ideal $\{0\}$ is maximal! Remember, by Corollary 9.1, a field only has two ideals.

In a commutative ring with identity, there is a nice test for maximality of ideals.

Theorem 9.20.

Let R be a commutative ring with identity, and M an ideal of R. Then M is maximal if and only if R / M is a field.

Proof.

Suppose that M is a maximal ideal. By Theorem 9.7, R / M is a commutative ring with identity. Furthermore, as $M\ne R$ , we know that R / M consists of more than one additive left coset. But the only ring in which $$0=1$$

is the ring consisting only of zero; thus, $0+M\ne 1+M$ . It remains to show that every nonzero element of R / M has an inverse. Let $a+M\ne 0+M$ . Now, define $I=\{m+ra:m\in M, r\in R\}$ . We claim that I is an ideal of R. Taking $$r=0$$

, we note that $m\in I$ for all $m\in M$ ; thus, $M\subseteq I$ and, in particular, $0\in I$ . If $m_i\in M$ , $r_i\in R$ , then

$(m_1+r_1a)-(m_2+r_2a)=(m_1-m_2)+(r_1-r_2)a\in I.$

Also, for any $s\in R$ , $$s(m_1+r_1a)=sm_1+sr_1a$$

. As $sm_1\in M$ and $sr_1\in R$ , we see that I has the absorption property and is, therefore, an ideal. But we noted above that $M\subseteq I$ . Furthermore, $a=0+1a\in I\backslash M$ . By the maximality of M, we have $$I=R$$

. In particular, $1\in I$ , so there exist $m\in M$ and $r\in R$ such that $$m+ra=1$$

. But then $$(r+M)(a+M)=1-m+M=1+M$$

, since $m\in M$ . That is, $$r+M$$

is the inverse of $$a+M$$

, and R / M is a field.

Conversely, suppose that R / M is a field. We must show that M is maximal. If $$M=R$$ , then R / M consists only of a single additive left coset, contradicting the fact that a field must have a distinct 0 and 1. Thus, $M\ne R$ . Suppose that I is an ideal of R with $M\subsetneq I\subsetneq R$ . Take $a\in I\backslash M$ . Now, $a+M\ne 0+M$ , so $$a+M$$ has an inverse, say $$b+M$$ . Then $$(a+M)(b+M)=1+M$$ ; in other words, $1-ab\in M\subseteq I$ . But also $a\in I$ , which means that $ab\in I$ by absorption, and therefore $1=(1-ab)+ab\in I$ . By Theorem 9.2, $$I=R$$ , giving us a contradiction and completing the proof. $\square$

Example 9.29.

This gives us another way to deal with Example 9.27. If $$n=0$$ , then we note that $\mathbb Z/\{0\}$ is simply $\mathbb Z$ , which is not a field. Thus, (0) is not maximal. If $$n=1$$ , then observe that $\mathbb Z/\mathbb Z$ is the ring with one element, which is not a field; hence, (1) is not maximal. For any $n\ge 2$ , we see from Example 9.24 that $\mathbb Z/n\mathbb Z$ is isomorphic to $\mathbb Z_n$ . But by Theorem 8.11, $\mathbb Z_n$ is a field if and only if n is prime. Thus, (n) is maximal if and only if n is prime.

Example 9.30.

By Example 9.25, $\mathbb R[x]/(x)$ is isomorphic to $\mathbb R$ , which is a field. Thus, (x) is a maximal ideal of $\mathbb R[x]$ .

Example 9.31.

In the same manner as Example 9.30, we see that $\mathbb Z[x]/(x)$ is isomorphic to $\mathbb Z$ . But $\mathbb Z$ is not a field, and hence (x) is not maximal. In fact, we can see this by noting that (x) is properly contained in the ideal M consisting of those polynomials whose constant terms are multiples of 5. We can use Theorem 9.20 to show that M is maximal. See Exercise 9.43.

It is worth mentioning that Theorem 9.20 only applies when R is a commutative ring with identity. For instance, the ideal containing only the zero matrix is maximal in $M_2(\mathbb R)$ ! See Exercise 9.54.

Definition 9.10.

Let R be a commutative ring and P an ideal of R. Then we say that P is a prime ideal² if

1.

$P\ne R$ ; and
2.

if $a, b\in R$ and $ab\in P$ , then either $a\in P$ or $b\in P$ .

Example 9.32.

In any integral domain, $\{0\}$ is a prime ideal. If $$ab=0$$ , then $$a=0$$ or $$b=0$$ .

Example 9.33.

Let us consider $R=\mathbb Z$ . By the preceding example, $\{0\}$ is prime, so we know immediately that maximal and prime are not the same thing. Of course, $$(1)=R$$ , so (1) is not prime. Suppose that $n\ge 2$ . If n is composite, say $$n=kl$$ with $$1<k, l<n$$ , we see that $kl\in (n)$ but neither k nor l lies in (n). Thus, (n) is not prime. But if n is prime, then (n) is a prime ideal. Indeed, if $ab\in (n)$ , then n|ab. Thus, by Theorem 2.7, n|a or n|b, and hence a or b is in (n).

Once again, there is another way to handle this last example.

Theorem 9.21.

Let R be a commutative ring with identity and P an ideal. Then P is prime if and only if R / P is an integral domain.

Proof.

Suppose that P is prime. Since R is a commutative ring with identity, so is R / P, by Theorem 9.7. Also, as $P\ne R$ , R / P has more than one element, and therefore $0+P\ne 1+P$ . Thus, it remains to show that R / P has no zero divisors. Suppose that $$(a+P)(b+P)=0+P$$ . Then $ab\in P$ , and hence $a\in P$ or $b\in P$ . That is, $$a+P=0+P$$ or $$b+P=0+P$$ , and R / P is an integral domain.

Conversely, let R / P be an integral domain. As R / P cannot be the ring with one element, $P\ne R$ . Suppose that $ab\in P$ . Then $$(a+P)(b+P)=0+P$$ . Since R / P has no zero divisors, $$a+P=0+P$$ or $$b+P=0+P$$ . That is, $a\in P$ or $b\in P$ , and P is prime. $\square$

Example 9.34.

Let us look at Example 9.33 again. We know that $\mathbb Z/(0)$ is just $\mathbb Z$ , which is an integral domain, and hence (0) is a prime ideal. If $n\ge 2$ , then $\mathbb Z/(n)$ is isomorphic to $\mathbb Z_n$ , by Example 9.24, and Theorem 8.11 tells us that this is an integral domain if and only if n is prime. Thus, for a nonnegative integer n, (n) is a prime ideal of $\mathbb Z$ if and only if n is 0 or prime.

Example 9.35.

Refer to Example 9.31. We see that (x) is a prime ideal in $\mathbb Z[x]$ , since $\mathbb Z[x]/(x)$ is isomorphic to $\mathbb Z$ .

Example 9.36.

Naturally, (x) is prime in $\mathbb R[x]$ , because we saw in Example 9.30 that $\mathbb R[x]/(x)$ is isomorphic to $\mathbb R$ , which is a field, hence an integral domain.

Of course, this last example can be generalized.

Theorem 9.22.

Let R be a commutative ring with identity. Then every maximal ideal of R is also a prime ideal.

Proof.

Use the last two theorems and the fact that every field is an integral domain. $\square$

As we have already seen, not every prime ideal is maximal. Also, this last theorem only applies to commutative rings with identity. In some commutative rings without an identity, it is possible to find maximal ideals that are not prime, and Exercise 9.51 asks for an example of this phenomenon.

Exercises

9.47.

Let R be the ring from Exercise 8.17. Is the ideal (2) prime? Is it maximal?

9.48.

Find all prime ideals in each of the following rings.

1.

$\mathbb Z_{10}$
2.

$\mathbb Z_{50}$

9.49.

Let R be a finite commutative ring with identity. Show that every prime ideal of R is maximal.

9.50.

Find every maximal ideal of $\mathbb Z_7\oplus \mathbb Z_7$ .

9.51.

Find an example of a commutative ring having an ideal that is maximal but not prime.

9.52.

Suppose that R is a commutative ring with identity in which the elements of R that are not units form an ideal. Show that this ideal is the unique maximal ideal of R.

9.53.

Show that every field has the property described in the preceding exercise. Also show that $\mathbb Z_{p^n}$ has this property, for every prime p and positive integer n.

9.54.

Show that the ideal containing only the zero matrix is maximal in $M_2(\mathbb R)$ .

9.55.

Let R be a commutative ring with identity having a prime ideal I. Find a prime ideal in $R\oplus R$ .

9.56.

Let $R\ne \{0\}$ be a commutative ring with identity. Suppose that every proper ideal of R is prime. Show that R is an integral domain, and then use this information to show that R is, in fact, a field.

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