A field is a place where addition, subtraction, multiplication, and division behave as expected, but polynomial equations ask for more than arithmetic. The equation $x^2 - 2 = 0$ can be written over $\mathbb{Q}$, yet its solutions do not live in $\mathbb{Q}$. The equation $x^2 + 1 = 0$ can be written over $\mathbb{R}$, yet its solutions do not live in $\mathbb{R}$. In both cases the field is closed under its own arithmetic operations, but not closed under solving the algebraic equations its arithmetic naturally produces.

For any field $F$, the notation $F[x]$ means the ring of polynomials in one variable $x$ with coefficients in $F$. Thus $\mathbb{Q}[x]$, $\mathbb{R}[x]$, $\mathbb{C}[x]$, and later $k[x]$ all name polynomial rings over the indicated field.

[example: Two Polynomial Equations That Escape Their Fields] Let $f \in \mathbb{Q}[x]$ and $g \in \mathbb{R}[x]$ be given by \begin{align*} f(x)=x^2-2 \end{align*} and \begin{align*} g(x)=x^2+1. \end{align*} We first show that $f(x)=0$ has no solution in $\mathbb{Q}$. Suppose, for contradiction, that $r \in \mathbb{Q}$ and $r^2=2$. Write $r=p/q$ with $p,q \in \mathbb{Z}$, $q \ne 0$, and $\gcd(p,q)=1$. Then \begin{align*} \left(\frac{p}{q}\right)^2=2. \end{align*} Multiplying by $q^2$ gives \begin{align*} p^2=2q^2. \end{align*} Thus $p^2$ is even, so $p$ is even. Write $p=2m$. Substituting gives \begin{align*} (2m)^2=2q^2. \end{align*} Hence \begin{align*} 4m^2=2q^2. \end{align*} Dividing by $2$ gives \begin{align*} q^2=2m^2. \end{align*} Thus $q^2$ is even, so $q$ is even. This contradicts $\gcd(p,q)=1$, since both $p$ and $q$ are divisible by $2$. Therefore $x^2-2$ has no rational root. The equation $g(x)=0$ has no solution in $\mathbb{R}$. For every $x \in \mathbb{R}$, one has $x^2 \ge 0$, so \begin{align*} x^2+1 \ge 1. \end{align*} Therefore $x^2+1 \ne 0$ for every real $x$. After enlarging the fields, the missing roots appear. In $\mathbb{R}[x]$, \begin{align*} (x-\sqrt{2})(x+\sqrt{2})=x^2+\sqrt{2}x-\sqrt{2}x-(\sqrt{2})^2=x^2-2. \end{align*} Thus \begin{align*} x^2-2=(x-\sqrt{2})(x+\sqrt{2}) \quad \text{in } \mathbb{R}[x]. \end{align*} In $\mathbb{C}[x]$, using $i^2=-1$, \begin{align*} (x-i)(x+i)=x^2+ix-ix-i^2=x^2+1. \end{align*} Thus \begin{align*} x^2+1=(x-i)(x+i) \quad \text{in } \mathbb{C}[x]. \end{align*} The issue is not whether the polynomial can be written over the original field, but whether that field contains the elements needed to realize all of its roots. [/example]

An algebraically closed field is designed to end this process of adjoining roots. It is a field in which every nonconstant polynomial equation in one variable has a solution already inside the field. Once this happens, repeated division by linear factors forces every polynomial to decompose completely, so the field behaves as a complete algebraic universe for one-variable polynomial equations.

## Definition

The parent concept is a [field](/page/Field): a commutative ring in which every nonzero element is invertible. Algebraic closedness is an extra condition on a field, not a replacement for the field axioms. It asks whether the [polynomial ring](/page/Polynomial%20Ring) $k[x]$ creates equations whose solutions remain inside $k$.

[definition: Algebraically Closed Field] A field $k$ is algebraically closed if every nonconstant polynomial $f \in k[x]$ has at least one root in $k$. [/definition]

The definition asks only for one root, but the condition is much stronger than it first appears. Once a root $a \in k$ is found, the [factor theorem](/theorems/3235) removes the factor $x-a$. The central task is therefore to identify the other forms this same condition takes, because different arguments recognize algebraic closedness through different symptoms. In the theorem below, $k^\times$ denotes the set of nonzero elements of $k$, viewed as the multiplicative group of units of the field.

[quotetheorem:8422]

This theorem is the main working dictionary for algebraically closed fields. Depending on the context, algebraic closedness may appear as solvability of equations, factorisation of polynomials, absence of higher-degree irreducibles, or absence of nontrivial finite algebraic extensions. A dictionary is useful only if we have a model case, so the next theorem names the field in which this behaviour first becomes familiar.

[quotetheorem:347]

The theorem explains why complex numbers are the natural coefficient field for much of linear algebra, representation theory, and algebraic geometry. Characteristic polynomials, eigenvalues, and polynomial equations stop requiring repeated field extensions once the base field is $\mathbb{C}$. Comparing this with smaller fields shows which failures algebraic closedness repairs.

[example: Real and Rational Fields Are Not Algebraically Closed] The polynomial $x^2+1$ is nonconstant and belongs to $\mathbb{R}[x]$. For every $x \in \mathbb{R}$, the square $x^2$ is nonnegative, so \begin{align*} x^2+1 \ge 1. \end{align*} Hence $x^2+1 \ne 0$ for every $x \in \mathbb{R}$, so $\mathbb{R}$ is not algebraically closed. The polynomial $x^2-2$ is nonconstant and belongs to $\mathbb{Q}[x]$. We show that it has no rational root. Suppose, for contradiction, that $r \in \mathbb{Q}$ and \begin{align*} r^2-2=0. \end{align*} Then $r^2=2$. Write $r=p/q$ with $p,q \in \mathbb{Z}$, $q \ne 0$, and $\gcd(p,q)=1$. Substituting $r=p/q$ gives \begin{align*} \left(\frac{p}{q}\right)^2=2. \end{align*} Since $q \ne 0$, multiplying by $q^2$ gives \begin{align*} p^2=2q^2. \end{align*} Thus $p^2$ is even. If $p$ were odd, then $p=2n+1$ for some $n \in \mathbb{Z}$, and \begin{align*} p^2=(2n+1)^2=4n^2+4n+1=2(2n^2+2n)+1, \end{align*} which is odd. Therefore $p$ is even, so $p=2m$ for some $m \in \mathbb{Z}$. Substituting into $p^2=2q^2$ gives \begin{align*} (2m)^2=2q^2. \end{align*} Expanding the left side gives \begin{align*} 4m^2=2q^2. \end{align*} Dividing by $2$ gives \begin{align*} q^2=2m^2. \end{align*} Thus $q^2$ is even, and the same parity argument shows that $q$ is even. Hence both $p$ and $q$ are divisible by $2$, contradicting $\gcd(p,q)=1$. Therefore $x^2-2$ has no root in $\mathbb{Q}$, so $\mathbb{Q}$ is not algebraically closed. The two failures are different: $\mathbb{R}$ misses roots because its order forces every square to be nonnegative, while $\mathbb{Q}$ misses roots because solving $x^2=2$ requires a number outside the rational field. [/example]

Before the later geometric viewpoint, we will also use standard ideal notation. In a polynomial ring, an ideal is a subset closed under addition and closed under multiplication by arbitrary polynomials from the ring. The notation $(h)$ means the ideal generated by $h$, namely all polynomial multiples of $h$; the unit ideal is the whole ring, and a proper ideal is an ideal that is not the whole ring.

## Roots, Splitting, and Factorisation

### Roots as Linear Factors

Algebraic closedness is often used through factorisation rather than by directly solving equations. To move between roots and factors, we first need a precise notion of a root relative to the chosen field. This matters because the same formula may have roots after enlarging the field even when it has none in the original field.

[definition: Root of a Polynomial] Let $k$ be a field and let $f \in k[x]$. An element $a \in k$ is a root of $f$ if $f(a)=0$. [/definition]

A root by itself is a solution of an equation, but factorisation needs a polynomial object that records the same solution. The obstruction is that a numerical equality $f(a)=0$ does not by itself say how much of $f$ remains after the solution $a$ has been accounted for. The relevant algebraic object is the degree-one divisor $x-a$, which lets the root be removed from the polynomial ring by division.

[definition: Linear Factor] Let $k$ be a field. A linear factor of a polynomial $f \in k[x]$ is a polynomial of the form $x-a$ with $a \in k$ such that $f=(x-a)q$ for some $q \in k[x]$. [/definition]