Maximal Ideals and Fields

Maximal ideals are the engine behind the construction of finite fields: $\mathbb{F}_p[x]/(f(x))$ is a field of order $p^n$ whenever $f$ is an irreducible polynomial of degree $n$ over $\mathbb{F}_p$, because irreducibility of $f$ makes $(f)$ maximal in the PID $\mathbb{F}_p[x]$. These finite fields underlie error-correcting codes (Reed–Solomon codes used in CDs, QR codes, and satellite communication) and much of modern cryptography, including AES's use of $\mathbb{F}_{256} = \mathbb{F}_2[x]/(x^8+x^4+x^3+x+1)$ and elliptic curve cryptography over prime fields $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$. In algebraic geometry, the maximal ideals of a coordinate ring $k[x_1,\dots,x_n]/I$ correspond (via Hilbert's Nullstellensatz, when $k$ is algebraically closed) to actual points of the affine variety cut out by $I$ — turning geometric points into purely algebraic objects. In algebraic number theory, maximal ideals of rings of integers $\mathcal{O}_K$ correspond to the prime ideals lying above rational primes, and their behavior (splitting, ramification) governs deep questions about factorization of primes in number fields. In commutative algebra and scheme theory, the set of maximal ideals (the "maximal spectrum") is the starting point for constructing the Zariski topology and, more generally, affine schemes. This same correspondence-theorem machinery is the direct precursor to the Chinese Remainder Theorem, which handles quotients by products of pairwise comaximal ideals simultaneously, and to the general method of Irreducibility and Field Construction for engineering new fields (like $\mathbb{C}$ from $\mathbb{R}[x]/(x^2+1)$) as quotients by maximal ideals generated by irreducible polynomials.

:::quiz[Why is the polynomial ring quotient $\mathbb{Q}[x]/(x^2-2)$ a field?]

• Because $\mathbb{Q}[x]$ has no ideals
• Because $x^2-2$ is irreducible over $\mathbb{Q}$, making $(x^2-2)$ maximal ✓
• Because $2$ is a prime number
• Because $\mathbb{Q}$ already contains $\sqrt2$

Ask the tutor: Can you show me why $(x,2)$ is maximal in $\mathbb{Z}[x]$, and identify the field $\mathbb{Z}[x]/(x,2)$ explicitly?

Think of the ideals of $R$ ordered by inclusion, forming a lattice with $R$ at the top and the zero ideal $(0)$ at the bottom (assuming $R$ has no other special structure, $(0)$ need not be prime or maximal, but it is always the bottom of the ideal lattice). A maximal ideal is a "highest possible" proper ideal: it sits directly below $R$ in this lattice, with no ideal squeezed in between. The reason this lattice picture connects to fields is the correspondence (or lattice isomorphism) theorem: ideals of the quotient ring $R/M$ correspond exactly to ideals of $R$ that contain $M$. If $M$ is maximal, there are only two such ideals — $M$ itself (which becomes the zero ideal downstairs) and $R$ (which becomes the whole ring downstairs). A ring whose only ideals are $(0)$ and itself turns out to be exactly a field. So maximality of $M$ translates, through this dictionary, into the quotient ring having no ideals to speak of, which is the defining feature of a field.

Example 1: Maximal ideals of $\mathbb{Z}$. Consider $R = \mathbb{Z}$. We claim the ideal $M = (7)$ is maximal, while $N = (12)$ is not. Checking $(7)$ is maximal. Every ideal of $\mathbb{Z}$ is principal, say $(n)$ for some $n \geq 0$ (since $\mathbb{Z}$ is a PID). Suppose $(7) \subseteq (n) \subseteq \mathbb{Z}$. Then $7 \in (n)$ means $n \mid 7$. Since $7$ is prime in $\mathbb{Z}$, its only positive divisors are $1$ and $7$. If $n = 1$, then $(n) = \mathbb{Z}$. If $n = 7$, then $(n) = (7)$. So there is no ideal strictly between $(7)$ and $\mathbb{Z}$, confirming $(7)$ is maximal. By the theorem, $\mathbb{Z}/(7) = \mathbb{Z}/7\mathbb{Z}$ must be a field — and indeed this is the familiar field $\mathbb{F}_7$ with $7$ elements, where every nonzero residue class has a multiplicative inverse mod $7$ (for instance $3 \cdot 5 = 15 \equiv 1 \pmod 7$, so $3^{-1} = 5$). Checking $(12)$ is not maximal. We exhibit an ideal strictly between: $(12) \subsetneq (4) \subsetneq \mathbb{Z}$. Indeed $12 = 4 \cdot 3 \in (4)$ so $(12) \subseteq (4)$; the inclusion is strict since $4 \notin (12)$ (as $12 \nmid 4$); and $(4) \neq \mathbb{Z}$ since $1 \notin (4)$. Hence $(12)$ is not maximal. Consistent with the theorem, $\mathbb{Z}/12\mathbb{Z}$ is not a field: the element $4$ is nonzero but not invertible mod $12$, since $4x \equiv 1 \pmod{12}$ has no solution ($\gcd(4,12) = 4 \nmid 1$). In fact $4 \cdot 3 = 12 \equiv 0 \pmod{12}$ exhibits $4$ and $3$ as nonzero zero-divisors, so $\mathbb{Z}/12\mathbb{Z}$ is not even an integral domain — consistent with $(12)$ failing to be prime as well (since $4 \cdot 3 \in (12)$ but neither $4$ nor $3$ lies in $(12)$). This illustrates the general fact (not needed for the theorem's proof, but a useful sanity check): in $\mathbb{Z}$, the ideal $(n)$ for $n > 1$ is maximal exactly when $n$ is prime, matching the classical fact that $\mathbb{Z}/p\mathbb{Z}$ is a field precisely for prime $p$.

Example 2: A maximal ideal in a polynomial ring. Let $R = \mathbb{Q}[x]$, the ring of polynomials with rational coefficients, and let

We show $M$ is maximal, using the fact that $\mathbb{Q}[x]$ is a Euclidean domain (hence a PID) via polynomial division, and that $x^2 - 2$ is irreducible over $\mathbb{Q}$. Irreducibility of $x^2 - 2$ over $\mathbb{Q}$. If $x^2 - 2$ factored nontrivially in $\mathbb{Q}[x]$, since it has degree 2, the factorization would be into two linear factors, giving a rational root $r$ with $r^2 = 2$. But $\sqrt2$ is irrational (the classical proof: if $\sqrt 2 = p/q$ in lowest terms, then $p^2 = 2q^2$ forces $p$ even, $p = 2k$, then $4k^2 = 2q^2$, so $q^2 = 2k^2$ forces $q$ even too, contradicting lowest terms). Hence $x^2 - 2$ has no rational root and is irreducible over $\mathbb{Q}$. Maximality of $(x^2-2)$. Suppose $J$ is an ideal of $\mathbb{Q}[x]$ with $(x^2-2) \subseteq J \subseteq \mathbb{Q}[x]$. Since $\mathbb{Q}[x]$ is a PID, $J = (g(x))$ for some $g$. Then $x^2 - 2 \in (g)$ means $g \mid x^2-2$ in $\mathbb{Q}[x]$. Since $x^2-2$ is irreducible, its only divisors (up to units, i.e. nonzero rational scalar multiples) are units and associates of $x^2-2$ itself. If $g$ is a unit, $(g) = \mathbb{Q}[x]$. If $g$ is an associate of $x^2-2$, $(g) = (x^2-2)$. So there is no ideal strictly between $(x^2-2)$ and $\mathbb{Q}[x]$, and $(x^2-2)$ is proper since $x^2-2$ is not a unit (it has positive degree). Hence $M = (x^2-2)$ is maximal. By the theorem, $\mathbb{Q}[x]/(x^2-2)$ is a field. Concretely, every element of this quotient can be written uniquely as $a + b\bar x$ with $a, b \in \mathbb{Q}$ (using polynomial division to reduce any representative mod degree 2), where $\bar x^2 = 2$ in the quotient (since $x^2 - 2 \equiv 0$). This field is exactly $\mathbb{Q}(\sqrt2) = \{a + b\sqrt2 : a,b \in \mathbb{Q}\}$, with the isomorphism $\bar x \mapsto \sqrt2$. To verify a nonzero element is invertible directly: take $2 + 3\bar x$; rationalizing, $(2+3\bar x)(2 - 3\bar x) = 4 - 9\bar x^2 = 4 - 18 = -14$, so $(2+3\bar x)^{-1} = -\tfrac{1}{14}(2 - 3\bar x) = -\tfrac17 + \tfrac{3}{14}\bar x$, and this lies in $\mathbb{Q} + \mathbb{Q}\bar x$ as required. This construction — building a field by quotienting a polynomial ring by an ideal generated by an irreducible polynomial — is the general method underlying Irreducibility and Field Construction.