Behaviour of Simple Modules under Field Extension

Q: Why is the statement about composition factors rather than direct summands?

Because M^K need not be semisimple. Over a separable extension one can upgrade the conclusion to a direct sum decomposition, but the proposition is stated so that it also covers inseparable extensions, where an extended simple module can be indecomposable of length greater than one.

Q: Can the number of simple modules decrease when the field grows?

No. The families of composition factors attached to distinct simple R-modules are disjoint and nonempty, so choosing one from each already yields r distinct simple R^K-modules.

Q: Does (7.14) mean the simple modules never change again?

It means they change only in the trivial way: over L K the simple modules are exactly the extensions V_i^L, with the same labels, the same multiplicities in the regular module, and the same dimensions. Nothing new appears and nothing merges.

Q: Where exactly does the proof of (7.13)(2) use finite dimensionality?

In the existence of the decomposition R/rad R ≅ B_1 × … × B_r, which requires the quotient to be semisimple — guaranteed here because R is artinian. Without that, there is no element a_i to separate the modules.

Q: Is the equality rad(R^K) = (rad R)^K ever available without splitting?

Yes, whenever K/k is separable; that is the content of the standard result on radicals under separable field extension. The version proved here uses splitting instead, and is often more convenient because it does not restrict the extension at all.

Q: How does this relate to the Schur index?

The Schur index of a simple module measures the size of the central division algebra in its endomorphism ring, and therefore how many copies of a single simple module appear in M^K over a splitting field. The propositions here say which modules appear; the Schur index says with what multiplicity.

Executive Summary

When the ground field grows, simple modules can break apart. What $(7.13)$ says is that they break apart in an orderly way: the simple modules of $R^{K}$ are exactly the composition factors of the extended modules $M^{K}$ , and different simple $R$ -modules contribute disjoint sets of them.

Two consequences dominate everything downstream. First, the number of simple modules is a non-decreasing function of the ground field. Second, once $K$ splits $R$ , every larger field $L$ produces exactly the extended modules $V_{i}^{L}$ and nothing new — so a splitting field is a genuine stopping point, not merely a convenient one.

$r \leq r'$ Simples can only increase

disjointFactors of

M_{1}^{K}

and

M_{2}^{K}

$V_{i}^{L}$ Simples over

L \supseteq K

once

K

splits

$(rad R)^{K}$ Equals

rad (R^{K})

when

R

splits

Overview

Let $R$ be a $k$ -algebra with $\dim_{k} R < \infty$ and $K \supseteq k$ a field extension. Tensoring with $K$ over $k$ is exact, so a composition series of an $R$ -module $M$ becomes a filtration of $M^{K}$ whose factors are the extended modules $S^{K}$ for $S$ a composition factor of $M$ . Refining each $S^{K}$ to a composition series over $R^{K}$ is the only remaining step, and $(7.13)$ organises the result.

Statement (2) is what makes the count $r \leq r'$ work, and it is proved not by module theory but by a single well-chosen element of $R$ : an element that acts as the identity on one simple module and as zero on all the others. Such elements exist because $R / rad R$ is a product, and they survive base change untouched.

What is deliberately absent is any claim of semisimplicity for $M^{K}$ . Over an inseparable extension a simple module can extend to an indecomposable module with repeated composition factors. Statements about $M^{K}$ decomposing require separability and belong on Radical under Field Extension.

Learning Objectives

Prove that every simple $R^{K}$ -module occurs in some $M^{K}$ , using a composition series of the regular module.
Prove disjointness of composition factors using a lift of a central idempotent.
Deduce that the number of isomorphism classes of simple modules does not decrease under extension.
State and prove the stability of splitting fields under further extension.
Prove that $rad (R^{K}) = (rad R)^{K}$ whenever $R$ splits over $k$ .
Follow the composition factors of $M^{K}$ in a separable and in an inseparable example.

Definitions

$M^{K}$: The extended module $M \otimes_{k} K$ over $R^{K} = R \otimes_{k} K$ . Its $K$ -dimension equals $\dim_{k} M$ .
Composition factor: A simple subquotient in a composition series. Every finite-dimensional module over a finite-dimensional algebra has one, and Jordan–Hölder makes the multiset canonical.
$a_{i}$: Any element of $R$ whose image in $R / rad R ≅ B_{1} \times \dots \times B_{r}$ is the identity of $B_{i}$ . It acts as the identity on $M_{i}$ and as zero on $M_{j}$ for $j \neq i$ .
$r$ , $r'$: The number of isomorphism classes of simple left modules over $R$ and over $R^{K}$ respectively.

The element $a_{i}$ need not be idempotent in $R$ — only its image modulo the radical is. That is enough for every argument on this page, because only actions on semisimple subquotients are used.

Core Concepts

Flatness turns series into filtrations

$K$ is free as a $k$ -module, so $- \otimes_{k} K$ is exact. A composition series $0 = I_{0} ⊊ I_{1} ⊊ \dots ⊊ I_{s} = R$ of the left regular module $_{R} R$ therefore extends to a chain of $R^{K}$ -submodules of $R^{K}$ whose successive quotients are $(I_{j} / I_{j - 1})^{K}$ — extensions of simple $R$ -modules.

composition series of

_{R} R

filtration of

_{R^{K}} R^{K}

factors

M^{K}

refine to simples over

R^{K}

Separating elements survive base change

Choose $a_{i} \in R$ mapping to the identity of the $i$ -th simple component of $R / rad R$ . Then $a_{i}$ acts as the identity on all of $M_{i}$ , hence as the identity on $M_{i}^{K}$ and on every subquotient of it; and it acts as zero on $M_{j}$ , hence on every subquotient of $M_{j}^{K}$ . A module cannot have the same element acting both as $1$ and as $0$ unless it is zero, so the two families of composition factors are disjoint.

Key Results

Proposition(7.13)Simple modules over an extended algebra

Let $R$ be a $k$ -algebra with $\dim_{k} R < \infty$ and let $K \supseteq k$ be a field extension. Then:

every simple left $R^{K}$ -module $V$ is a composition factor of $M^{K}$ for some simple left $R$ -module $M$ ;
if $M_{1}, M_{2}$ are non-isomorphic simple left $R$ -modules, then $M_{1}^{K}$ and $M_{2}^{K}$ have no composition factor in common.

Proof

(2) Write $R / rad R ≅ B_{1} \times \dots \times B_{r}$ and arrange that $M_{i}$ is the unique simple module of $B_{i}$ for $i = 1, 2$ . Choose $a_{i} \in R$ whose image in $R / rad R$ is the identity element of $B_{i}$ .

Then $a_{1}$ acts as the identity on $M_{1}$ , hence as the identity on $M_{1}^{K}$ under $a_{1} \otimes 1$ , hence as the identity on every subquotient of $M_{1}^{K}$ ; and $a_{1}$ annihilates $M_{2}$ , hence annihilates $M_{2}^{K}$ and all of its subquotients. If some simple $R^{K}$ -module $V$ occurred in both, $a_{1} \otimes 1$ would act on $V$ both as the identity and as zero, forcing $V = 0$ — impossible.

(1) Take a composition series $0 = I_{0} ⊊ I_{1} ⊊ \dots ⊊ I_{s} = R$ of the left regular module $_{R} R$ , which exists because $\dim_{k} R < \infty$ . Applying the exact functor $- \otimes_{k} K$ gives a chain of $R^{K}$ -submodules

0 = I_{0}^{K} ⊊ I_{1}^{K} ⊊ \dots ⊊ I_{s}^{K} = R^{K}, I_{j}^{K} / I_{j - 1}^{K} ≅ (I_{j} / I_{j - 1})^{K}

Each factor is $M^{K}$ for a simple left $R$ -module $M$ . Now let $V$ be any simple left $R^{K}$ -module. Being simple it is cyclic, so it is a quotient of $_{R^{K}} R^{K}$ , and therefore occurs as a composition factor of the regular module. Refining the displayed chain to a composition series and applying Jordan–Hölder, $V$ must occur as a composition factor of some $I_{j}^{K} / I_{j - 1}^{K} ≅ M^{K}$ .

CorollaryThe count does not decrease

With $r$ the number of simple left $R$ -modules and $r'$ the number of simple left $R^{K}$ -modules, $r \leq r'$ .

Indeed each $M_{i}^{K}$ is nonzero and so has at least one composition factor, and by $(7.13) (2)$ these sets of factors are pairwise disjoint for $i = 1, \dots, r$ . Choosing one factor from each produces $r$ pairwise non-isomorphic simple $R^{K}$ -modules.

Proposition(7.14)Splitting fields are stable under extension

Let $R$ be a $k$ -algebra with $\dim_{k} R < \infty$ and let $K \supseteq k$ be a splitting field for $R$ . Let $V_{1}, \dots, V_{m}$ be a full set of simple left $R^{K}$ -modules. Then for every field $L \supseteq K$ , the modules $V_{1}^{L}, \dots, V_{m}^{L}$ form a full set of simple left $R^{L}$ -modules. In particular $L$ is again a splitting field for $R$ .

Proof

Each $V_{i}$ is absolutely irreducible over $K$ by the definition of a splitting field, so $V_{i}^{L}$ is a simple $R^{L}$ -module by $(7.5) (3)$ .

They are pairwise non-isomorphic: applying $(7.13) (2)$ to the algebra $R^{K}$ over the field $K$ and the extension $L \supseteq K$ , the modules $V_{i}^{L}$ and $V_{j}^{L}$ have no common composition factor for $i \neq j$ ; since both are simple, they cannot be isomorphic.

They exhaust the simple modules: by $(7.13) (1)$ , applied again over $K$ , every simple $R^{L}$ -module is a composition factor of some $V_{i}^{L}$ , and $V_{i}^{L}$ is simple, so it equals $V_{i}^{L}$ up to isomorphism.

Finally each $V_{i}^{L}$ is absolutely irreducible over $L$ , because for any $L' \supseteq L$ we have $(V_{i}^{L})^{L'} ≅ V_{i}^{L'}$ , which is simple since $V_{i}$ is absolutely irreducible over $K$ . Hence $L$ splits $R$ .

CorollaryRadical and base change for split algebras

If the $k$ -algebra $R$ with $\dim_{k} R < \infty$ splits over $k$ , then $rad (R^{K}) = (rad R)^{K}$ for every field extension $K \supseteq k$ .

Proof

The inclusion $(rad R)^{K} \subseteq rad (R^{K})$ holds because $rad R$ is nilpotent, so its extension is a nilpotent ideal of $R^{K}$ .

For the reverse inclusion, note that $R / rad R ≅ \prod_{i} M_{n_{i}} (k)$ because $R$ splits over $k$ , whence

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

which is semisimple. A quotient of $R^{K}$ by an ideal is semisimple only if that ideal contains $rad (R^{K})$ , so $rad (R^{K}) \subseteq (rad R)^{K}$ and the two agree.

RemarkCommon factors persist upwards

If two finite-dimensional $R$ -modules $M$ and $N$ share a composition factor $S$ , then $M^{K}$ and $N^{K}$ share a composition factor for every $K \supseteq k$ : the composition factors of $S^{K}$ occur among those of both $M^{K}$ and $N^{K}$ , and $S^{K} \neq 0$ . The converse requires extra hypotheses — it holds, for instance, when $M^{K}$ and $N^{K}$ are semisimple.

Proof Techniques and Method

How these proofs work, and which move to reuse.

The two halves of $(7.13)$ use opposite strategies, and both are worth keeping.

Existence half

Everything lives in the regular module

A simple module is cyclic, hence a quotient of the regular module, hence one of its composition factors. Extending a composition series of $_{R} R$ therefore captures every simple $R^{K}$ -module at once.

Disjointness half

Separate by an element, not by a submodule

One element acting as $1$ on one family and as $0$ on the other separates all their subquotients simultaneously. Submodule arguments would not survive base change; element arguments do.

Bootstrapping

Apply the result to the extended algebra

$(7.14)$ is proved by applying $(7.13)$ with $R^{K}$ in place of $R$ and $L$ in place of $K$ . Base-change statements are almost always used this way — twice, in a tower.

Semisimplicity by quotient

Identify the radical from above

To show an ideal contains the radical, show the quotient by it is semisimple. This is how the split case of $rad (R^{K})$ is settled without computing anything.

Note the asymmetry of the tower argument: it needs $K$ to split $R$ before the induction begins. Without splitting, passing from $K$ to $L$ can genuinely produce new simple modules, and $(7.14)$ is false.

Worked Example

A separable example: $R = ℚ C_{3}$

Let $G = ⟨ g ⟩$ be cyclic of order $3$ , $R = ℚ G ≅ ℚ \times ℚ (ω)$ with $ω$ a primitive cube root of unity, and $K = ℚ (ω)$ . Then $rad R = 0$ and there are $r = 2$ simple modules: the trivial module $M_{1} = ℚ$ , and $M_{2} = ℚ (ω)$ of $ℚ$ -dimension $2$ .

Extending, $R^{K} ≅ K \times (ℚ (ω) \otimes_{ℚ} ℚ (ω)) ≅ K \times K \times K$ , so $r' = 3$ and every simple $R^{K}$ -module is one-dimensional. The two extended modules split as

Composition factors over $K = ℚ (ω)$
Simple $R$ -module	$\dim_{ℚ}$	Composition factors of $M^{K}$	Semisimple?
$M_{1} = ℚ$ (trivial)	$1$	$V_{1}$ (trivial character)	yes
$M_{2} = ℚ (ω)$	$2$	$V_{2} \oplus V_{3}$ (the two nontrivial characters)	yes

Both parts of $(7.13)$ are visible: the three simple $R^{K}$ -modules all occur among the factors, and the factor sets ${V_{1}}$ and ${V_{2}, V_{3}}$ are disjoint. The count rose from $r = 2$ to $r' = 3$ , and $K$ splits $R$ , so by $(7.14)$ no further extension changes anything: over $ℂ$ the simple modules are still three one-dimensional characters.

An inseparable example: repeated factors

Let $p$ be prime, $k = 𝔽_{p} (u)$ and $R = k [t] / (t^{p} - u)$ , which is the field $K : = k (u^{1 / p})$ . So $R$ is simple, $r = 1$ , $M = R$ is the unique simple module, and $rad R = 0$ .

Extend to $K$ . Then $R^{K} ≅ K [t] / ((t - u^{1 / p})^{p})$ is local with radical $(t - u^{1 / p})$ , and $r' = 1$ as well. The extended module $M^{K} = R^{K}$ is the regular module: indecomposable, of $K$ -dimension $p$ , with the unique simple module $V = K$ occurring $p$ times as a composition factor.

M^{K} = R^{K} ⊋ rad (R^{K}) ⊋ rad (R^{K})^{2} ⊋ \dots ⊋ rad (R^{K})^{p} = 0

A composition series with $p$ factors, all isomorphic to $V = K$ .

Comparison and Classification

What survives scalar extension $k \to K$
	Any $K$	$K / k$ separable	$K$ splits $R$
$\dim_{K} X^{K} = \dim_{k} X$	yes	yes	yes
$(rad R)^{K} \subseteq rad (R^{K})$	yes	yes	yes
$rad (R^{K}) = (rad R)^{K}$	no	yes	partial
$M$ simple $\Rightarrow M^{K}$ simple	no	no	partial
$M$ simple $\Rightarrow M^{K}$ semisimple	no	yes	partial
$M$ decomposable $\Rightarrow M^{K}$ decomposable	yes	yes	yes
Number of simple modules	grows	grows	final
$Hom$ commutes with extension	yes	yes	yes

What survives scalar extension $k \to K$

Counting simple modules across a tower
Algebra	$k$	$r$ over $k$	$K$	$r'$ over $K$
$ℚ C_{3}$	$ℚ$	$2$	$ℚ (ω)$	$3$
$ℝ C_{3}$	$ℝ$	$2$	$ℂ$	$3$
$ℍ$	$ℝ$	$1$	$ℂ$	$1$
$𝔽_{p} (u) [t] / (t^{p} - u)$	$𝔽_{p} (u)$	$1$	$𝔽_{p} (u^{1 / p})$	$1$
$M_{n} (k)$	$k$	$1$	any	$1$

Relationship Map

Scalar extension organises the simple modules into a tree: each simple $R$ -module sits above the composition factors it produces, and the branches never meet.

Simple $R$ -modules

M_{1}, \dots, M_{r}

Extended modules

M_{1}^{K}, \dots, M_{r}^{K}

, of the same

K

-dimensions

Composition factorspairwise disjoint families, exhausting the simple

R^{K}

-modules

After a splitting fieldeach family is a single absolutely irreducible module, and further extension changes nothing

M

simple over

R

M^{K}

over

R^{K}

composition factors

V

V^{L}

over

R^{L}

Applications and Industry Use

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

Modular representation theory. Decomposition matrices compare composition factors of a module before and after a change of coefficient ring; $(7.13) (2)$ is the statement that the matrix is block-structured with respect to distinct simples.
Computational module theory. The MeatAxe reports the field of definition of a simple module over a finite field; passing to that field is exactly the extension in $(7.13)$ , and $(7.14)$ guarantees the answer is stable once a splitting field is reached.
Coding theory. The minimal cyclic codes over $𝔽_{q}$ correspond to simple modules; extending to $𝔽_{q^{m}}$ splits them into the conjugate components indexed by cyclotomic cosets, a concrete instance of $(7.13) (1)$ .
Number theory. For a group algebra over a number field, the way $M^{K}$ decomposes as $K$ varies is governed by the Schur index and by Galois conjugation of characters — the arithmetic refinement of the qualitative statements here.

The uniform benefit is that computations may be carried out over the most convenient field and the results transported, with $(7.13)$ controlling exactly what is lost or gained in transit.

Computational Notes

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

To find the simple $R^{K}$ -modules from those of $R$ : extend each $M_{i}$ , compute the radical of the extended algebra restricted to $M_{i}^{K}$ , and split the semisimple quotient. Only $r$ modules need to be processed, by $(7.13) (1)$ .
Over a finite field the field of definition of a simple module is $𝔽_{q^{d}}$ with $d = \dim_{𝔽_{q}} End$ of the module; extending to $𝔽_{q^{d}}$ splits it into $d$ Galois-conjugate absolutely irreducible modules.
$(7.14)$ is the practical justification for computing once over a splitting field and reusing the answer; without it every new coefficient field would require a fresh decomposition.
The corollary $rad (R^{K}) = (rad R)^{K}$ for split algebras means a radical computed over the small field can be reused verbatim — a substantial saving when $K$ is large.

Failure Modes and Common Mistakes

Do not assume $r < r'$ strictly; equality is common, as for $ℍ$ over $ℝ$ extended to $ℂ$ .
Do not apply the radical corollary without checking that $R$ splits over $k$ — that hypothesis is doing all the work.
Do not conclude from a common composition factor of $M^{K}$ and $N^{K}$ that $M$ and $N$ share one; the converse direction needs $M^{K}$ and $N^{K}$ semisimple.

Quick Reference

Setting

\dim_{k} R < \infty

K \supseteq k

R^{K} = R \otimes_{k} K

(7.13)(1)every simple

R^{K}

-module occurs in some

M^{K}

(7.13)(2)

M_{1} \neq M_{2}

simple

\Rightarrow

M_{1}^{K}

M_{2}^{K}

share no factor

Count

r \leq r'

(7.14)

K

splits

R

L \supseteq K

\Rightarrow

simples over

L

are the

V_{i}^{L}

Split radical

R

splits over

k

\Rightarrow

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

Always

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

Dimension

\dim_{K} M^{K} = \dim_{k} M

Statement finder
Result	Content	Reference
Existence of factors	simple $R^{K}$ -modules occur inside extended simples	(7.13)(1)
Disjointness	distinct simples give disjoint factor families	(7.13)(2)
Monotone count	$r \leq r'$	corollary of (7.13)(2)
Stability of splitting	$V_{i}^{L}$ is a full set of simples over $L$	(7.14)
Radical base change	equality for split algebras	Exercise 7.2
Persistence of shared factors	common factor of $M$ , $N$ persists to $M^{K}$ , $N^{K}$	Exercise 7.1

Frequently Asked Questions

Why is the statement about composition factors rather than direct summands?

Because $M^{K}$ need not be semisimple. Over a separable extension one can upgrade the conclusion to a direct sum decomposition, but the proposition is stated so that it also covers inseparable extensions, where an extended simple module can be indecomposable of length greater than one.

Can the number of simple modules decrease when the field grows?

No. The families of composition factors attached to distinct simple $R$ -modules are disjoint and nonempty, so choosing one from each already yields $r$ distinct simple $R^{K}$ -modules.

Does

(7.14)

mean the simple modules never change again?

It means they change only in the trivial way: over $L \supseteq K$ the simple modules are exactly the extensions $V_{i}^{L}$ , with the same labels, the same multiplicities in the regular module, and the same dimensions. Nothing new appears and nothing merges.

Where exactly does the proof of (7.13)(2) use finite dimensionality?

In the existence of the decomposition $R / rad R ≅ B_{1} \times \dots \times B_{r}$ , which requires the quotient to be semisimple — guaranteed here because $R$ is artinian. Without that, there is no element $a_{i}$ to separate the modules.

Is the equality

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

ever available without splitting?

Yes, whenever $K / k$ is separable; that is the content of the standard result on radicals under separable field extension. The version proved here uses splitting instead, and is often more convenient because it does not restrict the extension at all.

How does this relate to the Schur index?

The Schur index of a simple module measures the size of the central division algebra in its endomorphism ring, and therefore how many copies of a single simple module appear in $M^{K}$ over a splitting field. The propositions here say which modules appear; the Schur index says with what multiplicity.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §7 (pp. 116–118), including Exercises 7.1 and 7.2.
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley-Interscience, 1981, §7.
R. S. Pierce, Associative Algebras, Graduate Texts in Mathematics 88, Springer-Verlag, 1982, Chapter 11.
N. Jacobson, Basic Algebra II, 2nd edition, W. H. Freeman, 1989, Chapter 4.
W. Feit, The Representation Theory of Finite Groups, North-Holland, 1982, Chapter I.

AI Suggested Questions

Prove that if $K / k$ is separable then $M^{K}$ is semisimple for every semisimple $R$ -module $M$ .
Compute the composition factors of $M^{K}$ for $R = ℚ S_{3}$ and $K = ℚ (\sqrt{- 3})$ .
Give an example where $r = r'$ but the simple modules genuinely change shape under extension.
How do decomposition matrices in modular representation theory use (7.13)(2)?
What is the largest possible multiplicity of a composition factor in $M^{K}$ for $M$ simple, in terms of the inseparability degree of $K$ over $k$ ?
Show that the converse of the last remark holds when $M^{K}$ and $N^{K}$ are semisimple.
How does one compute the field of definition of a simple module over a finite field in practice?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Flatness turns series into filtrations

Separating elements survive base change

06Key Results

07Proof Techniques and Method

Everything lives in the regular module

Separate by an element, not by a submodule

Apply the result to the extended algebra

Identify the radical from above

08Worked Example

A separable example: R=ℚC3

An inseparable example: repeated factors

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

12Computational Notes

13Failure Modes and Common Mistakes

14Quick Reference

15Frequently Asked Questions

16Related KEVOS Topics

17References

18AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

A separable example: $R = ℚ C_{3}$

Comparison and Classification

Relationship Map

Applications and Industry Use

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions