“Fate rarely calls upon us at a moment of our choosing.”
―Optimus Prime
Definition11.1.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 \(y \in P\text{.}\)
Example11.2.Prime Ideals in \(\Z\).
The prime ideals in \(\mathbb{Z}\) are those of the form \((p)\) for \(p\) a prime integer, and \((0)\text{.}\)
Example11.3.Prime Ideals in \(k[x]\).
When \(k\) is a field, prime ideals in \(k[x]\) are easy to describe: \(k[x]\) is a principal ideal domain, and \((f)\neq 0\) is prime if and only if \(f\) is an irreducible polynomial. Moreover, \((0)\) is also a prime ideal, since \(k[x]\) is a domain.
Example11.4.\((x^3-y^2)\) in \(k[x,y]\) is Prime.
Let \(k\) be a field. The ideal \(P = (x^3-y^2)\) in \(R=k[x,y]\) is prime: one can show that \(R/P \cong k[t^2,t^3] \subseteq k[t]\text{,}\) which is a domain.
Example11.5.Primes, Presentations, and Macaulay2.
The \(k\)-algebra \(R = k[t^3,t^4,t^5] \subseteq k[t]\) is a domain, so its defining ideal \(P\) in \(k[x,y,z,]\) is prime. This is the kernel of the presentation of \(R\) sending \(x,y,z\) to each of our 3 algebra generators, which we can compute with Macaulay2:
i1 : k = QQ
o1 = QQ
o1 : Ring
i2 : f = map(k[t],k[x,y,z],{t^3,t^4,t^5})
3 4 5
o2 = map (QQ[t], QQ[x..z], {t , t , t })
o2 : RingMap QQ[t] <--- QQ[x..z]
i3 : P = ker f
2 2 2 3
o3 = ideal (y - x*z, x y - z , x - y*z)
o3 : Ideal of QQ[x..z]
Definition11.6.Maximal Ideal.
A maximal ideal of an arbitrary ring \(R\) is a proper ideal \(M\) such that the only ideals of \(R\) containing \(M\) are \(M\) and \(R\text{.}\)
Definition11.7.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 \(\fm\text{.}\)
Remark11.8.
A ring may have many different residue fields. For example, the residue fields of \(\mathbb{Z}\) are all the finite fields with a prime numbers of elements, \(\mathbb{F}_p \cong \mathbb{Z}/p\text{.}\)
Theorem11.9.Prime, Maximal, Domain, Field.
Let \(R\) be a commutative ring with \(1\neq 0\text{,}\) and let \(I\) be an ideal of \(R\text{.}\) 1. The ideal \(I\) is maximal if and only if \(R/I\) is a field. 2. The ideal \(I\) is prime if and only if \(R/I\) is an integral domain. 3. Every maximal ideal of \(R\) is prime.
Proof.
The first assertion follows immediately from the Lattice Isomorphism Theorem and the fact that \(R/I\) is a field if and only if its only ideals are \(0\) and \(R/I\text{.}\)
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.
If \(I\) is maximal, then \(R/I\) is a field, which in particular implies that \(R/I\) is a domain, so \(I\) is prime.
Theorem11.10.Every Ideal Contained in a Maximal Ideal.
Proof.
Fix a ring \(R\) and a proper ideal \(I\text{.}\) Let \(S\) be the set of all proper ideals \(J\) in \(R\) such that \(J \supseteq I\text{,}\) which is partially ordered with the inclusion order \(\subseteq\text{.}\) We claim that [provisional cross-reference: cite] [[Mathematics/Foundations/Results/Theorem - Zorn’s Lemma|Zorn’s Lemma]] applies to \(S\text{.}\) First, \(S\) is nonempty, since it contains \(I\text{.}\) Now consider a chain of proper ideals in \(R\text{,}\) say \(\{ J_i \}_i\text{,}\) all of which contain \(I\text{.}\) Notice that \(J := \bigcup_i J_i\) is an ideal as well (exercise!), and moreover \(J \neq R\) since \(1 \notin J_i\) for all \(i\text{.}\) Since each \(J_i \supseteq I\text{,}\) we conclude that \(J \supseteq I\text{.}\) Thus we have checked that \(J \in S\text{.}\) Now this ideal \(J \in S\) is an upper bound for our chain \(\{ J_i \}_i\text{,}\) and thus applies to \(S\text{.}\) We conclude that \(S\) has a maximal element.
There is a subtle point missing: we have shown that there is a maximal element \(M\) in \(S\) containing \(I\text{,}\) but we have yet to show that this maximal element is a maximal ideal of \(R\text{.}\) Finally, suppose that \(L\) is an ideal in \(R\) with \(L \supseteq M\text{.}\) Since \(M\) contains \(J\text{,}\) so does \(L\text{.}\) If \(L \in S\text{,}\) by the maximality of \(M\) we must have \(L = M\text{.}\) Since \(L\) already satisfies \(L \supseteq J\text{,}\) if \(L \notin S\) then we must have \(L = R\text{.}\)
Theorem11.11.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.
