Prime and maximal ideals generalize prime numbers and points. Their quotient characterizations are the fastest way to use them.
Integral domains and fields through quotients
Definition. A commutative ring \(R\) is an integral domain if \(ab=0\) implies \(a=0\) or \(b=0\). It is a field if every nonzero element is invertible.
Example. \(\mathbb Z\) and \(F[x]\) are integral domains. The ring \(\mathbb Z/6\mathbb Z\) is not, because \(2\cdot3=0\). The quotient \(\mathbb Z/p\mathbb Z\) is a field exactly when \(p\) is prime.
Prime ideals
Definition. A proper ideal \(\mathfrak p\subset R\) is prime if \(ab\in\mathfrak p\) implies \(a\in\mathfrak p\) or \(b\in\mathfrak p\).
Proposition. An ideal \(\mathfrak p\) is prime iff \(R/\mathfrak p\) is an integral domain.
Proof. In \(R/\mathfrak p\), the product \((a+\mathfrak p)(b+\mathfrak p)\) is zero iff \(ab\in\mathfrak p\). The absence of zero divisors is exactly the defining implication for a prime ideal.
Maximal ideals
Definition. A proper ideal \(\mathfrak m\subset R\) is maximal if there is no ideal strictly between \(\mathfrak m\) and \(R\).
Proposition. An ideal \(\mathfrak m\) is maximal iff \(R/\mathfrak m\) is a field. Hence every maximal ideal is prime.
Proof. Ideals of \(R/\mathfrak m\) correspond to ideals of \(R\) containing \(\mathfrak m\). Thus \(\mathfrak m\) is maximal iff the quotient has only the zero ideal and the whole ring, which is equivalent to being a field.
Examples in polynomial rings
Example. In \(F[x]\), a nonzero prime ideal is generated by an irreducible polynomial. Since \(F[x]\) is a PID, such ideals are also maximal. Thus \(F[x]/(f)\) is a field exactly when \(f\) is irreducible.
Counterexample. In \(F[x,y]\), the ideal \((x)\) is prime because \(F[x,y]/(x)\cong F[y]\) is a domain, but it is not maximal because \(F[y]\) is not a field.
Radicals and nilpotents
Definition. The radical of an ideal \(I\) is \[ \sqrt I=\{r\in R:r^n\in I\text{ for some }n\ge1\}. \] An ideal is radical if \(I=\sqrt I\).
Proposition. The nilpotent elements of \(R/I\) correspond exactly to \(\sqrt I/I\). Thus \(I\) is radical iff \(R/I\) has no nonzero nilpotents.
Geometric intuition
Remark. In algebraic geometry, prime ideals behave like irreducible pieces and maximal ideals behave like points. Even when one is not doing geometry, this intuition helps organize quotient rings.
Spectrum, localization, and geometry
Proposition. For a multiplicative set \(S\subset R\), prime ideals of \(S^{-1}R\) correspond bijectively to prime ideals \(\mathfrak p\subset R\) disjoint from \(S\), through extension and contraction.
Proof. If \(\mathfrak p\cap S\) is empty, then \[ S^{-1}\mathfrak p =\{a/s:a\in\mathfrak p,\ s\in S\} \] is proper and prime. If \(\mathfrak q\subset S^{-1}R\) is prime, its contraction is prime and misses \(S\), because every \(s/1\) is a unit. The membership criterion \[ a/1\in S^{-1}\mathfrak p \Longleftrightarrow sa\in\mathfrak p\text{ for some }s\in S \] and primality recover the original ideals.
The prime spectrum \[ \operatorname{Spec}R \] becomes a topological space with closed sets \[ V(I)=\{\mathfrak p:I\subseteq\mathfrak p\}. \] The identities \[ V(IJ)=V(I)\cup V(J),\qquad V\!\left(\sum_\alpha I_\alpha\right)=\bigcap_\alpha V(I_\alpha) \] follow directly from primality. Localizing at \(\mathfrak p\) magnifies the neighborhood of that point and turns every element outside \(\mathfrak p\) into a unit.
Nullstellensatz viewpoint
For an algebraically closed field \(k\), maximal ideals of \(k[x_1,\ldots,x_n]\) are exactly \[ (x_1-a_1,\ldots,x_n-a_n). \] Hilbert's Nullstellensatz further states \[ I(V(I))=\sqrt I. \] Prime ideals therefore represent irreducible algebraic sets, while maximal ideals represent points. This is the precise bridge from ideal theory to algebraic geometry.