Splitting Fields for Finite-Dimensional Algebras

Q: Does every finite-dimensional algebra have a splitting field?

Yes: the algebraic closure of the ground field always works, because over an algebraically closed field every finite-dimensional division algebra is trivial. The substantive question is whether a finite extension suffices, which (7.10) answers affirmatively over perfect fields.

Q: Why is the definition left-right symmetric when absolute irreducibility is not?

Because the criterion (7.7) is a statement about the ring R^K/rad(R^K), and that ring satisfies the condition if and only if its opposite does — transposition gives M_n(K)^op ≅ M_n(K). A single module has no opposite, so the individual notion has no side symmetry to lose.

Q: If K splits R, is R^K semisimple?

No. Splitting constrains only the semisimple quotient. k[x]/(x^2) splits over every field, radical and all. Semisimplicity of R^K is a separate question, governed by separability of the extension and of the algebra.

Q: Can the number of simple modules stay the same when passing to a splitting field?

Yes, and it does exactly when the endomorphism algebras were already central of degree greater than one. For R = H over R there is one simple module before and one after extending to C; for R = C over R the count goes from one to two.

Q: How small can a splitting field be?

Over a finite field the minimal splitting field is the compositum of the endomorphism fields F_q^d_i and is unique. Over Q minimality is subtle: a quaternion algebra is split by infinitely many quadratic fields, none canonical.

Q: What is the relationship between (7.12) and elementary field theory?

It says the algebra-theoretic notion restricts correctly. The classical splitting field of f is the smallest field over which f has only linear factors; (7.12) says those are exactly the fields splitting the algebra k[t]/(f), with minimality being an extra condition the algebraic notion does not impose.

Executive Summary

Absolute irreducibility is a property of one module. Demanding it of every simple module at once is a property of the ground field, and the field is then called a splitting field for the algebra. The point of the definition is stability: once $R$ splits over $K$ , the list of simple modules is final and no further extension changes it.

The workable form of the definition is the matrix-algebra criterion $(7.7)$ : $K$ splits $R$ if and only if $R^{K} / rad (R^{K})$ is a finite direct product of full matrix algebras over $K$ . In that form the condition is visibly left-right symmetric, immediately checkable by a dimension count, and reducible to the semisimple quotient.

$\prod M_{n_{i}} (K)$ Split semisimple quotient

$\sum (\dim_{K} V_{i})^{2}$ Dimension test

$\bar{k}$ Always a splitting field

finiteSuffices over a perfect field

Overview

Throughout this page $R$ is a $k$ -algebra with $\dim_{k} R < \infty$ , and $R^{K} = R \otimes_{k} K$ for a field extension $K \supseteq k$ . Two obstructions can stop $R^{K}$ from looking like a product of matrix algebras: division algebras that have not yet split, and a radical. The definition of a splitting field addresses only the first.

Three questions follow immediately: does a splitting field always exist (yes — the algebraic closure), can it be taken finite over $k$ (yes when $k$ is perfect, by $(7.10)$ ), and does the property persist under further extension (yes, by $(7.14)$ , treated on Simple Modules under Field Extension). Together these make the splitting field a canonical place to do representation theory.

For the special class of commutative algebras $k [t] / (f)$ , the notion recovers the elementary one: $K$ splits the algebra exactly when $f$ factors into linear factors over $K$ . That is the sense in which $(7.6)$ generalises the splitting field of a polynomial.

Learning Objectives

State $(7.6)$ and explain what would change if right modules were used instead.
Prove the matrix-algebra criterion $(7.7)$ from the characterisation of absolute irreducibility.
Derive the dimension test $(7.8)$ and use it as a practical check.
Prove $(7.9)$ , reducing splitting to the semisimple quotient, and identify where nilpotence of the radical is used.
Prove that a finite splitting field exists over a perfect field, and say precisely what perfectness is used for.
Apply $(7.12)$ to compute splitting fields of algebras of the form $k [t] / (f)$ .

Definitions

Definition(7.6)Splitting field for an algebra

Let $R$ be a $k$ -algebra with $\dim_{k} R < \infty$ . A field extension $K \supseteq k$ is a **splitting field for $R$ ** — one also says $R$ splits over $K$ — if every simple left $R^{K}$ -module is absolutely irreducible as an $R^{K}$ -module over $K$ .

Strictly this defines a left splitting field. The criterion $(7.7)$ shows the condition is unchanged if right modules are used, so the qualifier is dropped.

$R^{K}$: The $K$ -algebra $R \otimes_{k} K$ ; $\dim_{K} R^{K} = \dim_{k} R$ .
$rad (R^{K})$: The Jacobson radical of the extended algebra. It always contains $(rad R)^{K}$ and may be strictly larger.
Separable $k$ -algebra: One for which $R \otimes_{k} \bar{k}$ is semisimple. Every semisimple algebra over a perfect field is separable.
Perfect field: Characteristic $0$ , or characteristic $p$ with $k = k^{p}$ . Finite fields and algebraically closed fields are perfect.
$M_{n} (K)$: The full ring of $n \times n$ matrices over $K$ ; the split simple $K$ -algebras are exactly these.

The definition quantifies over simple modules of $R^{K}$ , not of $R$ . Testing whether $k$ itself splits $R$ therefore means testing the simple $R$ -modules — which is what $(7.7)$ and $(7.8)$ make practical.

Core Concepts

Two obstructions, one of them addressed

Write $\bar{R^{K}} = R^{K} / rad (R^{K}) ≅ \prod_{i} M_{m_{i}} (Δ_{i})$ by Wedderburn–Artin. Splitting says every $Δ_{i} = K$ . It says nothing about $rad (R^{K})$ , which may be nonzero and may even be strictly larger than $(rad R)^{K}$ when $K / k$ is inseparable.

R^{K}

\bar{R^{K}}

\prod M_{m_{i}} (Δ_{i})

\prod M_{m_{i}} (K)

The last arrow is what a splitting field achieves; the first two are automatic. This is why a splitting field is the right place to do character theory — over it, the algebra sees only matrices and dimensions, never division algebra arithmetic.

Why the notion is symmetric

Transposition gives $M_{n} (K)^{op} ≅ M_{n} (K)$ , and $rad$ commutes with passing to the opposite ring. So the condition in $(7.7)$ holds for $R$ if and only if it holds for $R^{op}$ — that is, the left and right notions of splitting field coincide, even though absolute irreducibility of an individual module obviously does not transfer between sides.

Key Results

Theorem(7.7)Matrix-algebra criterion

Let $R$ be a $k$ -algebra with $\dim_{k} R < \infty$ and let $K \supseteq k$ be a field extension. Then $K$ is a splitting field for $R$ if and only if $R^{K} / rad (R^{K})$ is a finite direct product of full matrix algebras over $K$ .

Proof

Replacing $R$ by $R^{K}$ and $k$ by $K$ , it suffices to treat $K = k$ . Adopt the notation of $(7.1)$ : $\bar{R} ≅ M_{n_{1}} (D_{1}) \times \dots \times M_{n_{r}} (D_{r})$ with $M_{i}$ the simple modules and $D_{i} = End (_{R} M_{i})$ .

If every $M_{i}$ is absolutely irreducible, then $D_{i} = k$ for each $i$ by $(7.5)$ , so $\bar{R} ≅ \prod_{i} M_{n_{i}} (k)$ , a product of matrix algebras over $k$ .

Conversely, suppose $\bar{R} ≅ \prod_{i} M_{n_{i}} (k)$ . The simple left modules of this ring are the column spaces $k^{n_{i}}$ , one for each factor, and the endomorphism ring of $k^{n_{i}}$ as a module over $M_{n_{i}} (k)$ consists of the scalars. So $D_{i} = k$ for every $i$ , and $(7.5)$ makes each $M_{i}$ absolutely irreducible.

Corollary(7.8)Dimension test

Let $R$ be a $k$ -algebra with $\dim_{k} R < \infty$ and let $M_{1}, \dots, M_{r}$ be a full set of simple left $R$ -modules. Then $R$ splits over $k$ if and only if

\dim_{k} R = \dim_{k} rad R + \sum_{i = 1}^{r} (\dim_{k} M_{i})^{2}

(7.8)

Proof

By $(7.2)$ , $\dim_{k} R = \dim_{k} rad R + \sum_{i} n_{i}^{2} \dim_{k} D_{i}$ and $\dim_{k} M_{i} = n_{i} \dim_{k} D_{i}$ , so $(\dim_{k} M_{i})^{2} = n_{i}^{2} (\dim_{k} D_{i})^{2} \geq n_{i}^{2} \dim_{k} D_{i}$ , with equality precisely when $\dim_{k} D_{i} = 1$ .

Summing, the right-hand side of $(7.8)$ is always at least $\dim_{k} R$ , with equality if and only if every $D_{i}$ equals $k$ — that is, by $(7.7)$ , if and only if $R$ splits over $k$ .

Proposition(7.9)Reduction to the semisimple quotient

Let $R$ be a $k$ -algebra with $\dim_{k} R < \infty$ and $\bar{R} = R / rad R$ . A field extension $K \supseteq k$ is a splitting field for $R$ if and only if it is a splitting field for $\bar{R}$ .

Proof

Since $\dim_{k} R < \infty$ , the ideal $rad R$ is nilpotent, so $(rad R)^{K}$ is a nilpotent ideal of $R^{K}$ and therefore $(rad R)^{K} \subseteq rad (R^{K})$ .

Consequently ${\bar{R}}^{K} = R^{K} / (rad R)^{K}$ has radical $rad (R^{K}) / (rad R)^{K}$ , and

{\bar{R}}^{K} / rad ({\bar{R}}^{K}) ≅ R^{K} / rad (R^{K})

The two algebras have the same simple modules and the same semisimple quotient, so by $(7.7)$ one splits over $K$ exactly when the other does.

Proposition(7.10)Finite splitting fields over perfect fields

Let $k$ be a perfect field and $R$ a $k$ -algebra with $\dim_{k} R < \infty$ . Then $R$ splits over some finite extension $K$ of $k$ .

Proof

By $(7.9)$ we may assume $R$ is semisimple. Let $E = \bar{k}$ . Since $k$ is perfect, $E / k$ is separable, and therefore $R^{E}$ is again semisimple. As $E$ is algebraically closed, Wedderburn–Artin gives $R^{E} ≅ M_{n_{1}} (E) \times \dots \times M_{n_{r}} (E)$ .

Fix the finitely many matrix units realising this decomposition. Each is a finite sum $\sum_{j} r_{j} \otimes e_{j}$ with $r_{j} \in R$ and $e_{j} \in E$ , so all of them lie in $R^{K}$ for the subfield $K \subseteq E$ generated over $k$ by the finitely many coefficients $e_{j}$ ; being generated by finitely many algebraic elements, $K$ is finite over $k$ .

The $K$ -span of these matrix units is a $K$ -subalgebra of $R^{K}$ isomorphic to $\prod_{i} M_{n_{i}} (K)$ , of $K$ -dimension $\sum_{i} n_{i}^{2} = \dim_{E} R^{E} = \dim_{k} R = \dim_{K} R^{K}$ . A subspace of full dimension is everything, so $R^{K} ≅ \prod_{i} M_{n_{i}} (K)$ and $K$ splits $R$ by $(7.7)$ .

Remark(7.11)What perfectness was for

Perfectness entered the proof at exactly one point: to guarantee that $R^{\bar{k}}$ is semisimple. The proposition therefore holds verbatim for any semisimple $k$ -algebra $R$ with $R^{\bar{k}}$ semisimple — such algebras are called separable $k$ -algebras. Over an imperfect field a semisimple algebra can fail to be separable; the standard example is a purely inseparable field extension viewed as an algebra over its base.

Proposition(7.12)Splitting fields of

k [t] / (f)

Let $f \in k [t]$ be nonconstant and $R = k [t] / (f)$ , a commutative $k$ -algebra of dimension $\deg f$ . A field extension $K \supseteq k$ is a splitting field for the algebra $R$ if and only if $f$ factors into linear factors in $K [t]$ .

Proof

Write $f = f_{1}^{e_{1}} \dots f_{r}^{e_{r}}$ with the $f_{i}$ distinct monic irreducibles in $K [t]$ . The Chinese Remainder Theorem gives

R^{K} ≅ K [t] / (f) ≅ \prod_{i = 1}^{r} K [t] / (f_{i}^{e_{i}})

Each factor is a local ring whose maximal ideal $(f_{i}) / (f_{i}^{e_{i}})$ is nilpotent, so its radical is that ideal and its residue algebra is the field $K_{i} : = K [t] / (f_{i})$ , of degree $\deg f_{i}$ over $K$ . Hence $R^{K} / rad (R^{K}) ≅ \prod_{i} K_{i}$ .

By $(7.7)$ , $K$ splits $R$ if and only if each $K_{i}$ is a full matrix algebra over $K$ . Each $K_{i}$ is commutative, and $M_{n} (K)$ is commutative only for $n = 1$ , so the condition is $K_{i} = K$ for all $i$ , i.e. $\deg f_{i} = 1$ for all $i$ — precisely that $f$ splits into linear factors over $K$ .

Proof Techniques and Method

How these proofs work, and which move to reuse.

Four moves carry this material, and each reappears elsewhere in the chapter.

Move 1

Rename the base field

Almost every proof begins "we may as well take $K = k$ ", replacing $R$ by $R^{K}$ . This is legitimate because all the hypotheses are about the pair (algebra, its own ground field).

Move 2

Nilpotent ideals go up

$(rad R)^{K}$ is nilpotent because $rad R$ is, hence sits inside $rad (R^{K})$ . Finite dimensionality is what makes the radical nilpotent, and this containment is the only general statement available — equality can fail.

Move 3

Descend finitely many coefficients

Any finite configuration of elements of $R^{\bar{k}}$ already lives over a finite extension. This is the standard way to convert an algebraically closed statement into a finite one.

Move 4

Dimension forces equality

A subalgebra of the right dimension is the whole algebra. Used to upgrade "contains a product of matrix algebras" to "is a product of matrix algebras".

Move 2 is also the source of the main hazard on this page: $(rad R)^{K} \subseteq rad (R^{K})$ is an inclusion, not an identity. Equality holds when $K / k$ is separable, and also whenever $R$ already splits over $k$ .

Worked Example

$R = ℝ C_{3}$ , and the polynomial view

Let $G = ⟨ g ⟩$ be cyclic of order $3$ and $R = ℝ G ≅ ℝ [t] / (t^{3} - 1)$ , of dimension $3$ . Since $char ℝ = 0$ , Maschke gives $rad R = 0$ and $R ≅ ℝ \times ℂ$ .

**By $(7.7)$ :** $ℂ$ is not a matrix algebra over $ℝ$ , so $ℝ$ does not split $R$ .

**By $(7.8)$ :** the simple modules have $ℝ$ -dimensions $1$ and $2$ , and $1^{2} + 2^{2} = 5 \neq 3 = \dim_{ℝ} R$ . The test fails, confirming the same conclusion.

**By $(7.12)$ :** $t^{3} - 1 = (t - 1) (t^{2} + t + 1)$ over $ℝ$ , which does not split into linear factors. All three criteria agree.

Over $K = ℂ$ the polynomial splits, $R^{ℂ} ≅ ℂ \times ℂ \times ℂ$ , the three simple modules are one-dimensional, and $1 + 1 + 1 = 3 = \dim_{ℂ} R^{ℂ}$ . In fact $K = ℚ (ω)$ already splits $ℚ G$ , a finite extension of degree $2$ — as $(7.10)$ promises, since $ℚ$ is perfect.

A splitting field that leaves a radical behind

Let $p$ be a prime, $k = 𝔽_{p} (u)$ — an imperfect field — and $R = k [t] / (t^{p} - u)$ . The polynomial $t^{p} - u$ is irreducible over $k$ , so $R$ is a field, hence semisimple, and $k$ splits $R$ only if $R = k$ , which it is not: $\dim_{k} R = p$ and $(7.8)$ reads $p = 0 + p^{2}$ , false.

Put $K = k (u^{1 / p}) = R$ . In $K [t]$ we have $t^{p} - u = (t - u^{1 / p})^{p}$ , so by $(7.12)$ $K$ is a splitting field. Concretely

R^{K} ≅ K [t] / ((t - u^{1 / p})^{p}), rad (R^{K}) = (t - u^{1 / p}), R^{K} / rad (R^{K}) ≅ K

Process and Workflow

Compute $rad R$ and pass to $\bar{R}$ Legitimate by

(7.9)

: splitting fields of

R

and of

\bar{R}

coincide.

Decompose $\bar{R}$ Wedderburn–Artin gives

\prod M_{n_{i}} (D_{i})

; the

D_{i}

are the only obstruction.

Read off the obstructionIf every

D_{i} = k

, done —

k

splits

R

. Otherwise, list the division algebras that must be split.

Adjoin what is neededFor

D_{i}

a field, adjoin its normal closure; for

D_{i}

central of degree

m

, adjoin a maximal subfield, of degree

m

over the centre.

Verify with $(7.8)$ Check

\dim_{K} R^{K} = \dim_{K} rad (R^{K}) + \sum (\dim_{K} V_{i})^{2}

Does a convenient splitting field exist?

$k$ algebraically closed

k

itself splits every finite-dimensional

k

-algebra.

$k$ finiteFinite fields are perfect, so a finite splitting field exists; by Wedderburn's Little Theorem the

D_{i}

are fields, and their compositum splits

R

$k$ perfect, infinite

(7.10)

gives a finite splitting field. For

R = k G

in characteristic

0

, Brauer's theorem lets one take

K = k (ζ_{m})

with

m

the exponent of

G

$k$ imperfect

(7.10)

does not apply, since

R^{\bar{k}}

may fail to be semisimple. A finite splitting field may still exist — check directly whether

R^{K} / rad (R^{K})

becomes a product of matrix algebras.

Comparison and Classification

Splitting behaviour of small algebras
$R$ over $k$	$\bar{R}$	Does $k$ split $R$ ?	A splitting field
$M_{n} (k)$	$M_{n} (k)$	yes	$k$
$k [x] / (x^{n})$	$k$	yes	$k$
$T_{2} (k)$	$k \times k$	yes	$k$
$ℂ$ over $ℝ$	$ℂ$	no	$ℂ$
$ℍ$ over $ℝ$	$ℍ$	no	$ℂ$
$ℚ C_{3}$	$ℚ \times ℚ (ω)$	no	$ℚ (ω)$
$ℚ C_{5}$	$ℚ \times ℚ (ζ_{5})$	no	$ℚ (ζ_{5})$
$𝔽_{p} (u) [t] / (t^{p} - u)$	itself, a field	no	$𝔽_{p} (u^{1 / p})$

Three criteria for splitting, and what each costs
Criterion	Statement	Cost
Definition $(7.6)$	every simple $R^{K}$ -module is absolutely irreducible	requires knowing all simple modules
Matrix criterion $(7.7)$	$R^{K} / rad (R^{K}) ≅ \prod M_{n_{i}} (K)$	one radical computation plus a Wedderburn decomposition
Dimension test $(7.8)$	$\dim_{k} R = \dim_{k} rad R + \sum (\dim_{k} M_{i})^{2}$	arithmetic only, once dimensions are known
Polynomial case $(7.12)$	$f$ splits into linear factors over $K$	one factorisation

Relationship Map

The splitting condition is the algebra-level analogue of absolute irreducibility, and it sits inside a chain of increasingly strong requirements on the pair $(R, K)$ .

Conditions on $K \supseteq k$
- $K$ splits $R$
  - $R^{K} / rad (R^{K})$ is a product of matrix algebras over $K$
  - every simple $R^{K}$ -module is absolutely irreducible
  - the dimension identity (7.8) holds over $K$
- $K$ splits $R$ and $R^{K}$ is semisimple
  - $R^{K} ≅ \prod M_{n_{i}} (K)$
  - holds automatically if $R$ is separable and semisimple
- $K$ splits $R$ and $rad (R^{K}) = (rad R)^{K}$
  - automatic when $R$ splits over $k$ already
  - automatic when $K / k$ is separable

k

splits

R

K

splits

R

for all

K \supseteq k

characters determine simple modules

\dim_{k} R / T (R)

counts simples

Applications and Industry Use

Applications here means where this structure is used — inside mathematics and in the engineering and computing disciplines that consume it.

Group representation theory

Choosing the coefficient field

Ordinary character theory is done over a splitting field so that the number of irreducibles equals the number of conjugacy classes and characters separate modules. Brauer's theorem identifies $ℚ (ζ_{m})$ as a concrete choice in characteristic $0$ .

Brauer group theory

Splitting central simple algebras

For $A$ central simple over $k$ , $K$ splits $A$ in the sense of $(7.6)$ exactly when $A \otimes_{k} K ≅ M_{n} (K)$ — the definition used in the Brauer group. Maximal subfields of a division algebra are the minimal splitting fields.

Coding theory

Idempotents over the right field

The primitive idempotents of $𝔽_{q} G$ that generate minimal cyclic codes are visible only over a field containing the appropriate roots of unity; that field is a splitting field for the group algebra.

Computer algebra

Canonical form of an algebra

Systems that decompose algebras report the $D_{i}$ explicitly, and offer to extend the base field until they vanish. That extension is exactly a splitting field.

The unifying point is normalisation: over a splitting field the answer stops depending on arithmetic accidents of the ground field, so results proved there can be transported back by descent.

Standards and Notation

Standards here covers notation, symbol and markup standards, and reference implementations, rather than material or design codes.

Terminologysplitting field for

R

;

R

splits over

K

;

K

is a splitting field of

R

Central simple case

K

splits

A

iff

A \otimes_{k} K ≅ M_{n} (K)

— same notion

Extended algebra

R^{K}

(Lam);

R_{K}

K R

elsewhere

Roots of unity

ζ_{m}

for a primitive

m

-th root of unity;

ℚ (ζ_{m})

is the

m

-th cyclotomic field

GAPWedderburnDecomposition (Wedderga), FieldOfDefinition

MagmaAbsolutelyIrreducibleModules, SplittingField

MarkupPresentation MathML per ISO/IEC 40314; symbol conventions per ISO 80000-2

Computational Notes

Computational notes cover algorithms, cost and library behaviour rather than manufacturing process.

Over a finite field the whole problem is effective: compute $rad R$ , decompose $\bar{R}$ into simple components, and for each compute the endomorphism field $𝔽_{q^{d_{i}}}$ . The compositum $𝔽_{q^{ℓ}}$ with $ℓ = lcm (d_{i})$ is a splitting field, and it is the smallest one.
Over $ℚ$ the same recipe needs the centres of the $D_{i}$ as explicit number fields and then a splitting field for each division algebra; the second step is where the cost concentrates, since recognising a matrix algebra over $ℚ$ is at least as hard as factoring integers.
For group algebras in characteristic $0$ one bypasses all of this: $ℚ (ζ_{m})$ with $m$ the exponent of the group is a splitting field, and often a much smaller subfield already suffices.
The dimension test $(7.8)$ is the cheapest verification available and should be run on any machine-computed decomposition.

Failure Modes and Common Mistakes

Do not test the definition on the simple $R$ -modules when asking whether some larger $K$ splits $R$ ; the definition refers to the simple $R^{K}$ -modules.
Do not assume the number of simple modules is unchanged by passing to a splitting field — it typically grows.
Do not confuse the minimal splitting field of a polynomial with a splitting field of the algebra it defines; $(7.12)$ says the conditions match, not that the fields are canonically the same.

Historical Notes and Lessons Learned

1893MolienMolien determines the structure of group algebras over $ℂ$ , implicitly working over a field that splits everything.
1907WedderburnThe structure theorem for finite-dimensional algebras isolates the division algebra factors $D_{i}$ — the objects a splitting field is designed to remove.
1929NoetherNoether's work on crossed products identifies the splitting fields of a central division algebra with its maximal subfields, giving the Brauer-group formulation.
1932Albert–Brauer–Hasse–NoetherCentral simple algebras over number fields are shown to be cyclic, so splitting fields can be taken cyclic — a strong arithmetic refinement of mere existence.
1945–1947BrauerBrauer's induction theorem yields that a field of characteristic $0$ containing a primitive $m$ -th root of unity, $m$ the exponent of $G$ , is a splitting field for $G$ .

The methodological lesson is that the useful definition was the invariant one. Defining a splitting field by a property of the modules $(7.6)$ , rather than by generation by particular elements, is what makes $(7.9)$ , $(7.10)$ and $(7.14)$ available at all.

Quick Reference

Definitionevery simple

R^{K}

-module is absolutely irreducible

Criterion

R^{K} / rad (R^{K}) ≅ \prod_{i} M_{n_{i}} (K)

Dimension test

\dim_{k} R = \dim_{k} rad R + \sum_{i} (\dim_{k} M_{i})^{2}

Reduction

K

splits

R

iff

K

splits

R / rad R

Existence

\bar{k}

always splits

R

Finiteness

k

perfect, or

R

separable, gives a finite splitting field

Polynomial case

K

splits

k [t] / (f)

iff

f

splits into linear factors over

K

Symmetryleft and right splitting fields coincide

Caution

(rad R)^{K} \subseteq rad (R^{K})

, possibly strictly

Statement finder
Result	Content	Reference
Definition	splitting field of a finite-dimensional algebra	(7.6)
Matrix criterion	semisimple quotient is a product of matrix algebras	(7.7)
Dimension test	equality of the two dimension counts	(7.8)
Reduction to $\bar{R}$	splitting depends only on $R / rad R$	(7.9)
Finite splitting field	exists over a perfect field	(7.10)
Separable algebras	perfectness only needed for semisimplicity of $R^{\bar{k}}$	(7.11)
Polynomial algebras	splitting the algebra equals splitting $f$	(7.12)

Frequently Asked Questions

Does every finite-dimensional algebra have a splitting field?

Yes: the algebraic closure of the ground field always works, because over an algebraically closed field every finite-dimensional division algebra is trivial. The substantive question is whether a finite extension suffices, which (7.10) answers affirmatively over perfect fields.

Why is the definition left-right symmetric when absolute irreducibility is not?

Because the criterion (7.7) is a statement about the ring $R^{K} / rad (R^{K})$ , and that ring satisfies the condition if and only if its opposite does — transposition gives $M_{n} (K)^{op} ≅ M_{n} (K)$ . A single module has no opposite, so the individual notion has no side symmetry to lose.

K

splits

R

, is

R^{K}

semisimple?

No. Splitting constrains only the semisimple quotient. $k [x] / (x^{2})$ splits over every field, radical and all. Semisimplicity of $R^{K}$ is a separate question, governed by separability of the extension and of the algebra.

Can the number of simple modules stay the same when passing to a splitting field?

Yes, and it does exactly when the endomorphism algebras were already central of degree greater than one. For $R = ℍ$ over $ℝ$ there is one simple module before and one after extending to $ℂ$ ; for $R = ℂ$ over $ℝ$ the count goes from one to two.

How small can a splitting field be?

Over a finite field the minimal splitting field is the compositum of the endomorphism fields $𝔽_{q^{d_{i}}}$ and is unique. Over $ℚ$ minimality is subtle: a quaternion algebra is split by infinitely many quadratic fields, none canonical.

What is the relationship between (7.12) and elementary field theory?

It says the algebra-theoretic notion restricts correctly. The classical splitting field of $f$ is the smallest field over which $f$ has only linear factors; (7.12) says those are exactly the fields splitting the algebra $k [t] / (f)$ , with minimality being an extra condition the algebraic notion does not impose.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §7 (pp. 112–116).
C. W. Curtis and I. Reiner, Representation Theory of Finite Groups and Associative Algebras, Wiley-Interscience, 1962, §§29 and 70.
R. S. Pierce, Associative Algebras, Graduate Texts in Mathematics 88, Springer-Verlag, 1982, Chapters 12–13.
N. Jacobson, Basic Algebra II, 2nd edition, W. H. Freeman, 1989, Chapter 4.
A. A. Albert, Structure of Algebras, American Mathematical Society Colloquium Publications 24, 1939.

AI Suggested Questions

Compute a minimal splitting field for $ℚ S_{4}$ and for $ℚ Q_{8}$ , and explain the difference.
State and prove Brauer's theorem that $ℚ (ζ_{m})$ splits $ℚ G$ when $m$ is the exponent of $G$ .
Give an example of a finite-dimensional algebra over an imperfect field with no finite splitting field, or explain why none exists.
How do splitting fields of an algebra relate to the Schur index of its simple modules?
Show that if $R$ splits over $k$ then $rad (R^{K}) = (rad R)^{K}$ for every extension $K$ .
Which separable algebras over $k$ are split by a cyclic extension, and what does that say about the Brauer group?
How would one certify computationally that a given finite field is the minimal splitting field of a group algebra?

Splitting Fields for Algebras

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Two obstructions, one of them addressed

Why the notion is symmetric

Key Results

Proof Techniques and Method

Rename the base field

Nilpotent ideals go up

Descend finitely many coefficients

Dimension forces equality

Worked Example

$R = ℝ C_{3}$ , and the polynomial view

A splitting field that leaves a radical behind

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Choosing the coefficient field

Splitting central simple algebras

Idempotents over the right field

Canonical form of an algebra

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Two obstructions, one of them addressed

Why the notion is symmetric

06Key Results

07Proof Techniques and Method

Rename the base field

Nilpotent ideals go up

Descend finitely many coefficients

Dimension forces equality

08Worked Example

R=ℝC3, and the polynomial view

A splitting field that leaves a radical behind

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Choosing the coefficient field

Splitting central simple algebras

Idempotents over the right field

Canonical form of an algebra

13Standards and Notation

14Computational Notes

15Failure Modes and Common Mistakes

16Historical Notes and Lessons Learned

17Quick Reference

18Frequently Asked Questions

19Related KEVOS Topics

20References

21AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

$R = ℝ C_{3}$ , and the polynomial view

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions