Fields

Fields are the base structure for Vector Spaces and all of linear algebra: "vector space over a field" is the setting where dimension, bases, and linear maps behave predictably, because scalar division is always available. In Galois theory, field extensions of $\mathbb{Q}$ or finite fields $\mathbb{F}_q$ (which exist exactly when $q$ is a prime power, generalizing the $\mathbb{Z}/p\mathbb{Z}$ case above) control the solvability of polynomial equations by radicals. In cryptography, finite fields $\mathbb{F}_p$ and $\mathbb{F}_{2^n}$ underlie the Diffie–Hellman key exchange, elliptic curve cryptography, and the AES encryption standard, all of which crucially use that every nonzero element of a finite field is invertible. In coding theory, Reed–Solomon codes (used in QR codes, CDs, and satellite communication) are built directly from polynomials over finite fields. In physics and engineering, fields such as $\mathbb{C}$ underlie Fourier analysis and quantum mechanics, where the ability to divide by nonzero complex numbers is used constantly in normalizing wavefunctions and solving linear systems. In statistics and numerical computation, floating-point arithmetic is designed to approximate field operations over $\mathbb{R}$, and exact rational arithmetic over $\mathbb{Q}$ is used when floating-point rounding is unacceptable.

:::quiz[Which statement correctly explains why $\mathbb{Z}[x]$ is a domain but not a field?]

• It is not commutative
• It fails to have a multiplicative identity
• It is a domain (degrees of nonzero polynomials add under multiplication) but is infinite, so $x$ has no inverse ✓
• It contains zero divisors like $2 \cdot 3$

Ask the tutor: How does the field-of-fractions construction formally turn any integral domain into the smallest field containing it?

Think of a commutative ring as a world with addition, subtraction, and multiplication, but where division is a privilege, not a right. In $\mathbb{Z}$, you can multiply $2$ and $3$ to get $6$, but you cannot "undo" multiplication by $2$ in general — there is no integer that, multiplied by $2$, gives $1$. A field is a world where this privilege is granted universally: every nonzero citizen has a multiplicative inverse, so division by any nonzero element is always possible. The link between fields and domains comes from a simple cancellation idea. If $a$ has an inverse $a^{-1}$ and $ab = 0$, multiply both sides by $a^{-1}$: this forces $b = 0$. So invertible elements can never be zero divisors, and in a field every nonzero element is invertible — hence no nonzero element can be a zero divisor at all. This is why fields are always domains. The converse — domain implies field — is false in general because a domain like $\mathbb{Z}$ can have "room to grow": the map $x \mapsto 2x$ is injective but its image is only the even integers, not all of $\mathbb{Z}$, so $2$ has no inverse. But if the ring is finite, there is no room to grow: an injective map from a finite set to itself must be surjective. This pigeonhole phenomenon is the heart of why finite domains are automatically fields, and it is illustrated below for $\mathbb{Z}/5\mathbb{Z}$, where multiplication by $3$ permutes the nonzero residues.

Example 1: Verifying $\mathbb{Z}/7\mathbb{Z}$ is a field directly from the theorem, without exhaustive inverse-checking. Let $R = \mathbb{Z}/7\mathbb{Z} = \{0,1,2,3,4,5,6\}$, a commutative ring with identity under addition and multiplication mod $7$, and $1 \neq 0$ since $7 > 1$. Claim: $R$ has no zero divisors. Suppose $\bar a \bar b = \bar 0$ in $R$ with $\bar a \neq \bar 0$. This means $7 \mid ab$ for integers $a, b$ representing $\bar a, \bar b$, with $7 \nmid a$. Since $7$ is prime, Euclid's lemma gives $7 \mid a$ or $7 \mid b$; since $7 \nmid a$, we get $7 \mid b$, i.e. $\bar b = \bar 0$. So $R$ is an integral domain. (This step uses primality of $7$ essentially — it is exactly where the argument would fail for a composite modulus, as the counterexample below shows.) $R$ is manifestly finite, with $|R| = 7$. By Part 2 of the theorem, $R$ is a field. To see the inverse-construction machinery concretely, take $a = 3$. The map $\varphi_3(x) = 3x \bmod 7$ sends

a bijection of $\{1,\dots,6\}$ onto itself, exactly as the proof predicts. Reading off $\varphi_3(5) = 1$ tells us $3^{-1} = 5$ in $\mathbb{Z}/7\mathbb{Z}$, and indeed $3 \cdot 5 = 15 = 2\cdot 7 + 1$.

Example 2: An infinite domain that is genuinely not a field, and where the finite hypothesis is essential. Let $R = \mathbb{Z}[x]$, the ring of polynomials in one variable with integer coefficients. This is a commutative ring with identity $1 \neq 0$. Claim: $R$ is an integral domain. Suppose $f, g \in \mathbb{Z}[x]$ are both nonzero, with leading terms $a_n x^n$ and $b_m x^m$ respectively ($a_n, b_m \neq 0$ integers, $n = \deg f$, $m = \deg g$). The leading term of $fg$ is $a_n b_m x^{n+m}$. Since $a_n, b_m$ are nonzero integers and $\mathbb{Z}$ is an integral domain (a fact already established since $\mathbb{Z}$ has no zero divisors: $a_nb_m = 0$ in $\mathbb{Z}$ would force $a_n=0$ or $b_m=0$), the product $a_n b_m \neq 0$. Hence $fg$ has degree $n+m$ and is not the zero polynomial. So $fg \neq 0$ whenever $f,g\neq 0$, proving $R$ has no zero divisors. Claim: $R$ is not a field. Consider $a = x \in \mathbb{Z}[x]$, which is nonzero. Suppose toward contradiction that $x$ had an inverse $g(x) \in \mathbb{Z}[x]$ with $x \cdot g(x) = 1$. Comparing degrees: $\deg(x) = 1$ and $\deg(1) = 0$, while $\deg(x \cdot g) = 1 + \deg(g) \geq 1$ (using that $\mathbb{Z}[x]$ is a domain, so degrees add exactly with no cancellation). This contradicts $\deg(x\cdot g)=0$. Hence no such $g$ exists, and $x$ has no multiplicative inverse in $\mathbb{Z}[x]$. So $R$ is a domain but not a field. This does not contradict Part 2 of the theorem, because $\mathbb{Z}[x]$ is infinite: there is no upper bound on the degree of polynomials, so the pigeonhole argument (Step 2 of the proof) has no traction here — the map $\varphi_x(g) = xg$ is injective but its image (polynomials with zero constant term) is a proper subset of $\mathbb{Z}[x]$, exactly mirroring the failure of surjectivity that finiteness rules out.