Definition 3.2.1.
Let \(R\) be commutative with identity and \(P\subsetneq R\) a nonzero ideal. We say \(P\) is prime if whenever \(a,b\in R\) such that \(ab\in P\text{,}\) we have \(a\in P\) or \(b\in P\text{.}\)
Section 3.2 Prime and Maximal Ideals
In this section, we continue our exploration of quotient rings by looking more closely at properties of ideals. We focus on particular properties of ideals that ensure that the quotient \(R/I\) is either a domain or a field.
Theorem 3.2.2.
Let \(R\) be a domain and \(p\in R\) be prime. Then \(\ideal{p}\) is a prime ideal.
Activity 3.2.3.
Which of the following ideals are prime?
- \(\ideal{9}\) in \(\Z\)
- \(\ideal{11}\) in \(\Z\)
- \(\ideal{x^2+1}\) in \(\R[x]\)
- \(\ideal{x^2-1}\) in \(\R[x]\)
- \(\ideal{x^2-5x+6, x^4+2x^3-10x^2+5x-2}\) in \(\R[x]\)
It is this precise condition that guarantees that the resulting quotient is a domain.
Theorem 3.2.4.
Let \(R\) be commutative with identity and \(I\) an ideal of \(R\text{.}\) Then \(I\) is prime if and only if \(R/I\) is an integral domain.
We now consider another important class of ideals: the maximal ideals.
Definition 3.2.5.
Let \(R\) be commutative with identity and let \(M\subsetneq R\) be a nonzero ideal. We say that \(M\) is a maximal ideal if no proper ideal of \(R\) properly contains \(M\text{.}\) That is, if \(J\) is an ideal satisfying \(M\subseteq J\subseteq R\text{,}\) either \(J=M\) or \(J=R\text{.}\)
In other words, an ideal \(M\ne R\) is maximal if no “larger” ideal (with respect to inclusion) properly contains it. As we will see later, rings can have many maximal ideals.
It is a fact that any ring \(R\) with \(0_R\ne 1_R\) has a maximal ideal. This follows from Zorn’s Lemma; a rigorous exploration of Zorn’s Lemma lies outside of the scope of this text, but suffice it to say that Zorn’s Lemma is incredibly useful in all areas of algebra for proving existence theorems. For example, a proof that every vector space has a basis relies on Zorn’s Lemma.
Rings with only one maximal ideal are said to be local rings, and are actively studied in modern research in commutative algebra (the study of commutative rings and their properties).
The next two results demonstrate that the maximality of \(I\) is precisely the condition that guarantees that \(R/I\) is a field.
Lemma 3.2.6.
Let \(R\) be commutative with identity and \(M\) a maximal ideal of \(R\text{.}\) Let \(x\in R\setminus M\text{,}\) and set \(J = \setof{xr+y}{r\in R, \ y\in M}\text{.}\) Then: \(J\) is an ideal of \(R\text{;}\) \(M\subsetneq J\text{;}\) and thus there exist \(r'\in R\text{,}\) \(y'\in M\) such that \(1 = xr'+y'\text{.}\)
Theorem 3.2.7.
Let \(R\) be commutative with identity and \(I\) an ideal of \(R\text{.}\) Then \(I\) is maximal if and only if \(R/I\) is a field.
Hint.
For the forward direction, apply the previous lemma to construct an inverse for \(x+I\) given any \(x\in R\setminus I\text{.}\)
Theorem 3.2.8.
Every maximal ideal is prime.
In general, the converse is not true (see the Challenge below). However, it holds in sufficiently nice rings.
Theorem 3.2.9.
In a principal ideal domain, every prime ideal is maximal.
Exploration 3.2.10.
Describe the prime and maximal ideals of \(\Z\) and \(\Q[x]\text{.}\)
Hint.
For which ideals \(I\) is \(\Z/I\) a domain? A field? Similarly for \(\Q[x]\text{.}\) Or, use Theorem 3.2.9.
Challenge.
Find a commutative ring with identity, \(R\text{,}\) and a nonmaximal prime ideal \(P\) of \(R\text{.}\)
