Algebraic field extension, degree, splitting field

Intuition [Beginner]

Some equations have no solution in the number system you start with. The equation $x^2=2$ has no solution in the rational numbers. A field extension solves this by enlarging the number system: add a single new number called $\sqrt{2}$ whose square is $2$, together with all combinations like $1+3\sqrt{2}$ and $5-7\sqrt{2}$ that the new number generates. The result is written $\mathbb{Q}(\sqrt{2})$.

The degree of an extension counts how many new dimensions the enlarged system has over the original one. Think of $\mathbb{Q}(\sqrt{2})$ as a 2-dimensional space over $\mathbb{Q}$: every element is uniquely $a+b\sqrt{2}$ for rational $a$ and $b$, so the basis has two elements, $\{1,\sqrt{2}\}$. The degree is $2$.

A splitting field is the smallest extension in which a given polynomial breaks apart entirely into linear factors. The polynomial $x^2-2$ splits as $(x-\sqrt{2})(x+\sqrt{2})$ in $\mathbb{Q}(\sqrt{2})$, which is its splitting field.

Why does this concept exist? To understand which equations can be solved by formulas and which cannot, one must measure exactly how many new numbers are needed and how they relate.

Visual [Beginner]

A plane with the horizontal axis labelled $1$ and the vertical axis labelled $\sqrt{2}$. Every point in the plane represents one element $a+b\sqrt{2}$ of $\mathbb{Q}(\sqrt{2})$, with $a$ on the horizontal and $b$ on the vertical. The horizontal axis alone is the original field $\mathbb{Q}$.

The plane captures the degree: two independent directions over $\mathbb{Q}$, hence degree $2$.

Worked example [Beginner]

Consider the polynomial $x^2-2$ over the rational numbers $\mathbb{Q}$. Its roots are $\sqrt{2}$ and $-\sqrt{2}$, which are not rational.

Step 1. Form the extension $\mathbb{Q}(\sqrt{2})$ by adjoining $\sqrt{2}$ to $\mathbb{Q}$. Every element has the form $a+b\sqrt{2}$ with $a,b\in \mathbb{Q}$.

Step 2. Find the degree. The set $\{1,\sqrt{2}\}$ spans $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$ and is linearly independent (no rational multiple of $1$ equals a non-zero rational multiple of $\sqrt{2}$, because $\sqrt{2}$ is irrational). So the degree $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$.

Step 3. Factor the polynomial. In $\mathbb{Q}(\sqrt{2})$, the polynomial $x^2-2=(x-\sqrt{2})(x+\sqrt{2})$. Both roots are present, so this is the splitting field.

What this tells us: a degree-$2$ extension suffices to split a quadratic with irrational roots.

Check your understanding [Beginner]

Formal definition [Intermediate+]

Throughout, $F$ is a field ^{[Lang Algebra Ch. V]}. A field extension $E/F$ is an inclusion of fields $F\subseteq E$, or more generally a field homomorphism $\iota :F\to E$ (in which case $E$ is an $F$-algebra via $\iota$). We write $E/F$ and call $E$ an extension field of $F$.

Definition (Degree). The degree $[E:F]$ of an extension $E/F$ is the dimension of $E$ as an $F$-vector space ^{[Dummit-Foete 2004]}: $$ [E] = \dim_F E. $$ If $[E:F]<\infty$, the extension is finite; otherwise it is infinite.

Definition (Algebraic element). An element $\alpha \in E$ is algebraic over $F$ if there exists a nonzero polynomial $f\in F[x]$ with $f(\alpha )=0$. Otherwise $\alpha$ is transcendental over $F$. The extension $E/F$ is algebraic if every element of $E$ is algebraic over $F$.

Definition (Minimal polynomial). If $\alpha$ is algebraic over $F$, the minimal polynomial $m_{\alpha ,F}(x)\in F[x]$ is the unique monic polynomial of least degree satisfying $m_{\alpha ,F}(\alpha )=0$. It is irreducible over $F$, and $[F(\alpha ):F]=\mathrm{deg}{m}_{\alpha ,F}$.

Definition (Splitting field). Let $f\in F[x]$. An extension $E/F$ is a splitting field for $f$ over $F$ if:

1. $f$ factors into linear factors over $E$: $f(x)=c(x-{\alpha }_1)(x-{\alpha }_2)\cdots (x-{\alpha }_n)$ with ${\alpha }_i\in E$.
2. $E=F({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n)$, i.e., no proper subfield of $E$ containing $F$ also splits $f$.

Counterexamples to common slips [Intermediate+]

• A finite extension need not be a splitting field for any single polynomial. The extension $\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}$ has degree $4$ but does not split $x^4-2$, because the roots $\pm i\sqrt[4]{2}$ are missing. The splitting field of $x^4-2$ is $\mathbb{Q}(\sqrt[4]{2},i)$, which has degree $8$.

• Algebraic does not imply finite. The field of algebraic numbers $\overline{\mathbb{Q}}$ is algebraic over $\mathbb{Q}$ but has infinite degree $[\overline{\mathbb{Q}}:\mathbb{Q}]$.

• A transcendental extension can be finite. This never happens: if $\alpha$ is transcendental over $F$, then $[F(\alpha ):F]$ is infinite. The implication runs one way only — finite implies algebraic, but not conversely.

Key theorem with proof [Intermediate+]

Theorem (Tower law / Degree multiplicativity). Let $F\subseteq K\subseteq E$ be fields. Then $$ [E] = [E][K], $$ where the product of infinite cardinals is interpreted as $\infty$. In particular, if any two of the three degrees are finite, so is the third.

Proof. Let $\{e_i{\}}_{i\in I}$ be a basis for $E$ over $K$ and let $\{k_j{\}}_{j\in J}$ be a basis for $K$ over $F$. We prove that $\{e_i{k}_j{\}}_{(i,j)\in I\times J}$ is a basis for $E$ over $F$.

Spanning. Let $x\in E$. Since the $e_i$ span $E$ over $K$, there exist $c_i\in K$, only finitely many nonzero, with $x={\sum }_i{c}_i{e}_i$. Each $c_i\in K$ expands as $c_i={\sum }_j{a}_{ij}{k}_j$ for some $a_{ij}\in F$, finitely many nonzero. Substituting: $$ x = \sum_{i} \biggl(\sum_{j} a_{ij} k_j\biggr) e_i = \sum_{i,j} a_{ij} e_i k_j. $$ So the set $\{e_i{k}_j\}$ spans $E$ over $F$.

Independence. Suppose ${\sum }_{i,j}{a}_{ij}{e}_i{k}_j=0$ for some $a_{ij}\in F$. Rearranging: ${\sum }_i\left({\sum }_j{a}_{ij}{k}_j\right)e_i=0$. Since the $e_i$ are linearly independent over $K$, each coefficient ${\sum }_j{a}_{ij}{k}_j=0$ in $K$. Since the $k_j$ are linearly independent over $F$, each $a_{ij}=0$. So the set $\{e_i{k}_j\}$ is linearly independent over $F$.

Therefore ${\dim }_{F}E=|I|\cdot |J|=[E:K][K:F]$. $\square$

Bridge. The tower law builds toward 01.02.05 (solvable and nilpotent groups), where the degree multiplicativity underpins the criterion for solvability by radicals: a polynomial is solvable by radicals precisely when its Galois group is solvable, and the group-theoretic chain of subgroups mirrors a tower of radical extensions whose degrees multiply. The foundational reason is that the degree measures the "cost" of each adjunction in the tower, and this is exactly the arithmetic that constrains which polynomials admit radical solutions. The bridge is between the dimension-counting of linear algebra and the subgroup-chain structure of group theory, and the central insight is that both sides obey the same multiplicativity constraint. The pattern appears again in 01.01.04 (subspace, basis, dimension), where every finite-dimensional extension is a vector space whose dimension is the degree.

Exercises [Intermediate+]

Advanced results [Master]

Theorem 1 (Existence of splitting fields). Let $F$ be a field and $f\in F[x]$ a nonconstant polynomial. There exists a splitting field $E$ for $f$ over $F$. Moreover, $[E:F]\le (\mathrm{deg}f)!$ ^{[Lang Algebra Ch. V]}.

The proof constructs $E$ by iteratively adjoining roots of irreducible factors, using the Kronecker construction $F[x]/(p)$ for each irreducible factor $p$. Each adjunction increases the degree by at most $\mathrm{deg}f$, and the process terminates after at most $\mathrm{deg}f$ steps.

Theorem 2 (Uniqueness of splitting fields). Any two splitting fields of $f\in F[x]$ are $F$-isomorphic.

This is the isomorphism-extension result proved in Exercise 7, generalised by transfinite induction for the case of arbitrary-degree extensions.

Theorem 3 (Algebraic closure). Every field $F$ has an algebraic closure $\overline{F}$: an algebraically closed field that is algebraic over $F$. Any two algebraic closures of $F$ are $F$-isomorphic ^{[Steinitz 1910]}.

Steinitz 1910 proved existence by well-ordering and transfinite iteration of the Kronecker construction. Uniqueness up to $F$-isomorphism follows from the isomorphism extension theorem.

Theorem 4 (Separable and inseparable extensions). An algebraic element $\alpha$ over $F$ is separable if its minimal polynomial has no repeated roots in any extension. The extension $E/F$ is separable if every element of $E$ is separable over $F$. In characteristic zero, every algebraic extension is separable. In characteristic $p$, an element $\alpha$ is inseparable if and only if $m_{\alpha }(x)=g(x^p)$ for some polynomial $g$ ^{[Lang Algebra Ch. V]}.

Theorem 5 (Primitive element theorem). Let $E/F$ be a finite separable extension. Then there exists $\theta \in E$ such that $E=F(\theta )$. Equivalently, every finite separable extension is simple.

The classical proof constructs $\theta =\alpha +c\beta$ for generators $\alpha ,\beta$ and a suitable $c\in F$, then shows that only finitely many values of $c$ fail. The theorem fails for inseparable extensions: ${\mathbb{F}}_p(t,u)/{\mathbb{F}}_p(t^p,u^p)$ has degree $p^2$ but no primitive element.

Theorem 6 (Cyclotomic fields). The $n$-th cyclotomic field $\mathbb{Q}({\zeta }_n)$, where ${\zeta }_n=e^{2\pi i/n}$, has degree $[\mathbb{Q}({\zeta }_n):\mathbb{Q}]=\phi (n)$ (Euler's totient function). The minimal polynomial of ${\zeta }_n$ over $\mathbb{Q}$ is the $n$-th cyclotomic polynomial ${\Phi }_n(x)$, which has integer coefficients and is irreducible over $\mathbb{Q}$ ^{[Lang Algebra Ch. VI]}.

Theorem 7 (Degree bounds for splitting fields). If $f\in F[x]$ has degree $n$ and $E$ is the splitting field of $f$ over $F$, then $[E:F]$ divides $n!$. Equality holds when $f$ is "generic" — for instance, $[E:\mathbb{Q}]=n!$ for the generic polynomial $x^n-t$ over $\mathbb{Q}(t)$.

Synthesis. The tower law is the foundational reason that degree arithmetic controls the structure of all field extensions. The central insight is that the degree measures dimension as a vector space, and the multiplicativity $[E:F]=[E:K][K:F]$ identifies the degree with a multiplicative invariant that constrains how extensions compose. Putting these together with the existence and uniqueness of splitting fields, the bridge is between the linear-algebraic notion of dimension and the algebraic problem of polynomial factorisation.

This is exactly the structure that underlies Galois theory: the Galois group $\mathrm{Gal}(E/F)$ acts on the roots of a polynomial, and the degree $[E:F]$ equals the order $|\mathrm{Gal}(E/F)|$ for separable splitting fields. The pattern recurs in 01.01.08 (eigenvalues and the characteristic polynomial), where the eigenvalues are the roots of a polynomial over the base field and the splitting field of the characteristic polynomial is the ambient field extension that diagonalises the operator, and appears again in 01.02.05 (solvable and nilpotent groups), where the solvability of the Galois group determines whether the polynomial is solvable by radicals.

Full proof set [Master]

Proposition 1 (Existence of splitting fields). For every nonconstant $f\in F[x]$, there exists a splitting field $E$ for $f$ over $F$ with $[E:F]\le (\mathrm{deg}f)!$.

Proof. Proceed by induction on $n=\mathrm{deg}f$. If $n=1$, then $f$ already splits over $F$; take $E=F$.

Assume $n\ge 2$. If $f$ splits over $F$, take $E=F$. Otherwise, $f$ has an irreducible factor $p(x)$ of degree $d\ge 2$. Form the field $K=F[x]/(p)$ and let $\alpha =\bar{x}\in K$ be a root of $p$, hence of $f$. Then $[K:F]=d$ and $f(x)=(x-\alpha )g(x)$ for some $g\in K[x]$ with $\mathrm{deg}g=n-1$.

By the induction hypothesis, there exists a splitting field $E$ for $g$ over $K$, and $[E:K]\le (n-1)!$. Since $f(x)=(x-\alpha )g(x)$ splits in $E$, the field $E$ is a splitting field for $f$ over $F$. By the tower law, $[E:F]=[E:K][K:F]\le (n-1)!\cdot n=n!$. $\square$

Proposition 2 (Primitive element theorem). Let $E/F$ be a finite separable extension. Then $E=F(\theta )$ for some $\theta \in E$.

Proof. If $F$ is finite, then $E$ is finite, and the multiplicative group $E^{\ast }$ is cyclic; any generator $\theta$ of $E^{\ast }$ satisfies $E=F(\theta )$.

Assume $F$ is infinite. By induction on the number of generators, it suffices to prove: if $E=F(\alpha ,\beta )$ with both algebraic and separable over $F$, then $E=F(\theta )$ for some $\theta$.

Let $f=m_{\alpha }$ and $g=m_{\beta }$ be the minimal polynomials of $\alpha$ and $\beta$ over $F$, with roots $\alpha ={\alpha }_1,{\alpha }_2,\dots ,{\alpha }_r$ in some splitting field, and $\beta ={\beta }_1,{\beta }_2,\dots ,{\beta }_s$. Since $\beta$ is separable, the ${\beta }_j$ are distinct.

For each pair $(i,j)$ with $i\ge 1$ and $j\ge 2$, the equation ${\alpha }_i+c{\beta }_j={\alpha }_1+c{\beta }_1$ has at most one solution $c\in F$. Since $F$ is infinite and there are finitely many pairs, there exists $c\in F$ avoiding all these equations. Set $\theta =\alpha +c\beta$.

The polynomial $h(x)=f(\theta -cx)$ has $\beta$ as a root and shares no root with $g$ other than $\beta$ itself (by the choice of $c$). So $\gcd (h,g)$ in $E[x]$ is linear, hence $\beta \in F(\theta )$, and then $\alpha =\theta -c\beta \in F(\theta )$. Therefore $F(\alpha ,\beta )=F(\theta )$. $\square$

Proposition 3 (Cyclotomic field degree). The degree $[\mathbb{Q}({\zeta }_n):\mathbb{Q}]=\phi (n)$, where ${\zeta }_n=e^{2\pi i/n}$ and $\phi$ is Euler's totient function.

Proof. The $n$-th cyclotomic polynomial is ${\Phi }_n(x)={\prod }_{\gcd (k,n)=1}(x-{\zeta }_n^k)$, a product over the $\phi (n)$ primitive $n$-th roots of unity. One shows ${\Phi }_n\in \mathbb{Z}[x]$ by induction on $n$: $x^n-1={\prod }_{d\mid n}{\Phi }_d(x)$, and ${\Phi }_n$ is the quotient of $x^n-1$ by ${\prod }_{d\mid n,d<n}{\Phi }_d(x)$.

Irreducibility of ${\Phi }_n$ over $\mathbb{Q}$: suppose ${\Phi }_n=gh$ with $g$ monic and irreducible over $\mathbb{Q}$, and let $\zeta$ be a root of $g$. For any prime $p$ not dividing $n$, the element ${\zeta }^p$ is also a primitive $n$-th root of unity. If ${\zeta }^p$ is not a root of $g$, then $g$ and the polynomial $g(x^p)$ share no root. But $g(x^p)\equiv g(x{)}^p(\mathrm{mod}p)$, so ${\Phi }_n\equiv g(x{)}^{p}h(x{)}^p(\mathrm{mod}p)$, contradicting that ${\Phi }_n$ has no repeated roots modulo $p$ (since $p\nmid n$ implies $x^n-1$ is separable mod $p$). Hence ${\zeta }^p$ is a root of $g$ for every $p\nmid n$, and by iterating, every primitive $n$-th root is a root of $g$, so $g={\Phi }_n$.

Since ${\Phi }_n$ is monic and irreducible of degree $\phi (n)$, it is the minimal polynomial of ${\zeta }_n$ over $\mathbb{Q}$, giving $[\mathbb{Q}({\zeta }_n):\mathbb{Q}]=\phi (n)$. $\square$

Connections [Master]

• Solvable and nilpotent groups, Jordan-Holder theorem 01.02.05. The solvability of the Galois group of a polynomial determines whether the polynomial is solvable by radicals. A tower of radical extensions corresponds to a subnormal series of the Galois group, and the commutativity of successive quotients mirrors the radical nature of each adjunction. This is exactly the bridge between group-theoretic solvability and field-theoretic solvability-by-radicals.

• Subspace, basis, dimension 01.01.04. Every finite field extension $E/F$ is a vector space over $F$ whose dimension equals the degree $[E:F]$. The Steinitz exchange lemma that underpins the invariance of dimension is the same mechanism that guarantees the degree is well-defined. The identification of field-extension degree with vector-space dimension is the foundational reason that linear algebra applies to field theory.

• Eigenvalue, eigenvector, characteristic polynomial 01.01.08. The eigenvalues of a matrix are the roots of its characteristic polynomial ${\chi }_T(t)=\det (tI-T)$, and the splitting field of ${\chi }_T$ over the scalar field is the extension in which the operator diagonalises. The algebraic multiplicity of each eigenvalue equals the power of the corresponding linear factor in the splitting field, and the degree of this splitting field is bounded by $(\mathrm{deg}{\chi }_T)!$. The pattern recurs in the Jordan canonical form, where the splitting field determines the field over which the Jordan decomposition is defined.

Historical & philosophical context [Master]

Galois 1832 introduced the correspondence between field extensions and permutation groups in his memoir on the solvability of equations ^{[Galois 1832]}, establishing that the structure of the roots of a polynomial is governed by a group that permutes them. Abel 1826 had already proved the unsolvability of the general quintic by radicals ^{[Abel 1826]}, closing a problem open since Cardano and Ferrari in the sixteenth century. Galois's deeper contribution was identifying the group as the obstruction.

Steinitz 1910 created the abstract theory of field extensions in Algebraische Theorie der Körper ^{[Steinitz 1910]}, introducing algebraic closure, transcendence degree, and the classification of fields by characteristic and cardinality. Steinitz's work separated the general theory of field extensions from the specific study of polynomial equations, providing the axiomatic framework that underpins modern algebraic number theory and algebraic geometry.

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