“I have as much chance of becoming Prime Minister as of being decapitated by a frisbee or of finding Elvis.”
―Boris Johnson, former Prime Minister
Definition9.27.Prime Ideal.
A prime ideal of a commutative ring \(R\) is a proper ideal \(P\) such that whenever \(xy \in P\) for \(x, y \in R\text{,}\) we have \(x \in P\) or \(y \in P\text{.}\)
Remark9.28.
We often refer to prime ideals as just “primes”, this will become gently confusing later and we will avoid doing so when possible.
Example9.29.Prime Ideals.
In \(\Z\text{,}\) the prime ideals are \(\igen0\) and the ideals generated by prime integers \(P=\igen p\text{,}\) where \(p\) is a prime integer. The maximal ideals are the ideals generated by prime integers. In particular \(\igen0\) is prime but not maximal.
In \(\Z[i]\) the ideal \(\igen{13}\) is not prime, because \(13=(3+2i)(3-2i)\in\igen{13}\text{,}\) but \(3+2i\not \in \igen{13}\) and \(3-2i\not \in \igen{13}\) (because if \(3\pm2i=13\a\) then \(N(3\pm2i)=N(13)N(\a)\) so \(13=13^2N(\a)\text{,}\) a contradiction).
Here is an equivalent characterization of prime ideals that will become very useful in a later course on commutative algebra.
Proposition9.30.Prime iff Complement is Closed.
An ideal \(P\) is prime if and only if \(R \setminus P\) is closed under multiplication.
And here is another equivalent characterization 1
One might say it is an equivalent equivalent characterization... but we shant.
that will become very useful instantly and will never really stop.
Theorem9.31.Prime iff Quotient is Domain.
Let \(R\) be a commutative ring with \(1\neq 0\text{,}\) and let \(I\) be an ideal of \(R\text{.}\) The ideal \(I\) is prime if and only if \(R/I\) is an integral domain.
Proof.
Suppose \(I\) is prime. If \((r + I)(r' + I) = 0 + I\text{,}\) then \(rr' \in I\) and hence either \(r \in I\) or \(r' \in I\text{,}\) so that either \(r + I = 0\) or \(r'+ I = 0\text{.}\) This proves \(R/I\) is a domain. Suppose \(R/I\) is a domain and that \(xy \in I\text{.}\) Then \((x + I)(y + I) = 0\) in \(R/I\) and hence either \(x+ I = 0\) or \(y + I = 0\text{.}\) It follows \(x \in I\) or \(y \in I\text{,}\) so that \(I\) is prime.
Since we’re on such a roll of equivalent characterizations, let’s through in one more for good measure.
Let \(A\) be a nontrivial ring. Then \(A\) is an integral domain if and only if the ideal \(\igen0\) is prime.
Prime ideals and homomorphisms? I thought you’d never ask.
Theorem9.33.Prime Ideals and Ring Maps.
If \(R\) is a domain, \(S\) is a ring and \(\varphi:R\to S\) is a ring homomorphism, then \(\ker(\varphi)\) is a prime ideal.
If \(\varphi:A\to B\) is a ring homomorphism and \(P\) is a prime ideal in \(B\text{,}\) then \(\varphi\inv(P)\) is prime in \(A\text{.}\)
SubsectionMaximal Ideals
“It’s just like, we know how to push ourselves to the max.”
―Young Dolph
Definition9.34.Maximal Ideal.
A maximal ideal of an arbitrary ring \(R\) is a proper ideal \(\fm\) such that the only ideals of \(R\) containing \(\fm\) are \(\fm\) and \(R\text{.}\)
Exercise9.35.\(\igen{2,x}\) in \(\Z[x]\).
In \(\Z[x]\) the ideal \(\igen{2,x}\) is maximal and prime, the ideals \(\igen2\) and \(\igen x\) are prime but not maximal.
And, matching up quite nicely with Theorem 9.31, we have one of the spicier meatballs of the chapter.
Theorem9.36.Maximal Ideal iff Quotient is a Field.
Let \(R\) be a commutative ring with \(1\neq 0\text{,}\) and let \(I\) be an ideal of \(R\text{.}\) The ideal \(I\) is maximal if and only if \(R/I\) is a field.
Proof.
The first assertion follows immediately from the Lattice Isomorphism Theorem for Rings and the fact that \(R/I\) is a field if and only if its only ideals are \(0\) and \(R/I\text{.}\)
Definition9.37.Residue Field.
Given a maximal ideal \(\fm\) in \(R\text{,}\) the residue field of \(\fm\) is the field \(R/\fm\text{.}\) A field \(k\) is a residue field of \(R\) if \(k \cong R/\fm\) for some maximal ideal \(\fm\text{.}\)
Corollary9.38.Maximal Ideals are Prime.
Every maximal ideal is prime.
Proof.
If \(I\) is maximal, then \(R/I\) is a field by Theorem 9.36 which in particular implies that \(R/I\) is a domain, so \(I\) is prime by Theorem 9.31.
Qual Watch.
Proving Corollary 9.38 was [cross-reference to target(s) "jan-2021-4" missing or not unique] on the [cross-reference to target(s) "jan-2021" missing or not unique] qualifying exam.
Unlike their prime counterparts, maximal ideals are not preserved by preimages in general.
Lemma9.39.Zorn’s Lemma.
Let \(P\) be a non-empty family of sets. Suppose that for each chain \(C\) in \(P\text{,}\) the set \(\bigcup_{C_{i}\in C}C_{i}\) is in \(P\text{.}\) Then \(P\) has a maximal element.
Theorem9.40.All Ideals Contained in Maximal Ideal.
If \(R\) is a ring with \(1\neq 0\) and \(I\) is a proper ideal of \(R\text{,}\) then there is a maximal ideal of \(R\) containing \(I\text{.}\) In particular every ring \(R\) contains a maximal ideal.
Proof.
Let \(\cC\) be the set of proper ideals of \(R\) that contain \(I\) and view \(\cC\) as a poset under containment. We will apply Zorn’s Lemma. Suppose \(\cT\) is a totally ordered subset of \(\cC\text{.}\) We need to show \(\cT\) has an upper bound in \(\cC\text{.}\) If \(\cT\) is empty, \(I\) is such a bound. Otherwise, let \(U = \bigcup_{L \in \cT} L\text{.}\)
Since \(\cT\) is non-empty, we have \(I \subseteq U\) and so \(U\neq \emptyset\text{.}\)
Given \(x,y \in U\text{,}\) then \(x \in L, y \in L'\) for some \(L, L' \in \cT\text{.}\) Since \(\cT\) is totally ordered, either \(L \subseteq L'\) or \(L' \subseteq L\text{,}\) and hence \(x + y \in L\) or \(x +y \in L'\text{.}\) Either way, \(x + y \in U\text{.}\)
For \(x \in U\) and \(r \in R\text{,}\) we have \(x \in L\) for some \(L \in \cT\) and hence \(rx \in L \subseteq M\text{.}\)
This proves \(U\) is an ideal that contains \(I\text{.}\) Since every \(L\in\cT\) is a proper ideal, \(L\cap R^\times =\emptyset\text{,}\) so \(U\cap R^\times = \bigcup_{L \in \cT} L\cap R^\times=\emptyset\) and hence \(U\) is a proper ideal, so \(U\in\cC\text{.}\) By Zorn’s Lemma, we conclude \(\cC\) has at least one maximal element \(M\text{.}\) This is a maximal ideal in the sense of definition since if \(J\) is an ideal of \(R\) and \(M\subseteq J\) then either \(J=R\) or, if \(J\) is proper, then \(J\in \cC\text{,}\) which yields \(J=M\) by using that \(M\) is a maximal element of \(\cC\text{.}\)
The existence of a maximal ideal follows by applying the first part of the theorem for \(I=(0)\text{.}\)
