Page 197 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes

Note x R has a multiplicative inverse iff Rx = R.
Conversely, assume that Rp is a prime ideal. Then Rp  R, Thus, 1  Rp, and hence, p does not
have a multiplicative inverse. Now suppose p divides ab, where a, b  R. Then ab = rp far some
r  R, i.e., ab  Rp.

As Rp is a prime ideal, either a  Rp or b  Rp. Hence, either p | a or p | b. Thus, p is a prime
element in R.
Theorem 2 is very useful for checking whether an element is a prime element or not, or for
finding out when a principal ideal is a prime ideal.
Prime ideals have several useful properties.
Now consider the ideal 22 in Z. Suppose the ideal nZ in Z is such that 2Z  nZ Z. Then n | 2.
 n=  1or n = 2.  nZ = Z or nZ = 2Z.
This shows that no ideal can lie between 2Z and Z. That is, 22 is maximal among the proper
ideals of Z that contain it. So we say that it is a “maximal ideal”. Let us define this expression.
Definition: A proper ideal M of a ring R is called a maximal ideal if whenever I is an ideal of R
such that M  I  R, then either I = M or I = R.
Thus, a proper ideal M is a maximal ideal if there is no proper ideal of R which contains it. An
example that comes to mind immediately is the zero ideal in any field F. This is maximal
because you know that the only other ideal of F is F itself.
To generate more examples of maximal ideals, we can use the following characterisation of such
ideals.
Theorem 3: Let R be a ring with identity. An ideal M in R is maximal if and only if R/M is a field.
Proof: Let us first assume that M is a maximal ideal of R. We want to prove that R/M is a field.
For this, it is enough to prove that R/M has no non-zero proper ideals. So, let I be an ideal of
R/M. Consider the canonical homomorphism  : R  R/M :  (r) = r + M. Then, you know that
 (I) is an ideal of R containing M, the kernel of . Since M is a maximal ideal of R.  (I) = M or
1
-1
 (I) = R. Therefore, I = ( (I)) is either (M) or (R). That is, I = {0} or I = R/M, where; = O +M
-1
-1
= M. Thus, R/M is a field.
Conversely, let M be an ideal of R such that R/M is a field. Then the only ideals of R/M are
{0} and R/M. Let I be an ideal of R containing M. Then, as above (1) = {0} or, (I) = R/M.
 I =  ((1)) is M or R. Therefore, M is a maximal ideal of R.
-1
Corollary: Every maximal ideal of a ring with identity is a prime ideal.
Now, the corollary is a one-way statement. What about the converse? That is, is every prime
ideal maximal? What about the zero ideal in Z? Since Z is a domain but not a field and Z  Z/{0},
Z/{0} is a domain but not a field. Thus. (0) is a prime ideal but not a maximal ideal of Z.

Example: Show that an ideal mZ of Z is maximal iff m is a prime number.

Solution: You know that Z,,, is a field iff m is a prime number. You also know that Z/mZ = Z .
m
Z | mZ is a field iff m is prime. Hence, mZ is maximal in Z iff m is a prime number.

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