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Maximal and Prime Ideals

In this particular section we are especially interested in certain ideals of commutative rings. These ideals give us special types of factor rings.

Maximal and Prime Ideals

In this particular section we are especially interested in certain ideals of commutative rings. These ideals give us special types of factor rings. More specifically, we would like to characterize those ideals \(I\) of a commutative ring \(R\) such that \(R/I\) is an integral domain or a field.

A proper ideal \(M\) of a ring \(R\) is a maximal ideal of \(R\) if the ideal \(M\) is not a proper subset of any ideal of \(R\) except \(R\) itself. That is, \(M\) is a maximal ideal if for any ideal \(I\) properly containing \(M\), \(I = R\). The following theorem completely characterizes maximal ideals for commutative rings with identity in terms of their corresponding factor rings.

Theorem

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

Proof

Let \(M\) be a maximal ideal in \(R\). If \(R\) is a commutative ring, then \(R/M\) must also be a commutative ring. Clearly, \(1 + M\) acts as an identity for \(R/M\). We must also show that every nonzero element in \(R/M\) has an inverse. If \(a + M\) is a nonzero element in \(R/M\), then \(a \notin M\). Define \(I\) to be the set \(\{ ra + m : r \in R \text{ and } m \in M \}\). We will show that \(I\) is an ideal in \(R\). The set \(I\) is nonempty since \(0a+0=0\) is in \(I\). If \(r_1 a + m_1\) and \(r_2 a + m_2\) are two elements in \(I\), then \[\begin{aligned}\end{aligned}\] is in \(I\). Also, for any \(r \in R\) it is true that \(rI \subset I\); hence, \(I\) is closed under multiplication and satisfies the necessary conditions to be an ideal. Therefore, by and the definition of an ideal, \(I\) is an ideal properly containing \(M\). Since \(M\) is a maximal ideal, \(I=R\); consequently, by the definition of \(I\) there must be an \(m\) in \(M\) and an element \(b\) in \(R\) such that \(1=ab+m\). Therefore, \[\begin{aligned}\end{aligned}\].

Conversely, suppose that \(M\) is an ideal and \(R/M\) is a field. Since \(R/M\) is a field, it must contain at least two elements: \(0 + M = M\) and \(1 + M\). Hence, \(M\) is a proper ideal of \(R\). Let \(I\) be any ideal properly containing \(M\). We need to show that \(I = R\). Choose \(a\) in \(I\) but not in \(M\). Since \(a+ M\) is a nonzero element in a field, there exists an element \(b +M\) in \(R/M\) such that \((a+M)(b+M) = ab + M = 1+M\). Consequently, there exists an element \(m \in M\) such that \(ab + m = 1\) and \(1\) is in \(I\). Therefore, \(r1 =r \in I\) for all \(r \in R\). Consequently, \(I = R\).

Example

Let \(p{\mathbb Z}\) be an ideal in \({\mathbb Z}\), where \(p\) is prime. Then \(p{\mathbb Z}\) is a maximal ideal since \({\mathbb Z}/ p {\mathbb Z} \cong {\mathbb Z}_p\) is a field.

A proper ideal \(P\) in a commutative ring \(R\) is called a prime ideal if whenever \(ab \in P\), then either \(a \in P\) or \(b \in P\). It is possible to define prime ideals in a noncommutative ring. See [1] or [3].

Example

It is easy to check that the set \(P = \{ 0, 2, 4, 6, 8, 10 \}\) is an ideal in \({\mathbb Z}_{12}\). This ideal is prime. In fact, it is a maximal ideal.

Proposition

Let \(R\) be a commutative ring with identity \(1\), where \(1 \neq 0\). Then \(P\) is a prime ideal in \(R\) if and only if \(R/P\) is an integral domain.

Proof

First let us assume that \(P\) is an ideal in \(R\) and \(R/P\) is an integral domain. Suppose that \(ab \in P\). If \(a + P\) and \(b + P\) are two elements of \(R/P\) such that \((a + P)(b + P) = 0 + P = P\), then either \(a + P = P\) or \(b + P = P\). This means that either \(a\) is in \(P\) or \(b\) is in \(P\), which shows that \(P\) must be prime.

Conversely, suppose that \(P\) is prime and \[\begin{aligned}\end{aligned}\]. Then \(ab \in P\). If \(a \notin P\), then \(b\) must be in \(P\) by the definition of a prime ideal; hence, \(b + P = 0 + P\) and \(R/P\) is an integral domain.

Example

Every ideal in \({\mathbb Z}\) is of the form \(n {\mathbb Z}\). The factor ring \({\mathbb Z} / n{\mathbb Z} \cong {\mathbb Z}_n\) is an integral domain only when \(n\) is prime. It is actually a field. Hence, the nonzero prime ideals in \({\mathbb Z}\) are the ideals \(p{\mathbb Z}\), where \(p\) is prime. This example really justifies the use of the word prime in our definition of prime ideals.

Since every field is an integral domain, we have the following corollary.

Historical Note

Amalie Emmy Noether, one of the outstanding mathematicians of the twentieth century, was born in Erlangen, Germany in 1882. She was the daughter of Max Noether (18441921), a distinguished mathematician at the University of Erlangen. Together with Paul Gordon (18371912), Emmy Noether's father strongly influenced her early education. She entered the University of Erlangen at the age of 18. Although women had been admitted to universities in England, France, and Italy for decades, there was great resistance to their presence at universities in Germany. Noether was one of only two women among the university's 986 students. After completing her doctorate under Gordon in 1907, she continued to do research at Erlangen, occasionally lecturing when her father was ill.

Noether went to Göttingen to study in 1916. David Hilbert and Felix Klein tried unsuccessfully to secure her an appointment at Göttingen. Some of the faculty objected to women lecturers, saying, What will our soldiers think when they return to the university and are expected to learn at the feet of a woman? Hilbert, annoyed at the question, responded, Meine Herren, I do not see that the sex of a candidate is an argument against her admission as a Privatdozent. After all, the Senate is not a bathhouse. At the end of World War I, attitudes changed and conditions greatly improved for women. After Noether passed her habilitation examination in 1919, she was given a title and was paid a small sum for her lectures.

In 1922, Noether became a Privatdozent at Göttingen. Over the next 11 years she used axiomatic methods to develop an abstract theory of rings and ideals. Though she was not good at lecturing, Noether was an inspiring teacher. One of her many students was B. L. van der Waerden, author of the first text treating abstract algebra from a modern point of view. Some of the other mathematicians Noether influenced or closely worked with were Alexandroff, Artin, Brauer, Courant, Hasse, Hopf, Pontryagin, von Neumann, and Weyl. One of the high points of her career was an invitation to address the International Congress of Mathematicians in Zurich in 1932. In spite of all the recognition she received from her colleagues, Noether's abilities were never recognized as they should have been during her lifetime. She was never promoted to full professor by the Prussian academic bureaucracy.

Condensed — the full section is in Judson, Abstract Algebra: Theory and Applications.

Symbols used here

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

\neq

not equal
The two sides are different.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

\blacksquare\ \text{or}\ \square

end of proof (halmos)
Marks the point where the statement has been established.

a \equiv b \pmod n

congruent modulo n
n divides a − b; a and b have the same remainder.

a \mid b,\ \gcd(a,b)

divides, greatest common divisor
b is a multiple of a; the largest number dividing both.

(G, \cdot),\ e,\ g^{-1}

group, identity, inverse
A set with an operation; the do-nothing element; the element that undoes g.

G \cong H,\ G / N

isomorphic, quotient group
Same structure; the group of cosets of a normal subgroup N.

\mathbb{Z}/n\mathbb{Z},\ \mathbb{Z}_n

integers modulo n
The remainders 0…n−1 with clock arithmetic.

\operatorname{Hom}(A, B),\ f \circ g

arrows from A to B, composition
The set of morphisms; do g then f.

Questions people ask

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group — that is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

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Parts of this page are adapted from Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Condensed and re-explained here; errors are ours.

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