The first place where ordinary arithmetic begins to fracture is not with a complicated polynomial, but with the innocent equation $2x=1$. In $\mathbb{Q}$ it has the solution \begin{align*} x=\frac{1}{2}. \end{align*} In $\mathbb{Z}$ it has no solution. The difference is not that $\mathbb{Z}$ lacks addition or multiplication; it has both. The missing operation is division by a nonzero element. Fields are the algebraic worlds in which that obstruction has been removed. They are rings where every nonzero scalar can be inverted, and that single condition makes linear equations, polynomial division, vector spaces, determinants, and much of algebraic number theory behave with their familiar strength. A field is not merely a set with arithmetic. It is a place where multiplication by a nonzero element is reversible. [example: The Equation $2x=1$] In the ring $\mathbb{Z}$, the equation $2x=1$ asks for an integer $x$ whose double is $1$. If $x\in\mathbb{Z}$, then $2x$ is even, while $1$ is odd, so no integer $x$ satisfies $2x=1$. In $\mathbb{Q}$, the element $2$ has inverse $\frac{1}{2}$ because \begin{align*} 2\cdot \frac{1}{2}=1. \end{align*} Multiplying both sides of $2x=1$ by $\frac{1}{2}$ gives \begin{align*} \frac{1}{2}\cdot 2x=\frac{1}{2}\cdot 1. \end{align*} Using associativity and the identity property, \begin{align*} \left(\frac{1}{2}\cdot 2\right)x=\frac{1}{2}. \end{align*} Since $\frac{1}{2}\cdot 2=1$, this becomes \begin{align*} x=\frac{1}{2}. \end{align*} Substitution checks the solution: \begin{align*} 2\cdot \frac{1}{2}=1. \end{align*} This is the structural difference: multiplication by $2$ on $\mathbb{Q}$ has inverse multiplication by $\frac{1}{2}$, so every rational target $y$ is hit by $x=\frac{y}{2}$, while multiplication by $2$ on $\mathbb{Z}$ hits only the even integers and therefore misses $1$. [/example] The example raises the organizing question for the chapter: what algebraic hypotheses guarantee that every nonzero scalar can be divided by, and what consequences follow from that guarantee? ## Definition To make division into a permanent algebraic operation, we start with a ring and impose a condition on its nonzero elements. The question is not whether some elements have inverses, but whether every element except zero does. [definition: Field] A field is a nonzero commutative ring $k$ with identity $1_k$ such that every nonzero element has a multiplicative inverse in $k$. Thus for every $a \in k \setminus \{0\}$, there exists $b \in k$ such that \begin{align*} ab = 1_k. \end{align*} Such an invertible element is called a unit; the next section isolates that word in the broader setting of rings. [/definition] The letter $k$ is often used for a base field, while $K/k$ denotes a [field extension](/page/Field%20Extension). The slash in $K/k$ does not denote a quotient; it records that $K$ contains a copy of $k$ as a subfield. The definition also assumes the standard commutative-algebra convention that fields are commutative; division rings, where multiplication need not commute, are a separate noncommutative variant. ## Units and First Examples ### Units in Rings The field axiom depends on the word unit, and the next question is what it means for multiplication by an element of a ring to be reversible inside that same ring. Naming this concept lets us compare fields with rings such as $\mathbb{Z}$, where only a few elements can be undone multiplicatively. [definition: Unit] Let $R$ be a ring with identity $1_R$. An element $u \in R$ is a unit if there exists $v \in R$ such that \begin{align*} uv = vu = 1_R. \end{align*} The set of all units of $R$ is denoted $R^\times$. [/definition] Thus the defining condition for a field can be written compactly as $k^\times = k \setminus \{0\}$. This notation is useful because $k^\times$ carries a group structure under multiplication, while $k$ itself carries both addition and multiplication. ### First Examples A definition should immediately separate the familiar systems that support division from the familiar systems that do not. The next example asks where the field axiom is already present in standard arithmetic. [example: Standard Fields] The usual arithmetic laws make $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ commutative rings with identity, so the field axiom is the point to check. If $a \in \mathbb{Q}\setminus\{0\}$, write $a=\frac{m}{n}$ with $m,n\in\mathbb{Z}$ and $m\ne 0$, $n\ne 0$. Then $\frac{n}{m}\in\mathbb{Q}$, and \begin{align*} \frac{m}{n}\cdot\frac{n}{m}=\frac{mn}{nm}=1. \end{align*} Thus every nonzero rational number has a rational inverse. If $x\in\mathbb{R}\setminus\{0\}$, its real reciprocal $\frac{1}{x}$ satisfies \begin{align*} x\cdot\frac{1}{x}=1. \end{align*} If $z=a+bi\in\mathbb{C}\setminus\{0\}$, then $a,b\in\mathbb{R}$ are not both $0$, so $a^2+b^2\ne 0$. The inverse candidate \begin{align*} \frac{a-bi}{a^2+b^2} \end{align*} satisfies \begin{align*} (a+bi)\frac{a-bi}{a^2+b^2}=\frac{a^2-abi+abi-b^2i^2}{a^2+b^2}. \end{align*} Since $i^2=-1$, this becomes \begin{align*} \frac{a^2+b^2}{a^2+b^2}=1. \end{align*} Hence $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ are fields. For a prime number $p$, the residue class ring $\mathbb{Z}/p\mathbb{Z}$ is also a field. Let $\bar{a}\ne\bar{0}$. Then $p\nmid a$, and since $p$ is prime this gives $\gcd(a,p)=1$. By *Bezout's identity*, there exist integers $r,s\in\mathbb{Z}$ such that \begin{align*} ra+sp=1. \end{align*} Reducing this equality modulo $p$ gives \begin{align*} \overline{ra+sp}=\bar{1}. \end{align*} Using compatibility of reduction with addition and multiplication, \begin{align*} \bar{r}\bar{a}+\bar{s}\bar{p}=\bar{1}. \end{align*} Since $\bar{p}=\bar{0}$ in $\mathbb{Z}/p\mathbb{Z}$, we get \begin{align*} \bar{r}\bar{a}+\bar{s}\bar{0}=\bar{1}. \end{align*} Because $\bar{s}\bar{0}=\bar{0}$, this is \begin{align*} \bar{r}\bar{a}=\bar{1}. \end{align*} Thus every nonzero residue class modulo $p$ has a multiplicative inverse, so $\mathbb{Z}/p\mathbb{Z}$ is a field. [/example] The previous example suggests that fields are common, but the definition is restrictive. The next question is what fails in the most important ring that is not a field. [example: The Ring $\mathbb{Z}$ Is Not a Field] In $\mathbb{Z}$, a unit is an integer $n$ for which there is an integer $m$ satisfying $mn=1$. Since $mn=1\ne 0$, neither $m$ nor $n$ is $0$. If $|n|\ge 2$, then $|m|\ge 1$, so \begin{align*} |mn|=|m||n|\ge 2. \end{align*} But $mn=1$ gives $|mn|=1$, a contradiction. Therefore $|n|=1$, so $n=1$ or $n=-1$. Conversely, \begin{align*} 1\cdot 1=1. \end{align*} \begin{align*} (-1)(-1)=1. \end{align*} Thus the only units of $\mathbb{Z}$ are $1$ and $-1$. In particular, $2\in\mathbb{Z}\setminus\{0\}$ is not a unit, because it is neither $1$ nor $-1$. Hence $\mathbb{Z}$ is not a field: the nonzero element $2$ has no multiplicative inverse in $\mathbb{Z}$. The same obstruction appears in equations such as $2x=3$. If $x\in\mathbb{Z}$, then $2x$ is even, while $3$ is odd, so no integer $x$ satisfies $2x=3$. Polynomial division over $\mathbb{Z}[x]$ has the same defect: dividing by a leading coefficient such as $2$ would require the missing integer inverse of $2$. [/example] The zero ring is excluded because it erases the distinction between zero and nonzero elements. Requiring a nonzero ring keeps the additive identity and multiplicative identity separate enough for $k^\times$ to be meaningful. ## Division and Zero Divisors ### Multiplicative Structure Once all nonzero elements are invertible, they should not be viewed as isolated invertible elements. The next structural question is whether the nonzero elements form a group under multiplication. [quotetheorem:8300] [citeproof:8300] This group viewpoint explains why multiplying by a nonzero scalar preserves information: the multiplication map has an inverse multiplication map. The next question is how that reversibility appears in equations, where cancellation is the operation that lets us remove a common nonzero factor without changing the truth of the equation. The theorem is naturally stated for integral domains: nonzero commutative rings in which a product can be zero only if one factor is zero. [Fields are integral domains](/theorems/8276), so the result applies to fields, but the domain formulation isolates the exact hypothesis that makes cancellation possible. [quotetheorem:2577] Cancellation is one of the first places where fields differ from general rings. To see what the theorem prevents, we look at a nonfield where a nonzero multiplier can identify different elements. [example: Cancellation Fails in $\mathbb{Z}/6\mathbb{Z}$] In $\mathbb{Z}/6\mathbb{Z}$, the classes $\bar{1}$, $\bar{2}$, and $\bar{4}$ are residue classes modulo $6$. The class $\bar{2}$ is nonzero because $6\nmid 2$, and $\bar{1}\ne \bar{4}$ because $6\nmid(4-1)$. Multiplying by $\bar{2}$ gives \begin{align*} \bar{2}\bar{1}=\overline{2\cdot 1}=\bar{2}. \end{align*} For $\bar{4}$, we get \begin{align*} \bar{2}\bar{4}=\overline{2\cdot 4}=\bar{8}. \end{align*} Since $8-2=6$ is divisible by $6$, the residue classes $\bar{8}$ and $\bar{2}$ are equal, so \begin{align*} \bar{2}\bar{4}=\bar{2}. \end{align*} Therefore \begin{align*} \bar{2}\bar{1}=\bar{2}\bar{4}. \end{align*} But $\bar{1}\ne\bar{4}$, so cancellation by the nonzero element $\bar{2}$ fails in $\mathbb{Z}/6\mathbb{Z}$. The obstruction is that $\bar{2}$ has no multiplicative inverse modulo $6$: if $\bar{r}\bar{2}=\bar{1}$, then $2r\equiv 1 \pmod 6$, but $2r$ is even while every integer congruent to $1$ modulo $6$ is odd. Thus multiplication by $\bar{2}$ is not reversible. [/example] ### Domains and Fractions Cancellation is closely related to the absence of zero divisors. The next question is how much of field behavior survives if products vanish only for the expected reason, even when not every nonzero element is invertible. [definition: Integral Domain] An [integral domain](/page/Integral%20Domain) is a nonzero commutative ring $R$ with identity such that whenever $a,b \in R$ satisfy \begin{align*} ab=0, \end{align*} then $a=0$ or $b=0$. [/definition] The definition captures the part of field arithmetic that prevents accidental zero products. In a field, a product $ab=0$ can be tested by multiplying by the inverse of any nonzero factor, so a nonzero factor cannot hide the vanishing of the other one. This raises a structural check: every field should automatically satisfy the domain condition, because invertibility of nonzero elements is stronger than cancellation. The theorem records that fields are the basic source of integral domains. [quotetheorem:8301] [citeproof:8301] The converse fails: $\mathbb{Z}$ is an integral domain but not a field. The next construction asks how to repair that failure by formally adding the quotients that the domain is missing. [definition: Fraction Field] Let $R$ be an integral domain. A fraction field of $R$ is a field $\operatorname{Frac}(R)$ together with an injective ring homomorphism $\iota:R\to\operatorname{Frac}(R)$ such that every element of $\operatorname{Frac}(R)$ has the form \begin{align*} \iota(a)\iota(b)^{-1} \end{align*} for some $a,b \in R$ with $b \ne 0$. [/definition] The fraction field is the smallest field obtained by allowing division by nonzero elements of the original domain. To use fractions intrinsically, one must know that different constructions give the same field structure over $R$, rather than unrelated fields with similar-looking elements. [quotetheorem:8302] [citeproof:8302] For $R=\mathbb{Z}$, the fraction field is $\mathbb{Q}$. For $R=k[x]$, the fraction field is the rational function field $k(x)$. In both cases, the construction turns a domain where division is partial into a field where division by nonzero elements is always available. ## Characteristic and Prime Fields ### Additive Periodicity Fields can have different arithmetic clocks. In $\mathbb{Q}$, repeatedly adding $1$ never returns to $0$. In $\mathbb{Z}/p\mathbb{Z}$, adding $1$ exactly $p$ times gives $0$. A single invariant is needed to record whether the additive copy of the integers inside the field repeats, and if so where the first repetition occurs. [definition: Characteristic of a Field] Let $k$ be a field. The characteristic of $k$, denoted $\operatorname{char}(k)$, is the least positive integer $n \in \mathbb{N}$ such that \begin{align*} \underbrace{1_k+\cdots+1_k}_{n\text{ times}}=0, \end{align*} if such an integer exists. If no such integer exists, then $\operatorname{char}(k)=0$. [/definition] The characteristic is not an auxiliary label; it changes the algebra. The next example separates the two main worlds: characteristic zero and prime characteristic. [example: Characteristic $0$ and Characteristic $p$] For $\mathbb{Q}$, the identity is $1=\frac{1}{1}$. If $n\ge 1$, then the sum of $n$ copies of $1$ is \begin{align*} \underbrace{1+\cdots+1}_{n\text{ times}}=\frac{n}{1}. \end{align*} Since $n\ne 0$ in $\mathbb{Z}$, the rational number $\frac{n}{1}$ is not $0$, so no positive sum of copies of $1$ vanishes in $\mathbb{Q}$. The same integer sum is nonzero in $\mathbb{R}$, and in $\mathbb{C}$ it is $n+0i\ne 0+0i$. Hence $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ have characteristic $0$. Now let $p$ be prime and write $\mathbb{F}_p=\mathbb{Z}/p\mathbb{Z}$. The sum of $p$ copies of $\bar{1}$ is \begin{align*} \underbrace{\bar{1}+\cdots+\bar{1}}_{p\text{ times}}=\bar{p}. \end{align*} Since $p-p=0$ is divisible by $p$, we have $\bar{p}=\bar{0}$. If $1\le n<p$, then the sum of $n$ copies of $\bar{1}$ is $\bar{n}$. The equality $\bar{n}=\bar{0}$ would mean that $p$ divides $n$, but no positive integer smaller than $p$ is divisible by $p$. Therefore the least positive number of copies of $\bar{1}$ that sums to $\bar{0}$ is $p$, so $\operatorname{char}(\mathbb{F}_p)=p$. [/example] The example involves only characteristic $0$ and prime characteristics. The next question is whether a field could have composite positive characteristic, such as $6$ or $10$, and the answer uses the fact that fields have no zero divisors. [quotetheorem:8303] [citeproof:8303] This dichotomy is what prevents fields from having arbitrary additive clocks. Composite characteristics would force a product of two nonzero integer multiples of $1_k$ to vanish, contradicting the domain behavior inherited from field arithmetic. Thus every field belongs to exactly one of two regimes: characteristic $0$, where the integers embed without repetition, or characteristic $p$, where arithmetic contains a prime residue field as its basic clock. ### Smallest Subfields Every field contains the arithmetic generated by its own identity element. The next question is what the smallest possible subfield inside a given field looks like. [definition: Prime Subfield] Let $k$ be a field. The prime subfield of $k$ is the intersection of all subfields of $k$. [/definition] The prime subfield is the algebraic seed of the field. Once the characteristic is known, there is very little freedom in this seed: either the integer multiples of $1$ never repeat and generate rational arithmetic, or they repeat modulo a prime and generate finite prime-field arithmetic. The point is that the definition by intersection is abstract, while computations need an explicit model for this smallest subfield. The classification theorem identifies exactly which familiar field appears inside $k$. [quotetheorem:8304] [citeproof:8304] In characteristic $p$, binomial coefficients divisible by $p$ vanish. The next question is what universal self-map this creates on every field of characteristic $p$. [definition: Frobenius Endomorphism] Let $k$ be a field with $\operatorname{char}(k)=p$, where $p$ is prime. The Frobenius endomorphism is the function $\operatorname{Frob}:k\to k$ defined by \begin{align*} \operatorname{Frob}(a)=a^p \end{align*} for every $a\in k$. [/definition] The Frobenius map is useful only if taking $p$th powers is compatible with addition and multiplication. Multiplication is automatic, but addition is the obstruction: the mixed binomial terms must disappear, which is exactly what prime characteristic makes possible. [quotetheorem:8305] [citeproof:8305] For a field, Frobenius is automatically injective because $a^p=0$ forces $a=0$. It need not be surjective: in an imperfect field of characteristic $p$, some elements have no $p$th root inside the field. Over finite fields, however, injectivity already implies surjectivity, so Frobenius becomes an automorphism. This contrast between finite, perfect, and imperfect behavior is one of the main reasons the map appears repeatedly in field theory. ## Linear Algebra Over Fields ### Scalars and Vector Spaces The reason fields appear so early in linear algebra is that row reduction requires division by pivots. If a pivot is nonzero, a field guarantees it can be rescaled to $1$. To separate the scalars used for these rescalings from the objects being added, we need a structure where field elements act on an abelian group in a distributive and associative way. [definition: Vector Space Over a Field] Let $k$ be a field. A [vector space](/page/Vector%20Space) over $k$ is an abelian group $(V,+)$ together with scalar multiplication \begin{align*} k\times V \to V \end{align*} written $(a,v)\mapsto av$, such that for all $a,b\in k$ and all $v,w\in V$, \begin{align*} a(v+w)=av+aw. \end{align*} \begin{align*} (a+b)v=av+bv. \end{align*} \begin{align*} (ab)v=a(bv). \end{align*} \begin{align*} 1_kv=v. \end{align*} [/definition] The field condition lets nonzero scalars act reversibly on a vector space. The next question is how to recognize a list of vectors that contains no scalar redundancy. [definition: Linear Independence] Let $V$ be a vector space over a field $k$. A finite list $v_1,\dots,v_n \in V$ is linearly independent if whenever $a_1,\dots,a_n \in k$ satisfy \begin{align*} a_1v_1+\cdots+a_nv_n=0, \end{align*} then $a_1=\cdots=a_n=0$. A subset $S\subset V$ is linearly independent if every finite list of distinct elements of $S$ is linearly independent. [/definition] [Linear independence](/page/Linear%20Independence) is the algebraic expression of nonredundancy. The next question is how to combine nonredundancy with the ability to express every vector. [definition: Basis] Let $V$ be a vector space over a field $k$. A basis of $V$ is a linearly independent subset $B\subset V$ such that every $v\in V$ can be written as a finite linear combination of elements of $B$. [/definition] A basis is a coordinate system without geometry. The next question is whether the number of basis vectors is an invariant of the space or an accident of the chosen coordinates. [definition: Dimension] Let $V$ be a vector space over a field $k$. If $V$ has a basis $B$, then the dimension of $V$ over $k$ is the cardinality \begin{align*} \dim_k V=|B|. \end{align*} If $\dim_k V$ is finite, then $V$ is called finite-dimensional over $k$. [/definition] This definition depends on a theorem: different bases must have the same cardinality. Without that fact, $\dim_k V$ would depend on which coordinate system was chosen, so dimension would not be an invariant of the vector space itself. The obstruction is that a vector space may admit many unrelated-looking coordinate systems. To make dimension meaningful, one proves an exchange principle: in a finite-dimensional vector space, a linearly independent list cannot be longer than a spanning list. Applying this in both directions to two bases shows that any two finite bases have the same number of vectors. Thus dimension is not extra structure; it is an invariant recovered from the vector space operations. This invariant is what makes computations meaningful. Row operations may change the displayed equations, but they should not change the underlying solution space or the number of free coordinates needed to describe it. ### Row Reduction The computational engine behind finite-dimensional linear algebra is row reduction. Solving a linear system requires changing equations without changing their solution set, and the delicate step is pivot normalization: a nonzero coefficient must be divided out using its field inverse. [quotetheorem:8306] [citeproof:8306] The theorem hides a division step: when a nonzero pivot $a\in k$ appears, the row is multiplied by $a^{-1}$. The next example shows why the same procedure changes meaning over $\mathbb{Z}$. [example: A Pivot That Cannot Be Normalized Over $\mathbb{Z}$] Consider the integer equation $2x=1$. If this were a one-row linear system over a field containing $\mathbb{Z}$, the pivot $2$ would be normalized by multiplying the row by $2^{-1}$. In $\mathbb{Q}$, this inverse is $\frac{1}{2}$, since \begin{align*} \frac{1}{2}\cdot 2=1. \end{align*} Multiplying both sides of $2x=1$ by $\frac{1}{2}$ gives \begin{align*} \frac{1}{2}(2x)=\frac{1}{2}\cdot 1. \end{align*} By associativity of multiplication, \begin{align*} \left(\frac{1}{2}\cdot 2\right)x=\frac{1}{2}. \end{align*} Since $\frac{1}{2}\cdot 2=1$ and $1x=x$, this becomes \begin{align*} x=\frac{1}{2}. \end{align*} Substitution verifies the rational solution: \begin{align*} 2\cdot \frac{1}{2}=1. \end{align*} Over $\mathbb{Z}$, the same normalization step is not available. If $2$ had an inverse in $\mathbb{Z}$, there would be some $m\in\mathbb{Z}$ with \begin{align*} 2m=1. \end{align*} But $2m$ is even for every integer $m$, while $1$ is odd, so no such integer $m$ exists. Thus $2$ is a nonzero nonunit in $\mathbb{Z}$, and the equation $2x=1$ has no integer solution. The field hypothesis therefore changes the solvability problem: over $\mathbb{Q}$ the pivot can be inverted, while over $\mathbb{Z}$ it cannot. [/example] ## Polynomials and Roots ### Division of Polynomials Fields are the natural coefficient systems for polynomial algebra because polynomial long division must divide by the leading coefficient of the divisor. Before division can be discussed, the objects being divided must be fixed: finite formal sums whose coefficients come from a field and whose powers of $x$ are only bookkeeping symbols. [definition: Polynomial Ring Over a Field] Let $k$ be a field. The [polynomial ring](/page/Polynomial%20Ring) $k[x]$ is the ring of finite formal sums \begin{align*} f(x)=a_0+a_1x+\cdots+a_nx^n, \end{align*} where $n\ge 0$ and $a_0,a_1,\dots,a_n\in k$. [/definition] To run a Euclidean algorithm for polynomials, we need a measure that decreases during division. The relevant measure is controlled by the highest power of $x$ with nonzero coefficient, because canceling that term is what makes each division step progress. [definition: Degree of a Polynomial] Let $k$ be a field, and let $f\in k[x]$ be nonzero. If $f(x)=a_0+a_1x+\cdots+a_nx^n$ with $a_n\ne 0$, then the degree of $f$ is $\deg f=n$. [/definition] The degree function makes $k[x]$ resemble the integers. The key obstruction to polynomial division is the leading coefficient of the divisor: to cancel the leading term of the dividend, that coefficient must be invertible, which is why the coefficient system is a field. [quotetheorem:1706] The conclusion gives both a quotient and a remainder, and the remainder is the part that has become too small to keep dividing by the chosen divisor. This is the polynomial analogue of integer division, but it depends essentially on working over a field: over a general coefficient ring, the leading coefficient of the divisor may not be invertible, so the cancellation step can fail. The most important first use is division by a linear polynomial, where the remainder has degree less than $1$ and is therefore a constant. ### Roots and Factorisation Polynomial division becomes especially powerful when the divisor is linear. A value $a\in k$ matters because substituting it into $f$ tests whether the linear polynomial $x-a$ should divide $f$. [definition: Root of a Polynomial] Let $k$ be a field, let $f\in k[x]$, and let $a\in k$. The element $a$ is a root of $f$ if \begin{align*} f(a)=0. \end{align*} [/definition] A root is not only a vanishing point. The obstruction is that evaluation is a numerical test, while factorisation is an algebraic statement about divisibility in $k[x]$. Over a field, polynomial division by $x-a$ connects these two viewpoints and turns vanishing into the existence of a linear factor. [quotetheorem:3235] The [factor theorem](/theorems/3235) turns root-counting into degree-counting, but this only helps if repeated linear factors cannot accumulate beyond the polynomial's degree. The obstruction is a polynomial with many distinct roots: each root would force another linear divisor, so one needs a bound showing that this process must stop. [quotetheorem:1708] The root bound uses the integral-domain hypothesis: distinct roots give distinct linear factors without zero divisors interfering with cancellation. The next example shows the kind of failure that appears when zero divisors enter. [example: Too Many Roots Over $\mathbb{Z}/8\mathbb{Z}$] In the ring $\mathbb{Z}/8\mathbb{Z}$, consider $f(x)=x^2-\bar{1}$. We evaluate $f$ at the four residue classes $\bar{1}$, $\bar{3}$, $\bar{5}$, and $\bar{7}$. For $\bar{1}$, \begin{align*} f(\bar{1})=\bar{1}^2-\bar{1}. \end{align*} Since $\bar{1}^2=\overline{1\cdot 1}=\bar{1}$, we get \begin{align*} f(\bar{1})=\bar{1}-\bar{1}=\bar{0}. \end{align*} For $\bar{3}$, \begin{align*} f(\bar{3})=\bar{3}^2-\bar{1}. \end{align*} Now $\bar{3}^2=\overline{3\cdot 3}=\bar{9}$. Since $9-1=8$ is divisible by $8$, we have $\bar{9}=\bar{1}$ in $\mathbb{Z}/8\mathbb{Z}$. Hence \begin{align*} f(\bar{3})=\bar{9}-\bar{1}=\bar{1}-\bar{1}=\bar{0}. \end{align*} For $\bar{5}$, \begin{align*} f(\bar{5})=\bar{5}^2-\bar{1}. \end{align*} Now $\bar{5}^2=\overline{5\cdot 5}=\bar{25}$. Since $25-1=24$ is divisible by $8$, we have $\bar{25}=\bar{1}$. Therefore \begin{align*} f(\bar{5})=\bar{25}-\bar{1}=\bar{1}-\bar{1}=\bar{0}. \end{align*} For $\bar{7}$, \begin{align*} f(\bar{7})=\bar{7}^2-\bar{1}. \end{align*} Now $\bar{7}^2=\overline{7\cdot 7}=\bar{49}$. Since $49-1=48$ is divisible by $8$, we have $\bar{49}=\bar{1}$. Thus \begin{align*} f(\bar{7})=\bar{49}-\bar{1}=\bar{1}-\bar{1}=\bar{0}. \end{align*} The polynomial $x^2-\bar{1}$ has degree $2$ but has at least the four distinct roots $\bar{1},\bar{3},\bar{5},\bar{7}$ in $\mathbb{Z}/8\mathbb{Z}$. This shows why the field hypothesis in the root bound cannot be dropped: zero divisors allow a quadratic to have more roots than its degree. [/example] ### Algebraic Closure Some fields still do not contain enough roots for all polynomials. The polynomial $x^2+1$ has no root in $\mathbb{R}$, even though $\mathbb{R}$ is a field. The obstruction is not a failure of field arithmetic, but a shortage of solutions inside the field itself. We isolate the fields where every nonconstant polynomial equation can already be solved without enlarging the field. [definition: Algebraically Closed Field] A field $k$ is algebraically closed if every nonconstant polynomial $f\in k[x]$ has a root in $k$. [/definition] Algebraic closedness is not part of the definition of field; it is an additional completeness property for polynomial equations. The [real numbers](/page/Real%20Numbers) fail this test because $x^2+1$ has no real root, so the natural question is whether adjoining $i$ and forming $\mathbb{C}$ repairs all polynomial equations at once, not just that one quadratic. [quotetheorem:347] The [Fundamental Theorem of Algebra](/theorems/347) says precisely that $\mathbb{C}$ is algebraically closed: every nonconstant polynomial with complex coefficients has a complex root, and hence splits into linear factors over $\mathbb{C}$. It does not say the same about $\mathbb{R}$; for instance, $x^2+1$ has no real root. This contrast motivates field extensions. When a field does not contain enough roots, we enlarge it in a controlled way so that selected polynomials acquire solutions. ## Extensions and Constructions ### Enlarging Fields Many important fields arise by enlarging a known field to include a missing solution. The move from $\mathbb{R}$ to $\mathbb{C}$ adds a root of $x^2+1$; the move from $\mathbb{Q}$ to $\mathbb{Q}(\sqrt{2})$ adds a root of $x^2-2$. The next definition is the language for this enlargement. [definition: Field Extension] Let $k$ and $K$ be fields. A field extension $K/k$ is an inclusion of fields $k\subset K$ such that the operations of $k$ agree with the operations induced from $K$. [/definition] Once $K/k$ is a field extension, the larger field $K$ is automatically a vector space over the smaller field $k$. The issue is how to measure the size of an enlargement without counting its elements, which may be infinite in both fields. Vector-space dimension gives a scale-sensitive measure of how many independent directions were added over the base field. [definition: Degree of a Field Extension] Let $K/k$ be a field extension. The degree of $K/k$ is the dimension of $K$ as a vector space over $k$: \begin{align*} [K:k]=\dim_k K. \end{align*} If this dimension is finite, then $K/k$ is called a finite extension. [/definition] The degree measures how many independent scalar directions were added. The next example shows this measurement in the first quadratic extension. [example: The Field $\mathbb{Q}(\sqrt{2})$] Let $L=\{a+b\sqrt{2}:a,b\in\mathbb{Q}\}$. The rational numbers sit inside $L$ as $q=q+0\sqrt{2}$, and $\sqrt{2}=0+1\sqrt{2}$ belongs to $L$. The set is closed under the ring operations: for $a,b,c,d\in\mathbb{Q}$, \begin{align*} (a+b\sqrt{2})+(c+d\sqrt{2})=(a+c)+(b+d)\sqrt{2}. \end{align*} \begin{align*} -(a+b\sqrt{2})=(-a)+(-b)\sqrt{2}. \end{align*} \begin{align*} (a+b\sqrt{2})(c+d\sqrt{2})=ac+ad\sqrt{2}+bc\sqrt{2}+bd(\sqrt{2})^2=(ac+2bd)+(ad+bc)\sqrt{2}. \end{align*} Since $a+c$, $b+d$, $-a$, $-b$, $ac+2bd$, and $ad+bc$ are rational, the results remain in $L$. Now let $a+b\sqrt{2}\in L$ be nonzero. We first check that $a^2-2b^2\ne 0$. If $b=0$, then $a+b\sqrt{2}=a$, so nonzero means $a\ne 0$, and therefore $a^2-2b^2=a^2\ne 0$. If $b\ne 0$ and $a^2-2b^2=0$, then \begin{align*} a^2=2b^2. \end{align*} Dividing by $b^2$ gives \begin{align*} \left(\frac{a}{b}\right)^2=2. \end{align*} Thus $\frac{a}{b}$ would be a rational square root of $2$, contradicting the irrationality of $\sqrt{2}$. Hence $a^2-2b^2$ is nonzero, so the rational numbers $\frac{a}{a^2-2b^2}$ and $\frac{-b}{a^2-2b^2}$ are defined. The proposed inverse lies in $L$: \begin{align*} \frac{a-b\sqrt{2}}{a^2-2b^2}=\frac{a}{a^2-2b^2}+\frac{-b}{a^2-2b^2}\sqrt{2}. \end{align*} Multiplying it by $a+b\sqrt{2}$ gives \begin{align*} (a+b\sqrt{2})(a-b\sqrt{2})=a^2-ab\sqrt{2}+ab\sqrt{2}-b^2(\sqrt{2})^2=a^2-2b^2. \end{align*} Therefore \begin{align*} (a+b\sqrt{2})\frac{a-b\sqrt{2}}{a^2-2b^2}=\frac{a^2-2b^2}{a^2-2b^2}=1. \end{align*} So every nonzero element of $L$ has a multiplicative inverse in $L$, and $L=\mathbb{Q}(\sqrt{2})$ is a field. Finally, $1$ and $\sqrt{2}$ span $L$ over $\mathbb{Q}$ by the definition of $L$. They are linearly independent over $\mathbb{Q}$: if \begin{align*} r\cdot 1+s\sqrt{2}=0 \end{align*} with $r,s\in\mathbb{Q}$ and $s\ne 0$, then \begin{align*} \sqrt{2}=-\frac{r}{s}, \end{align*} which again contradicts the irrationality of $\sqrt{2}$. Hence $s=0$, and then $r=0$. Thus $\{1,\sqrt{2}\}$ is a basis of $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$, so $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$. [/example] ### Algebraic Generation A single element can generate an extension when all rational expressions in that element are allowed. The subtlety is that merely adjoining $\alpha$ to $k$ must also force closure under addition, multiplication, and division by nonzero elements. The construction therefore asks for the smallest field in the ambient extension that contains both $k$ and the chosen element. [definition: Simple Field Extension] Let $K/k$ be a field extension and let $\alpha\in K$. The simple field extension generated by $\alpha$ over $k$ is the smallest subfield of $K$ containing $k$ and $\alpha$. It is denoted $k(\alpha)$. [/definition] Simple extensions are controlled by whether the generating element satisfies a polynomial equation over the base field. This raises a new classification problem for elements of an extension. Some elements are constrained by a finite algebraic relation with coefficients in the base field, while others behave as if their powers introduce endlessly new independent data. The next definition names the constrained case, where a nonzero polynomial relation is present. [definition: Algebraic Element] Let $K/k$ be a field extension. An element $\alpha\in K$ is algebraic over $k$ if there exists a nonzero polynomial $f\in k[x]$ such that \begin{align*} f(\alpha)=0. \end{align*} [/definition] An algebraic element may satisfy many polynomial equations over the base field, since multiplying one relation by another polynomial gives more relations. The problem is to find the relation that is intrinsic rather than accidental. Choosing the monic relation of least positive degree gives a canonical polynomial that captures the first dependence among the powers of $\alpha$. [definition: Minimal Polynomial] Let $K/k$ be a field extension and let $\alpha\in K$ be algebraic over $k$. The [minimal polynomial](/page/Minimal%20Polynomial) of $\alpha$ over $k$ is the monic polynomial $m_{\alpha,k}\in k[x]$ of least positive degree such that \begin{align*} m_{\alpha,k}(\alpha)=0. \end{align*} [/definition] The minimal polynomial gives one relation among powers of $\alpha$, but it is not yet clear how much of the field $k(\alpha)$ this relation controls. The obstruction is that $k(\alpha)$ allows quotients of expressions in $\alpha$, while the minimal polynomial only speaks about polynomial expressions. The formal way to package this is by a [quotient ring](/page/Quotient%20Ring): $K[t]/\langle P_\alpha\rangle$ means the polynomial ring $K[t]$ with two polynomials identified when their difference is a multiple of $P_\alpha$. Here $\langle P_\alpha\rangle$ denotes the principal ideal generated by $P_\alpha$, the set of all multiples of $P_\alpha$ in $K[t]$. In the theorem card below, the base field denoted $K$ is the field we have been calling $k$, and the polynomial $P_\alpha$ is the minimal polynomial $m_{\alpha,k}$. With that translation, the key result shows that the relation is strong enough to make finitely many powers of $\alpha$ span the whole generated field. [quotetheorem:1251] The quotient presentation makes a simple algebraic extension concrete: elements of $k(\alpha)$ can be represented by polynomial expressions in $\alpha$, and the minimal polynomial gives the relations needed to reduce powers of $\alpha$. If the minimal polynomial has degree $d$, then $1,\alpha,\dots,\alpha^{d-1}$ form the standard $k$-basis. For instance, adjoining a root of an irreducible quadratic gives a two-dimensional vector space over the base field. This description depends on algebraicity; transcendental adjunctions do not collapse to a finite basis in this way. ### Splitting and Finite Fields Adjoining one root may not add all roots of a polynomial. For example, a polynomial may have several roots that live in different successive extensions, so solving one equation inside an extension does not by itself give a field where the polynomial has completely factored. The useful object is the field that removes exactly this obstruction: it contains all the roots needed for the polynomial to decompose into linear factors, and it is generated by those roots over the original field. [definition: Splitting Field] Let $k$ be a field and let $f\in k[x]$ be nonconstant. A [splitting field](/page/Splitting%20Field) of $f$ over $k$ is a field extension $K/k$ such that $f$ factors into linear factors in $K[x]$, and $K$ is generated over $k$ by the roots of $f$ in $K$. [/definition] Splitting fields are the entry point to Galois theory: once all roots are present, one studies the symmetries of the extension that preserve the base field. The next example shows the simplest case. [example: Splitting $x^2-2$ Over $\mathbb{Q}$] Let $f(x)=x^2-2\in\mathbb{Q}[x]$, and work in the field $\mathbb{Q}(\sqrt{2})$. Evaluating at $\sqrt{2}$ gives \begin{align*} f(\sqrt{2})=(\sqrt{2})^2-2=2-2=0. \end{align*} Evaluating at $-\sqrt{2}$ gives \begin{align*} f(-\sqrt{2})=(-\sqrt{2})^2-2=(\sqrt{2})^2-2=2-2=0. \end{align*} Thus $\sqrt{2}$ and $-\sqrt{2}$ are roots of $f$ in $\mathbb{Q}(\sqrt{2})$. The corresponding factorization is visible by expanding: \begin{align*} (x-\sqrt{2})(x+\sqrt{2})=x^2+x\sqrt{2}-\sqrt{2}x-(\sqrt{2})^2. \end{align*} Since coefficients commute with $x$ in $\mathbb{Q}(\sqrt{2})[x]$, we have $x\sqrt{2}=\sqrt{2}x$, so the middle terms cancel: \begin{align*} x^2+x\sqrt{2}-\sqrt{2}x-(\sqrt{2})^2=x^2-2. \end{align*} Therefore \begin{align*} x^2-2=(x-\sqrt{2})(x+\sqrt{2}) \end{align*} splits into linear factors over $\mathbb{Q}(\sqrt{2})$. It remains to check the generation condition. The field $\mathbb{Q}(\sqrt{2})$ contains $\sqrt{2}$ by definition, and it contains $-\sqrt{2}$ because fields are closed under additive inverses. Hence the field generated over $\mathbb{Q}$ by the two roots is contained in $\mathbb{Q}(\sqrt{2})$. Conversely, any field containing $\mathbb{Q}$ and the root $\sqrt{2}$ contains the simple extension $\mathbb{Q}(\sqrt{2})$, so \begin{align*} \mathbb{Q}(\sqrt{2},-\sqrt{2})=\mathbb{Q}(\sqrt{2}). \end{align*} Thus $\mathbb{Q}(\sqrt{2})$ contains all roots of $x^2-2$, the polynomial splits there, and the field is generated over $\mathbb{Q}$ by those roots. Hence $\mathbb{Q}(\sqrt{2})$ is the splitting field of $x^2-2$ over $\mathbb{Q}$. [/example] Field theory is not limited to infinite systems like $\mathbb{Q}$ and $\mathbb{C}$. There is also a separate finiteness question: can the field axioms hold when there are only finitely many scalars, even though every nonzero element must still have a multiplicative inverse? Such fields are rare enough to have rigid arithmetic, but common enough to appear throughout coding theory, finite geometry, cryptography, and arithmetic geometry. [definition: Finite Field] A finite field is a field with finitely many elements. A finite field with $q$ elements is denoted $\mathbb{F}_q$. [/definition] The notation $\mathbb{F}_q$ only makes sense after one knows which values of $q$ can occur. The central structural question is therefore not just how to name a finite field, but which finite cardinalities are compatible with the field axioms. Repeatedly adding $1$ produces a prime subfield, and the whole finite field must organize itself as a finite-dimensional vector space over that subfield. This rules out arbitrary cardinalities and forces the size to be a prime power. [quotetheorem:8307] [citeproof:8307] The converse is also true: for every prime power $p^n$, there is a field with $p^n$ elements, unique up to isomorphism. That construction uses splitting fields and belongs naturally on a dedicated finite fields page. ## Beyond and Connected Topics Fields are the scalar systems for [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Once the scalars form a field, bases have well-defined size, matrices represent linear maps cleanly, determinants control invertibility, and Gaussian elimination works without extra divisibility hypotheses. The next algebraic step is field extension theory. Extensions explain how to enlarge a field by adjoining roots, how degrees multiply in towers, and how polynomial equations create new scalar systems. Splitting fields and algebraically closed fields are refinements of this idea. Number theory studies fields that remember arithmetic. A number field is a finite extension of $\mathbb{Q}$, and its ring of integers recovers divisibility in a richer setting. This is the natural continuation in [Cambridge II Number Fields](/page/Cambridge%20II%20Number%20Fields). Commutative algebra studies fields from the viewpoint of rings and ideals. Fields appear as residue fields $R/\mathfrak{m}$ at maximal ideals, as fraction fields of integral domains, and as coefficient fields in local algebra. The systematic development belongs with [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Local fields add topology and valuation to field theory. The $p$-adic field $\mathbb{Q}_p$ is obtained by completing $\mathbb{Q}$ with respect to the $p$-adic absolute value, and its arithmetic mixes field extensions with analytic convergence. This is developed in [Cambridge III Local Fields](/page/Cambridge%20III%20Local%20Fields). Algebraically closed fields sit at the opposite end from arithmetic fields: they contain roots for every nonconstant polynomial, so polynomial equations can be studied geometrically through their solution sets. This is the field-theoretic gateway to algebraic geometry and model theory. ## References Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Cambridge II Number Fields](/page/Cambridge%20II%20Number%20Fields). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Androma, [Cambridge III Local Fields](/page/Cambridge%20III%20Local%20Fields). Michael Artin, *Algebra* (1991). Serge Lang, *Algebra* (2002). David S. Dummit and Richard M. Foote, *Abstract Algebra* (2004).