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] Linear factors matter because they reduce a polynomial equation one root at a time. Finding a root gives a division, and the quotient records the remaining roots still to be found. The basic theorem connecting these two viewpoints is the tool that makes algebraic closedness usable in computations. [quotetheorem:3235] After the factor theorem, the natural endpoint is a polynomial that has been decomposed as far as possible into linear factors. One root only removes one factor; the remaining quotient may still contain an irreducible piece of higher degree. We therefore need a term for the case where no such obstruction remains and every root needed for the full factorisation already lies in the chosen field. [definition: Splitting over a Field] Let $k$ be a field and let $f \in k[x]$ have degree $n \ge 1$. The polynomial $f$ splits over $k$ if there exist $c,a_1,\ldots,a_n \in k$ with $c \ne 0$ such that \begin{align*} f(x) = c\prod_{i=1}^{n}(x-a_i). \end{align*} [/definition] Splitting is stronger than having one root. A cubic over $k$ may have one root in $k$ and an irreducible quadratic factor left over. Algebraically closed fields rule out this half-success, and the following example shows the distinction in a small calculation. [example: A Polynomial With One Root But Not All Roots] In $\mathbb{Q}[x]$, consider \begin{align*} f(x)=x^3-2x^2+x-2. \end{align*} Group the first two terms and the last two terms: \begin{align*} x^3-2x^2+x-2=x^2(x-2)+1(x-2). \end{align*} Factoring out the common factor $x-2$ gives \begin{align*} x^2(x-2)+1(x-2)=(x^2+1)(x-2). \end{align*} Thus \begin{align*} f(x)=(x-2)(x^2+1). \end{align*} This factorization shows that $2$ is a rational root, because \begin{align*} f(2)=(2-2)(2^2+1)=0\cdot 5=0. \end{align*} The remaining factor has no rational root: if $r \in \mathbb{Q}$, then $r^2 \ge 0$, so \begin{align*} r^2+1\ge 1. \end{align*} Hence $r^2+1 \ne 0$ for every $r \in \mathbb{Q}$. Therefore $x^2+1$ does not split into linear factors over $\mathbb{Q}$, and $f$ does not split over $\mathbb{Q}$ even though it has the rational root $2$. Over $\mathbb{C}$, the element $i$ satisfies $i^2=-1$. Expanding the two complex linear factors gives \begin{align*} (x-i)(x+i)=x^2+ix-ix-i^2=x^2-(-1)=x^2+1. \end{align*} Substituting this into the factorization of $f$ gives \begin{align*} f(x)=(x-2)(x-i)(x+i). \end{align*} Thus the same polynomial has one visible root over $\mathbb{Q}$, but all of its roots appear only after enlarging the field to $\mathbb{C}$. [/example] ### Irreducibility as an Obstruction The irreducible polynomials over a field measure the ways in which factorisation can get stuck. If algebraic closedness means factorisation never gets stuck beyond degree $1$, then irreducibles provide a compact test for the whole condition. [quotetheorem:3319] This theorem is a useful practical test. If a calculation over $k$ produces an irreducible polynomial of degree greater than $1$, then $k$ was not algebraically closed. The next section turns that obstruction around: instead of staying inside $k$, we enlarge $k$ until those irreducibles disappear. ## Algebraic Closures and Minimal Enlargements Most fields encountered at first are not algebraically closed, so the next question is how to enlarge a field without adding unnecessary transcendental information. The goal is to add exactly the algebraic elements needed to solve all polynomial equations. To state this minimality condition, we first distinguish algebraic extensions from arbitrary extensions. [definition: Algebraic Extension] Let $K/k$ be a [field extension](/page/Field%20Extension). The extension $K/k$ is algebraic if every element $\alpha \in K$ is a root of some nonzero polynomial $f \in k[x]$. [/definition] Algebraic extensions are controlled by polynomial equations over the base field. They exclude elements such as an indeterminate $t$ over $k$, which is not forced by any nonzero polynomial relation with coefficients in $k$. The desired closure of $k$ should be algebraically closed while still remaining algebraic over $k$. [definition: Algebraic Closure of a Field] Let $k$ be a field. An [algebraic closure](/page/Algebraic%20Closure) of $k$ is a field extension $\overline{k}/k$ such that $\overline{k}$ is algebraically closed and $\overline{k}/k$ is algebraic. [/definition] The definition would be of limited use if such fields did not always exist. The next theorem answers the existence problem: every field, no matter how many polynomial equations it currently fails to solve, has an algebraic enlargement that solves them all without adding transcendental elements. [quotetheorem:1313] Existence is a promise that every field can be completed algebraically. The construction is not canonical on the nose, because choices are involved. For the notation $\overline{k}$ to be meaningful across different constructions, we also need to know that the answer is unique up to an isomorphism fixing $k$. [quotetheorem:1314] The uniqueness theorem justifies writing $\overline{k}$ as though there were a single object. Different constructions may produce different fields, but they are indistinguishable from the viewpoint of algebra over $k$ once an isomorphism fixing $k$ is chosen. The rational numbers provide the standard example of an algebraic closure that is large but still much smaller than the surrounding complex field. [example: The Algebraic Closure of $\mathbb{Q}$ inside $\mathbb{C}$] Inside $\mathbb{C}$, define \begin{align*} \overline{\mathbb{Q}}=\{z\in \mathbb{C}: f(z)=0 \text{ for some nonzero } f\in \mathbb{Q}[x]\}. \end{align*} The element $\sqrt{2}$ belongs to this set because it is a root of $x^2-2\in \mathbb{Q}[x]$: \begin{align*} (\sqrt{2})^2-2=2-2=0. \end{align*} The element $i$ belongs to this set because it is a root of $x^2+1\in \mathbb{Q}[x]$: \begin{align*} i^2+1=-1+1=0. \end{align*} More generally, if $\zeta\in \mathbb{C}$ is an $n$th root of unity, then $\zeta^n=1$, so \begin{align*} \zeta^n-1=0. \end{align*} Thus $\zeta$ is a root of $x^n-1\in \mathbb{Q}[x]$, and hence $\zeta\in \overline{\mathbb{Q}}$. This set is a field. If $\alpha,\beta\in \overline{\mathbb{Q}}$, then $\alpha$ and $\beta$ are algebraic over $\mathbb{Q}$ by the definition of the set. The field $\mathbb{Q}(\alpha,\beta)$ is a finite extension of $\mathbb{Q}$ by the [tower law](/theorems/1248) for finite algebraic extensions. Therefore $\alpha+\beta$, $\alpha-\beta$, $\alpha\beta$, and, if $\beta\ne 0$, $\alpha/\beta$ all lie in the finite-dimensional $\mathbb{Q}$-[vector space](/page/Vector%20Space) $\mathbb{Q}(\alpha,\beta)$. If $u\in \mathbb{Q}(\alpha,\beta)$ and $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]=n$, then the $n+1$ elements \begin{align*} 1,u,u^2,\ldots,u^n \end{align*} are linearly dependent over $\mathbb{Q}$, so there exist rational numbers $a_0,\ldots,a_n$, not all zero, such that \begin{align*} a_0+a_1u+a_2u^2+\cdots+a_nu^n=0. \end{align*} Hence $u$ is a root of the nonzero polynomial $a_0+a_1x+\cdots+a_nx^n\in \mathbb{Q}[x]$, so $u\in \overline{\mathbb{Q}}$. The extension $\overline{\mathbb{Q}}/\mathbb{Q}$ is algebraic because every element of $\overline{\mathbb{Q}}$ is, by definition, a root of some nonzero polynomial in $\mathbb{Q}[x]$. It is also algebraically closed. Let \begin{align*} h(x)=c_0+c_1x+\cdots+c_nx^n\in \overline{\mathbb{Q}}[x] \end{align*} with $n\ge 1$ and $c_n\ne 0$. Each coefficient $c_j$ is algebraic over $\mathbb{Q}$, so the field $F=\mathbb{Q}(c_0,\ldots,c_n)$ is a finite algebraic extension of $\mathbb{Q}$. By the *[Fundamental Theorem of Algebra](/theorems/347)*, the polynomial $h$ has a complex root $\alpha\in \mathbb{C}$. Since $h\in F[x]$, this root $\alpha$ is algebraic over $F$. Since $F/\mathbb{Q}$ is algebraic and $\alpha/F$ is algebraic, transitivity of algebraicity gives that $\alpha$ is algebraic over $\mathbb{Q}$. Therefore $\alpha\in \overline{\mathbb{Q}}$, so every nonconstant polynomial over $\overline{\mathbb{Q}}$ has a root in $\overline{\mathbb{Q}}$. The set still does not contain every complex number. If $z\in \mathbb{C}$ is transcendental over $\mathbb{Q}$, then no nonzero polynomial in $\mathbb{Q}[x]$ vanishes at $z$, so $z\notin \overline{\mathbb{Q}}$. Since $\pi$ and $e$ are transcendental over $\mathbb{Q}$, they are not elements of $\overline{\mathbb{Q}}$. Thus $\overline{\mathbb{Q}}$ contains exactly the complex numbers forced by algebraic equations with rational coefficients, not all complex numbers. [/example] The algebraic closure is usually infinite even when the original field is small. It gathers roots of all degrees and all irreducible polynomials at once, rather than solving a single equation. ## Finite Fields, Real Fields, and Standard Examples ### Finite Fields Algebraic closedness behaves differently in different characteristics. Finite fields give a sharp obstruction: a finite set cannot contain roots for every polynomial. The obstruction is not subtle; a single polynomial can be built to avoid every element of the field. [quotetheorem:8423] The theorem is a warning against confusing good factorisation behaviour for some polynomials with algebraic closedness for all polynomials. Finite fields have finite splitting fields for individual polynomials, but not a finite field that splits every polynomial. The construction behind the obstruction is explicit. [example: A Polynomial With No Root in a Finite Field] Let $k=\mathbb{F}_q$, so $k$ has exactly $q$ elements. Define \begin{align*} f(x)=1+\prod_{a \in k}(x-a)\in k[x]. \end{align*} The product is finite because $k$ is finite, and it has degree $q$, so $f$ is nonconstant. We show that no element of $k$ is a root of $f$. Let $b\in k$. In the product \begin{align*} \prod_{a\in k}(b-a), \end{align*} one of the factors is obtained by taking $a=b$. That factor is \begin{align*} b-b=0. \end{align*} Since a product with a zero factor is zero, \begin{align*} \prod_{a\in k}(b-a)=0. \end{align*} Therefore \begin{align*} f(b)=1+\prod_{a\in k}(b-a)=1+0=1. \end{align*} Thus $f(b)\ne 0$ for every $b\in k$, so the nonconstant polynomial $f\in k[x]$ has no root in $k$. Hence $k$ is not algebraically closed. [/example] ### Real Closed Fields The [real numbers](/page/Real%20Numbers) fail for a different reason. Their order is compatible with multiplication, so squares are nonnegative. This suggests a class of fields that are not algebraically closed but become algebraically closed after adjoining a square root of $-1$. [definition: Real Closed Field] A field $k$ is real closed if $k$ admits an ordering for which every positive element is a square and every polynomial in $k[x]$ of odd degree has a root in $k$. [/definition] This ordered-field definition captures the algebraic content of the real line without naming limits or completeness. It has a useful equivalent form: a field $k$ is real closed exactly when $k$ is not algebraically closed and $k(i)$ is algebraically closed, where $i$ is an element in an extension field satisfying $i^2=-1$. The next theorem uses this equivalent viewpoint by identifying $k(i)$ not merely as an algebraically closed extension, but as an algebraic closure of $k$. [quotetheorem:8424] This result explains why the passage from $\mathbb{R}$ to $\mathbb{C}$ is so efficient. For real closed fields, one element is enough to reach algebraic closedness. The usual complex numbers are the guiding example. [example: Complex Numbers as the Algebraic Closure of the Reals] Let $i$ be an element satisfying $i^2=-1$. The polynomial $x^2+1$ has no root in $\mathbb{R}$, because for every $x\in\mathbb{R}$ one has $x^2\ge 0$, hence \begin{align*} x^2+1\ge 1. \end{align*} Thus $x^2+1\ne 0$ for every real $x$, so adjoining $i$ produces a genuine extension $\mathbb{R}(i)/\mathbb{R}$. Every element of $\mathbb{R}(i)$ can be written as $a+bi$ with $a,b\in\mathbb{R}$. Indeed, since $i^2=-1$, every power of $i$ reduces to either a real multiple of $1$ or a real multiple of $i$: for example, \begin{align*} i^2=-1,\quad i^3=-i,\quad i^4=1. \end{align*} Multiplication in this form is \begin{align*} (a+bi)(c+di)=ac+adi+bci+bdi^2=(ac-bd)+(ad+bc)i. \end{align*} This is exactly the usual multiplication rule for complex numbers, so the map \begin{align*} a+bi\mapsto a+bi\in\mathbb{C} \end{align*} identifies $\mathbb{R}(i)$ with $\mathbb{C}$. The extension $\mathbb{R}(i)/\mathbb{R}$ is algebraic. If $b=0$, then $a+bi=a$ is a root of $x-a\in\mathbb{R}[x]$. If $b\ne 0$, put $z=a+bi$. Then \begin{align*} (z-a)^2+b^2=(bi)^2+b^2=b^2i^2+b^2=-b^2+b^2=0. \end{align*} Thus $z$ is a root of $(x-a)^2+b^2\in\mathbb{R}[x]$. By the *Fundamental Theorem of Algebra*, every nonconstant polynomial in $\mathbb{C}[x]$ has a root in $\mathbb{C}$. Since $\mathbb{R}(i)\cong\mathbb{C}$, every nonconstant polynomial over $\mathbb{R}(i)$ has a root in $\mathbb{R}(i)$. Therefore $\mathbb{R}(i)$ is algebraically closed and algebraic over $\mathbb{R}$, so it is an algebraic closure of $\mathbb{R}$. [/example] In contrast, the rational numbers need infinitely many algebraic adjunctions. Adding $\sqrt{2}$ solves $x^2-2$, but leaves $x^2-3$ unsolved; adding both still leaves many other polynomials unsplit. This shows why algebraic closures are generally much larger than a single simple extension. [example: No Single Quadratic Extension Closes $\mathbb{Q}$] The field $\mathbb{Q}(\sqrt{2})$ contains the two roots of $x^2-2$, because $\sqrt{2}\in \mathbb{Q}(\sqrt{2})$, $-\sqrt{2}\in \mathbb{Q}(\sqrt{2})$, and \begin{align*} (\sqrt{2})^2-2=2-2=0. \end{align*} Also \begin{align*} (-\sqrt{2})^2-2=2-2=0. \end{align*} However, adjoining these roots does not make the field algebraically closed. We show that $x^3-2$ has no root in $\mathbb{Q}(\sqrt{2})$. First, $x^3-2$ is irreducible over $\mathbb{Q}$ by *[Eisenstein's criterion](/theorems/859)* with the prime $2$: the prime $2$ divides the coefficients $0$, $0$, and $-2$, it does not divide the leading coefficient $1$, and $2^2=4$ does not divide $-2$. Therefore, if $\alpha$ is a root of $x^3-2$, the [minimal polynomial](/page/Minimal%20Polynomial) of $\alpha$ over $\mathbb{Q}$ is $x^3-2$, so \begin{align*} [\mathbb{Q}(\alpha):\mathbb{Q}]=3. \end{align*} On the other hand, \begin{align*} \mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}:a,b\in \mathbb{Q}\}. \end{align*} The two elements $1$ and $\sqrt{2}$ are linearly independent over $\mathbb{Q}$: if $a+b\sqrt{2}=0$ with $a,b\in\mathbb{Q}$ and $b\ne 0$, then \begin{align*} \sqrt{2}=-\frac{a}{b}\in\mathbb{Q}, \end{align*} contradicting the irrationality of $\sqrt{2}$. Hence $\{1,\sqrt{2}\}$ is a $\mathbb{Q}$-basis of $\mathbb{Q}(\sqrt{2})$, and \begin{align*} [\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2. \end{align*} Suppose, for contradiction, that $x^3-2$ had a root $\alpha\in \mathbb{Q}(\sqrt{2})$. Then \begin{align*} \mathbb{Q}\subseteq \mathbb{Q}(\alpha)\subseteq \mathbb{Q}(\sqrt{2}). \end{align*} By the *Tower Law*, \begin{align*} [\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2}):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]. \end{align*} Substituting the known degree $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$ gives \begin{align*} 2=3[\mathbb{Q}(\sqrt{2}):\mathbb{Q}(\alpha)]. \end{align*} The right side is a multiple of $3$, while the left side is $2$, which is impossible. Therefore $x^3-2$ has no root in $\mathbb{Q}(\sqrt{2})$. Thus $\mathbb{Q}(\sqrt{2})$ solves the equation $x^2-2=0$, but it still misses a root of the polynomial $x^3-2$; solving one polynomial usually does not solve all polynomial equations over the field. [/example] ## Polynomial Ideals and Algebraic Geometry ### Points from Polynomial Equations The reason algebraically closed fields dominate algebraic geometry is that polynomial equations are supposed to describe geometric sets. If the field is not algebraically closed, an equation can be nontrivial but have no visible points, and the dictionary between ideals and solution sets becomes distorted. We first name the ambient space in which those solutions are sought. [definition: Affine Space over a Field] Let $k$ be a field and let $n \in \mathbb{N}$. The affine $n$-space over $k$ is the set \begin{align*} \mathbb{A}^n_k = k^n. \end{align*} [/definition] Affine space is the ambient set in which polynomial equations are solved. A single polynomial equation usually imposes a condition on many possible points, and a system of equations asks for the points satisfying all of those conditions at once. The obstruction is that this common solution set may be empty or too small over the chosen field, so we need precise notation for the points actually visible in $k^n$. [definition: Zero Set of Polynomials] Let $k$ be a field and let $S \subset k[x_1,\ldots,x_n]$. The zero set of $S$ in $\mathbb{A}^n_k$ is \begin{align*} V(S)=\{a=(a_1,\ldots,a_n) \in k^n : f(a)=0 \text{ for every } f \in S\}. \end{align*} [/definition] The zero set construction can lose information over fields that are not algebraically closed. The polynomial $x^2+1$ defines an empty zero set over $\mathbb{R}$, but not because the polynomial is a unit in $\mathbb{R}[x]$. [example: Empty Real Zero Set Without Unit Ideal] In $\mathbb{R}[x]$, let $S=\{x^2+1\}$. By the definition of zero set, \begin{align*} V(S)=\{a\in \mathbb{R}:a^2+1=0\}. \end{align*} For every $a\in \mathbb{R}$, the square $a^2$ is nonnegative, so \begin{align*} a^2+1\ge 1. \end{align*} Hence $a^2+1\ne 0$ for every $a\in \mathbb{R}$, and therefore \begin{align*} V(S)=\varnothing \subsetneq \mathbb{A}^1_{\mathbb{R}}. \end{align*} The ideal $(x^2+1)$ is nevertheless a proper ideal of $\mathbb{R}[x]$. If it were all of $\mathbb{R}[x]$, then $1\in (x^2+1)$, so there would be some $h(x)\in \mathbb{R}[x]$ such that \begin{align*} 1=(x^2+1)h(x). \end{align*} If $h(x)=0$, the right side is $0$, not $1$. If $h(x)\ne 0$, then the leading term of $(x^2+1)h(x)$ is obtained by multiplying $x^2$ by the leading term of $h(x)$, so \begin{align*} \deg((x^2+1)h)=\deg h+2\ge 2. \end{align*} This contradicts $\deg 1=0$. Thus $1\notin (x^2+1)$, so $(x^2+1)$ is proper. Over $\mathbb{C}$, the same polynomial has visible roots. Since $i^2=-1$, \begin{align*} i^2+1=-1+1=0. \end{align*} Also, \begin{align*} (-i)^2+1=i^2+1=-1+1=0. \end{align*} If $z\in\mathbb{C}$ and $z^2+1=0$, then \begin{align*} z^2-i^2=z^2+1=0. \end{align*} Factoring the difference of squares gives \begin{align*} (z-i)(z+i)=0. \end{align*} Because $\mathbb{C}$ is a field, this implies $z-i=0$ or $z+i=0$, so $z=i$ or $z=-i$. Therefore the complex zero set is $\{i,-i\}$. The polynomial defines no real point, but after passing to $\mathbb{C}$ its algebraic content appears as two points. [/example] ### The Nullstellensatz Viewpoint For a field $k$, the notation $k[x]$ means the ring of polynomials in one variable with coefficients in $k$; more generally, $k[T_1,\dots,T_n]$ is the polynomial ring in $n$ variables. An ideal is a collection of polynomials closed under addition and under multiplication by arbitrary polynomials from the ring. A proper ideal is one that is not the whole ring, and a maximal ideal is a proper ideal that is not contained in any larger proper ideal. The notation $\operatorname{mspec}(R)$ denotes the set of maximal ideals of a ring $R$. These definitions let us state the geometric question precisely: if a system of polynomial equations is algebraically consistent, meaning its ideal is proper, must there be a point where all those polynomials vanish? The [weak Nullstellensatz](/theorems/2123) is the first formal statement that algebraically closed fields repair the gap between proper algebraic conditions and visible solutions. Over an algebraically closed field, a proper ideal cannot vanish nowhere. [quotetheorem:2123] This theorem is one of the main reasons algebraically closed fields are the default base fields in classical algebraic geometry. Proper algebraic conditions produce actual points, so geometry can see algebra. The next refinement asks not only whether solutions exist, but how the algebraic objects that behave like points are classified. In the language of ideals, maximal ideals of polynomial rings are the algebraic stand-ins for points. Over an algebraically closed field $\Omega$, the maximal ideals of $\Omega[T_1,\dots,T_n]$ are exactly the point ideals $(T_1-a_1,\dots,T_n-a_n)$ for tuples $(a_1,\dots,a_n)\in\Omega^n$. Equivalently, $\operatorname{mspec}(\Omega[T_1,\dots,T_n])$ can be identified with affine space $\Omega^n$. Thus maximal ideals are points when the base field is algebraically closed. Without that hypothesis, maximal ideals can also encode hidden algebraic extensions of the base field. ## Galois-Theoretic Consequences ### Algebraic Extensions Above Closed Fields From the viewpoint of field extensions, an algebraically closed field has no algebraic room left above it. This makes it a terminal object for algebraic extension problems and simplifies the behaviour of automorphism groups. The first consequence states this terminal property directly. [quotetheorem:3320] This statement does not say that $k$ has no larger fields at all. Transcendental extensions such as $k(t)$ still exist. It says that any larger field must contain elements not governed by polynomial equations over $k$. [example: A Transcendental Extension Still Exists] Let $k$ be algebraically closed and let $t$ be an indeterminate. We regard $k$ as a subfield of $k(t)$ by sending each $a\in k$ to the constant rational function $a/1$. This extension is proper because $t\notin k$: if $t=a$ for some $a\in k$, then the polynomial \begin{align*} t-a\in k[t] \end{align*} would be zero, but its coefficient of $t$ is $1$, so it is not the zero polynomial. We show that $k(t)/k$ is not algebraic by proving that $t$ is not algebraic over $k$. Suppose, for contradiction, that $t$ were algebraic over $k$. Then there would be a nonzero polynomial \begin{align*} f(x)=c_0+c_1x+\cdots+c_nx^n\in k[x] \end{align*} such that $f(t)=0$. Substituting $t$ gives \begin{align*} f(t)=c_0+c_1t+\cdots+c_nt^n\in k[t]. \end{align*} Since $t$ is an indeterminate, a polynomial in $k[t]$ is zero exactly when all of its coefficients are zero. Therefore \begin{align*} c_0=c_1=\cdots=c_n=0. \end{align*} This contradicts that $f$ was nonzero. Hence $t$ is not algebraic over $k$, so $k(t)/k$ is not an algebraic extension. Thus algebraic closedness says that no new algebraic elements are needed above $k$, but it does not prevent larger transcendental fields such as $k(t)$. [/example] ### Separability and Characteristic $p$ In positive characteristic, it is useful to separate two notions: having no separable algebraic extensions and having no algebraic extensions at all. Galois theory sees separable extensions most directly, so the separable analogue of algebraic closedness deserves its own name. [definition: Separably Closed Field] A field $k$ is separably closed if it has no nontrivial finite [separable extension](/page/Separable%20Extension). [/definition] The definition filters algebraic extensions through the part most visible to Galois theory. A possible source of confusion is that separable closedness only forbids separable finite extensions, while algebraic closedness forbids all finite algebraic extensions. This raises the first structural comparison between the two notions: does solving every finite algebraic equation automatically rule out the separable extensions detected by Galois theory? Since every finite separable extension is finite and algebraic, the next formal result records algebraic closedness as a sufficient condition for separable closedness. [quotetheorem:8425] The distinction matters most in characteristic $p>0$. A field can have all separable algebraic equations solved while still admitting purely inseparable elements above it. The following example exhibits the kind of polynomial responsible for that gap. [example: Purely Inseparable Obstruction] Let $k$ have characteristic $p>0$, and let $a\in k$ be an element which is not a $p$-th power in $k$. We first check that \begin{align*} x^p-a\in k[x] \end{align*} has no root in $k$. If $b\in k$ were a root, then \begin{align*} b^p-a=0. \end{align*} Hence \begin{align*} b^p=a, \end{align*} which says exactly that $a$ is a $p$-th power in $k$, contrary to the assumption. Now adjoin an element $\alpha$ satisfying $\alpha^p=a$. In the extension field containing $\alpha$, we compute \begin{align*} (x-\alpha)^p=\sum_{j=0}^{p}\binom{p}{j}x^{p-j}(-\alpha)^j. \end{align*} For $1\le j\le p-1$, the prime $p$ divides $\binom{p}{j}$, so these middle coefficients are $0$ in characteristic $p$. Therefore only the terms with $j=0$ and $j=p$ remain: \begin{align*} (x-\alpha)^p=x^p+(-\alpha)^p. \end{align*} Since $p$ is prime, $(-\alpha)^p=-\alpha^p$ when $p$ is odd, and the same identity also holds when $p=2$ because $-1=1$ in characteristic $2$. Thus \begin{align*} (x-\alpha)^p=x^p-\alpha^p. \end{align*} Using $\alpha^p=a$, this becomes \begin{align*} (x-\alpha)^p=x^p-a. \end{align*} Let $m_\alpha(x)\in k[x]$ be the minimal polynomial of $\alpha$ over $k$. Since $x^p-a$ vanishes at $\alpha$, the minimal polynomial $m_\alpha$ divides $x^p-a$ in $k[x]$. After passing to an extension containing $\alpha$, the identity above shows that \begin{align*} x^p-a=(x-\alpha)^p. \end{align*} Hence every root of $m_\alpha$ is equal to $\alpha$, so $m_\alpha$ has only one geometric root. Equivalently, the algebraic relation defining $\alpha$ is purely inseparable: it creates an algebraic extension $k(\alpha)/k$, but it does not create a new separable root. This is why controlling only separable extensions is weaker than algebraic closedness. [/example] For perfect fields, the distinction disappears because every algebraic extension is separable. This includes fields of characteristic $0$ and finite fields, though finite fields are not separably closed. In this setting there is no purely inseparable obstruction left over: any finite algebraic extension that exists is already visible to separability. Thus, for perfect fields, ruling out separable finite extensions rules out finite algebraic extensions altogether. [quotetheorem:8426] The theorem shows that algebraic closedness can sometimes be checked by looking only at separable extensions. In characteristic $0$, this is especially natural because algebraic extensions are separable. ## Beyond and Connected Topics Algebraically closed fields sit at the meeting point of field theory, linear algebra, commutative algebra, and geometry. In linear algebra, working over an algebraically closed field ensures that every matrix has at least one eigenvalue, because its [characteristic polynomial](/page/Characteristic%20Polynomial) has a root. This is the setting in which [Jordan normal form](/theorems/864) and many classification theorems take their cleanest form; see Androma's [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra) for the surrounding linear algebra. In field theory and Galois theory, algebraic closures provide the ambient field in which all algebraic extensions of $k$ may be compared. The [Galois group](/page/Galois%20Group) of $\overline{k}/k$ packages the symmetries of all algebraic equations over $k$ at once. The background on fields, rings, and modules is developed in [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). In commutative algebra, algebraically closed fields are the base case for the Nullstellensatz, where maximal ideals of polynomial rings become points of affine space. This is the bridge from ideal theory to geometry, and it is one reason algebraic geometry often begins over an algebraically closed field. For the ring-theoretic foundations, see [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). In representation theory and Lie theory, algebraically closed fields remove polynomial obstructions to diagonalisation and weight decompositions, although additional hypotheses such as characteristic $0$ or semisimplicity may still be needed. This perspective connects naturally with [Lie Algebras I: Foundations](/page/Lie%20Algebras%20I%3A%20Foundations), where the choice of base field affects the structure theory. ## References Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Androma, [Lie Algebras I: Foundations](/page/Lie%20Algebras%20I%3A%20Foundations). Atiyah and Macdonald, *Introduction to Commutative Algebra* (1969). Lang, *Algebra* (2002). Dummit and Foote, *Abstract Algebra* (2004).