The Radical under Field Extension of Scalars

Q: Why can scalar extension create radical at all?

Because R^K has more elements than R does, and new nilpotents can appear that were invisible over k. The mechanism is concrete: if K = k() with ^p = a ∈ k, then 1 ⊗ - ⊗ 1 is a nonzero element of K ⊗_k K whose p-th power is 1 ⊗ a - a ⊗ 1 = 0. Nothing like it exists over k.

Q: Is rad R ⊆ rad R^K always true?

No. It holds for algebraic K/k and for finite-dimensional R, both by (5.14)(2). It fails for R = k[[x]] and K = k(t), where rad R = (x) meets rad R^K in 0 by (5.15).

Q: How does the finite-degree bound interact with (5.17)?

When K/k is separable of degree n, (5.17) gives equality, so the bound in (5.14)(3) is not needed. The bound earns its keep in the inseparable case, where it says the discrepancy between rad R^K and (rad R)^K is nilpotent of index at most n — and the degree-p example shows the index n is attained.

Q: Does (5.16) need the extension to be Galois?

The trace argument is run over a finite Galois extension, but the statement only assumes separable algebraic. The proof reduces first to a finite subextension, then to its normal hull, which is Galois because a normal separable extension is Galois. Separability is thus used twice: to make the hull Galois and to make the trace form nondegenerate.

Q: What does this say about semisimple algebras?

If R is finite-dimensional and semisimple and K/k is separable algebraic, then R^K is again semisimple, since rad R^K = (rad R)^K = 0 and _K R^K = _k R < ∈fty. Over a perfect field this extends to arbitrary extensions; over an imperfect field the inseparable counterexample shows it can fail.

Q: Where is (5.15) used?

It supplies the missing half of Amitsur's rational function field theorem (5.13): the assertion that N' = R rad R(T) is nil. Since k(T)/k is transcendental, (5.15) applies verbatim.

Executive Summary

Enlarging the scalars of a $k$ -algebra $R$ to a field extension $K / k$ produces $R^{K} = R \otimes_{k} K$ . Four facts govern the radical, and they are sensitive to quite different features of $K / k$ .

Always: $R \cap rad R^{K} \subseteq rad R$ , because $R R$ is a direct summand of $R R^{K}$ .
**Algebraic $K / k$ , or $\dim_{k} R < \infty$ :** equality, $R \cap rad R^{K} = rad R$ .
**Transcendental $K / k$ :** $R \cap rad R^{K}$ is a nil ideal, so the inclusion is strict whenever $rad R$ is not nil.
**Separable algebraic $K / k$ :** $rad (R^{K}) = (rad R)^{K}$ exactly.

Separability is not a technical convenience: for a purely inseparable extension $K = k (α)$ with $α^{p} = a \in k$ , the algebra $K \otimes_{k} K$ has nonzero nilpotent radical even though $K$ is a field. That single example blocks any hope of extending $(5.17)$ .

$\subseteq rad R$ Contraction, always

NilContraction, transcendental

K / k

$(rad R)^{K}$

rad R^{K}

, separable algebraic

$n$ Nilpotence bound when

[K : k] = n

Overview

Scalar extension is the standard way to simplify an algebra: pass to a splitting field, decompose, and descend. The question is what happens to the obstruction — the radical — along the way. Unlike a polynomial extension, a scalar extension can create radical: $K \otimes_{k} K$ is the smallest example, and it is the reason the theory of separable algebras exists.

R^{K} = R \otimes_{k} K, (r \otimes λ) (r' \otimes λ') = r r' \otimes λ λ' .

(5.S)

No hypothesis is imposed on $K / k$ : it may be finite, infinite algebraic, or transcendental. $R \otimes 1$ is identified with $R$ .

Choosing a $k$ -basis ${e_{i}}$ of $K$ with $e_{1} = 1$ decomposes $R^{K} = ⨁_{i} R e_{i}$ as a left $R$ -module, so $R$ is a direct summand and the change-of-rings machinery of Behaviour of the Radical under Ring Extensions applies immediately. When the basis is finite, its elements also centralise $R$ , so ascent applies too — which is why finiteness of $[K : k]$ changes the answer.

Learning Objectives

State $(5.14)$ as three separate assertions with three separate hypotheses.
Prove the contraction inclusion from the direct summand decomposition of $R^{K}$ .
Prove $(rad R^{K})^{n} \subseteq (rad R)^{K}$ for $[K : k] = n$ using composition length.
Show the contraction is nil for transcendental $K / k$ by producing an algebraic element of the radical.
Run the Galois trace argument proving that $rad R = 0$ implies $rad R^{K} = 0$ for separable algebraic $K / k$ .
Exhibit the purely inseparable counterexample and identify the nilpotent generator of its radical.

Definitions

$R^{K}$: The K-algebra R tensor over k with K. Its dimension over K equals the dimension of R over k when that is finite.
$(rad R)^{K}$: The image of rad R tensor K inside R tensor K; a two-sided ideal, since K is central.
Semiprimitive: Zero Jacobson radical; the property whose stability under scalar extension is at issue here.
Normal hull: For a finite extension K of k, the smallest normal extension of k containing K; separable together with normal means Galois.
Separable algebra: A finite-dimensional k-algebra whose scalar extension to every field extension of k is semisimple.

$k$ is always a field and $R$ an associative $k$ -algebra with identity, not assumed commutative or finite-dimensional unless stated.

Core Concepts

Two module structures, two theorems

Fix a $k$ -basis ${e_{i}}_{i \in I}$ of $K$ with $e_{1} = 1$ . Then

R^{K} = R \oplus ⨁_{i \neq 1} R e_{i} as left R -modules .

(5.14a)

This is a splitting, so descent $(5.6) (1)$ applies for any $K / k$ and gives the contraction inclusion. The $e_{i}$ also centralise $R$ , since $R \otimes 1$ and $1 \otimes K$ commute in a tensor product of algebras over a central field. But ascent $(5.7)$ additionally demands finitely many generators, so it applies only when $[K : k] < \infty$ — and then a direct limit extends the conclusion to all algebraic extensions.

[K : k] < \infty

ascent and descent both apply

rad R = R \cap rad R^{K}

direct limit: same for all algebraic

K / k

Why separability is the dividing line

For a finite Galois extension $E / k$ with group $G$ , the action of $G$ on $E$ extends to $R^{E}$ by $1 \otimes σ$ , and $rad R^{E}$ is invariant because it is invariant under every ring automorphism. Averaging an element of the radical over $G$ produces field traces, and the trace form of $E / k$ is nondegenerate **precisely when $E / k$ is separable**. That nondegeneracy is what forces the averaged element to vanish coordinatewise.

({tr}_{E / k} (e_{i} e_{j}))_{i, j} invertible over k iff E / k separable .

(5.16a)

In the purely inseparable situation the trace map is identically zero, the averaging produces nothing, and indeed the conclusion is false.

Key Results

Theorem(5.14)Contraction, equality and the finite-degree bound

Let $k$ be a field, $R$ a $k$ -algebra and $K / k$ a field extension. Then:

$R \cap rad R^{K} \subseteq rad R$ , with no hypothesis on $K / k$ ;
if $K / k$ is algebraic, or if $\dim_{k} R < \infty$ , then $R \cap rad R^{K} = rad R$ ;
if $[K : k] = n < \infty$ , then $(rad R^{K})^{n} \subseteq (rad R)^{K} = (rad R) \otimes_{k} K$ .

Proof

(1) By $(5.14 a)$ , $R R$ is a direct summand of $R R^{K}$ , so descent $(5.6) (1)$ gives the inclusion.

(2), finite-dimensional case. If $\dim_{k} R < \infty$ then $R$ is artinian, so $rad R$ is nilpotent by $(4.12)$ , say $(rad R)^{m} = 0$ . Then $((rad R)^{K})^{m} = ((rad R)^{m})^{K} = 0$ , so $(rad R)^{K}$ is a nilpotent ideal of $R^{K}$ and $(4.11)$ places it inside $rad R^{K}$ . Hence $rad R \subseteq R \cap rad R^{K}$ , and (1) gives equality.

(2), algebraic case. First let $[K : k] = n < \infty$ . Then $R^{K} = \sum_{i = 1}^{n} R e_{i}$ is generated as a left $R$ -module by finitely many elements centralising $R$ , so ascent $(5.7)$ gives $rad R \subseteq rad R^{K}$ , hence $rad R \subseteq R \cap rad R^{K}$ ; with (1) this is equality. For general algebraic $K / k$ , let $a \in rad R$ and $z \in R^{K}$ . Then $z$ involves finitely many elements of $K$ , all algebraic over $k$ , so $z \in R^{L}$ for a finite subextension $L / k$ . By the finite case $a \in rad R^{L}$ , so $1 - z a$ is invertible in $R^{L} \subseteq R^{K}$ . As $z$ was arbitrary, $a \in rad R^{K}$ .

(3) Let $[K : k] = n$ with $k$ -basis $e_{1}, \dots, e_{n}$ , and let $V$ be any simple right $R$ -module. Then $V^{K} = V \otimes_{k} K$ is a right $R^{K}$ -module, and as a right $R$ -module $V^{K} = ⨁_{i = 1}^{n} V \otimes e_{i}$ has composition length exactly $n$ . Every $R^{K}$ -submodule is in particular an $R$ -submodule, so the composition length of $V^{K}$ over $R^{K}$ is at most $n$ . The radical annihilates each composition factor, so $V^{K} \cdot (rad R^{K})^{n} = 0$ .

Now take $z \in (rad R^{K})^{n}$ and write $z = \sum_{i = 1}^{n} r_{i} \otimes e_{i}$ with $r_{i} \in R$ . For every $v \in V$ , $0 = (v \otimes 1) z = \sum_{i} (v r_{i}) \otimes e_{i}$ , and since $V^{K} = ⨁_{i} V \otimes e_{i}$ this forces $v r_{i} = 0$ for each $i$ . Thus $V r_{i} = 0$ for every simple right $R$ -module $V$ , i.e. $r_{i} \in rad R$ , and $z \in (rad R) \otimes_{k} K$ .

Proposition(5.15)Transcendental extensions give a nil contraction

Let $K / k$ be a field extension that is not algebraic. Then for every $k$ -algebra $R$ , the ideal $R \cap rad R^{K}$ is nil. Consequently, if $rad R$ is not nil, the inclusion in $(5.14) (1)$ is strict.

Proof

Let $a \in R \cap rad R^{K}$ and choose $ξ \in K$ transcendental over $k$ . Applying $(5.14) (1)$ to the $k (ξ)$ -algebra $R^{k (ξ)}$ and the extension $K / k (ξ)$ gives $R^{k (ξ)} \cap rad R^{K} \subseteq rad R^{k (ξ)}$ , so $a \in rad R^{k (ξ)}$ . We may therefore assume $K = k (ξ)$ , and we write $ξ = t$ .

Since $a t \in rad R^{K}$ , the element $1 - a t$ is invertible, and its inverse has the form $f (t) / g (t)$ with $f \in R [t]$ and $0 \neq g \in k [t]$ central. Write $f (t) = b_{0} + b_{1} t + \dots + b_{m} t^{m}$ with $b_{m} \neq 0$ and $g (t) = c_{0} + c_{1} t + \dots + c_{m + 1} t^{m + 1}$ with $c_{j} \in k$ . Then $(1 - a t) f (t) = g (t)$ , and comparing coefficients (with the convention $b_{m + 1} = 0$ ) gives $c_{0} = b_{0}$ and $c_{i} = b_{i} - a b_{i - 1}$ for $1 \leq i \leq m + 1$ . Solving recursively,

b_{i} = c_{i} + a c_{i - 1} + a^{2} c_{i - 2} + \dots + a^{i} c_{0} (0 \leq i \leq m + 1) .

(5.15a)

Because $b_{m} \neq 0$ , the scalars $c_{0}, \dots, c_{m}$ are not all zero. Taking $i = m + 1$ in $(5.15 a)$ and using $b_{m + 1} = 0$ yields

c_{0} a^{m + 1} + c_{1} a^{m} + \dots + c_{m} a + c_{m + 1} = 0,

(5.15b)

A nonzero polynomial relation over $k$ satisfied by $a$ .

so $a$ is algebraic over $k$ . An element of the radical of a $k$ -algebra that is algebraic over $k$ is nilpotent: if $a$ satisfies a polynomial of least degree, write it as $a^{d} q (a) = 0$ with $q (0) = c \neq 0$ ; since $a \in rad R^{K}$ , $q (a) \in c + rad R^{K}$ is a unit, so $a^{d} = 0$ . This is $(4.18)$ . Hence $a$ is nilpotent.

Lemma(5.16)Semiprimitivity survives separable algebraic extension

Let $R$ be a $k$ -algebra and $K / k$ a separable algebraic field extension. If $rad R = 0$ then $rad R^{K} = 0$ .

Proof

Reduction to a finite extension. Any $z \in rad R^{K}$ lies in $R^{L}$ for some finite subextension $L / k$ , and $(5.14) (1)$ applied over $L$ gives $R^{L} \cap rad R^{K} \subseteq rad R^{L}$ . So it suffices to show $rad R^{L} = 0$ for every finite separable $L / k$ .

Reduction to a Galois extension. Let $E$ be the normal hull of $L / k$ ; then $E / k$ is finite Galois. By $(5.14) (2)$ applied to the algebraic extension $E / L$ , $rad R^{L} \subseteq rad ((R^{L})^{E}) = rad R^{E}$ , so it is enough to prove $rad R^{E} = 0$ .

The trace argument. Let $e_{1}, \dots, e_{n}$ be a $k$ -basis of $E$ and $G = Gal (E / k)$ , acting on $R^{E}$ by $1 \otimes σ$ . Take $z = \sum_{i} r_{i} \otimes e_{i} \in rad R^{E}$ and fix an index $j$ . For each $σ \in G$ the element $σ (z (1 \otimes e_{j})) = \sum_{i} r_{i} \otimes σ (e_{i} e_{j})$ again lies in $rad R^{E}$ , because that ideal is invariant under all ring automorphisms. Summing over $σ \in G$ ,

\sum_{σ \in G} σ (z (1 \otimes e_{j})) = \sum_{i} r_{i} \otimes {tr}_{E / k} (e_{i} e_{j}) = (\sum_{i} r_{i} {tr}_{E / k} (e_{i} e_{j})) \otimes 1,

(5.16b)

which lies in $R \cap rad R^{E} \subseteq rad R = 0$ by $(5.14) (1)$ . Hence $\sum_{i} r_{i} {tr}_{E / k} (e_{i} e_{j}) = 0$ for every $j$ . Since $E / k$ is separable, the trace form is nondegenerate, so the matrix $({tr}_{E / k} (e_{i} e_{j}))$ is invertible over $k$ ; solving the linear system gives $r_{i} = 0$ for all $i$ , i.e. $z = 0$ .

Theorem(5.17)The radical commutes with separable algebraic extension

Let $R$ be a $k$ -algebra and $K / k$ a separable algebraic field extension. Then $rad (R^{K}) = (rad R)^{K} = (rad R) \otimes_{k} K$ .

Proof

By $(5.14) (2)$ , $rad R \subseteq rad R^{K}$ , and since $rad R^{K}$ is an ideal, $(rad R)^{K} = (rad R) \cdot R^{K} \subseteq rad R^{K}$ . Tensoring over the field $k$ is exact, so

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

(5.17a)

The algebra $R / rad R$ is semiprimitive, so $(5.16)$ makes the right-hand side semiprimitive. Since $(rad R)^{K}$ is contained in $rad R^{K}$ , the quotient rule $(4.6)$ gives $rad (R^{K}) / (rad R)^{K} = rad (R^{K} / (rad R)^{K}) = 0$ , i.e. $rad (R^{K}) = (rad R)^{K}$ .

CounterexampleInseparability breaks (5.16) and (5.17)

Let $char k = p > 0$ and $a \in k ∖ k^{p}$ ; put $K = k (α)$ with $α^{p} = a$ , a purely inseparable extension of degree $p$ . Take $R = K$ , regarded as a $k$ -algebra: it is a field, hence semisimple, and $rad R = 0$ . But

R^{K} = K \otimes_{k} K ≅ K [x] / (x^{p} - a) = K [x] / ((x - α)^{p}),

(E.0)

a local $K$ -algebra of dimension $p$ whose radical is the nonzero nilpotent ideal generated by $x - α$ ; tracing back through the isomorphism, $rad (K \otimes_{k} K)$ is generated by $1 \otimes α - α \otimes 1$ . More generally, if $A$ is any $k$ -algebra with $K \subseteq Z (A)$ , then $A^{K}$ has the nonzero central nilpotent element $1 \otimes α - α \otimes 1$ and is never semiprimitive.

RemarkPerfect base fields

Over a perfect field — in particular in characteristic $0$ and over every finite field — all algebraic extensions are separable, so $(5.17)$ applies to every algebraic $K / k$ . For finite-dimensional algebras over a perfect field the stronger classical statement holds: semisimplicity is preserved by every scalar extension, so such algebras are separable in the sense defined above.

Proof Techniques and Method

The reusable moves behind (5.14)–(5.17).

Move 1

Read the tensor product as a module

Choosing a basis of $K$ with $1$ among it turns $R^{K}$ into a free left $R$ -module containing $R$ as a summand. Descent then costs nothing, and finiteness of the basis is precisely what buys ascent.

Move 2

Bound composition length

A module of finite length over the smaller ring has at most that length over the bigger one, because bigger-ring submodules are smaller-ring submodules. That converts $[K : k] = n$ into a nilpotence exponent $n$ .

Move 3

Average with the Galois group and use the trace form

Summing $σ (z e_{j})$ over $σ \in G$ lands in the fixed ring, where the contraction inclusion applies. Nondegeneracy of the trace form then converts the resulting linear system into $z = 0$ .

Move 3 is the only place separability is used, and it is used through a linear-algebra fact rather than a field-theoretic one: the Gram matrix of the trace form must be invertible. In the purely inseparable case that matrix is zero.

Which statement applies to my extension $K / k$ ?

Separable algebraic

(5.17)

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

. Semiprimitivity and semisimplicity both transfer, in both directions.

Algebraic but not separable

(5.14) (2)

still gives

R \cap rad R^{K} = rad R

, and for

[K : k] = n

you get

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

— but equality can fail, as

K \otimes_{k} K

shows.

Transcendental, $R$ arbitrary

(5.15)

: the contraction is nil. Expect strict inclusion whenever

rad R

is not nil, e.g.

R = k [[x]]

Transcendental, $\dim_{k} R < \infty$

(5.14) (2)

gives equality of the contraction with

rad R

, consistent with

(5.15)

because a finite-dimensional algebra has nilpotent radical.

Worked Example

A separable extension: quaternions complexified

Let $k = ℝ$ , $K = ℂ$ (separable, degree $2$ ) and $R = ℍ$ , the real quaternions. $ℍ$ is a division ring, so $rad ℍ = 0$ , and $(5.17)$ predicts $rad (ℍ \otimes_{ℝ} ℂ) = 0$ . Indeed

ℍ \otimes_{ℝ} ℂ ≅ M_{2} (ℂ),

(E.1)

A simple artinian algebra, radical zero — the prediction is confirmed.

Here scalar extension destroys the division ring structure but not semiprimitivity; $ℂ$ is a splitting field for $ℍ$ .

An inseparable extension in characteristic $2$

Let $k = 𝔽_{2} (u)$ with $u$ transcendental over $𝔽_{2}$ , and $K = k (α)$ with $α^{2} = u$ . Then $[K : k] = 2$ and $K / k$ is purely inseparable. Take $R = K$ , so $rad R = 0$ and $(rad R)^{K} = 0$ . Computing the scalar extension:

R^{K} = K \otimes_{k} K ≅ K [x] / (x^{2} - u) = K [x] / ((x - α)^{2}),

(E.2)

a $2$ -dimensional local $K$ -algebra. Its radical is $K \cdot (x - α)$ , of dimension $1$ , and squares to zero. So $rad R^{K} \neq (rad R)^{K}$ and $(5.17)$ genuinely fails. The nilpotent generator is $z = 1 \otimes α - α \otimes 1$ , and one checks $z^{2} = 1 \otimes α^{2} - α^{2} \otimes 1 = 1 \otimes u - u \otimes 1 = 0$ because $u \in k$ .

A transcendental extension where the contraction shrinks

Let $R = k [[x]]$ , a $k$ -algebra with $rad R = (x)$ , which contains no nonzero nilpotent element. Let $K = k (t)$ , transcendental over $k$ . By $(5.15)$ , $R \cap rad R^{K}$ is nil, hence zero. So

R \cap rad R^{k (t)} = 0 ⊊ (x) = rad R,

(E.3)

The equality in $(5.14) (2)$ fails as soon as the extension is transcendental and the radical is not nil.

Comparison and Classification

What holds for which type of extension
Type of $K / k$	$R \cap rad R^{K}$	$rad R^{K}$ vs $(rad R)^{K}$	Reference
Arbitrary	$\subseteq rad R$	no relation in general	(5.14)(1)
Separable algebraic	$= rad R$	equal	(5.17)
Algebraic, possibly inseparable	$= rad R$	$\supseteq$ , can be strict	(5.14)(2)
Finite of degree $n$	$= rad R$	$(rad R^{K})^{n} \subseteq (rad R)^{K}$	(5.14)(3)
Transcendental, general $R$	nil	no relation in general	(5.15)
Transcendental, $\dim_{k} R < \infty$	$= rad R$ , nilpotent	$\supseteq$	(5.14)(2)
$K = k (T)$ , $R$ any $k$ -algebra	nil	$rad R (T) = N' (T)$	(5.13), (5.15)

Which property of $K / k$ each conclusion consumes
	Splitting of $R^{K}$ over $R$	Finitely many basis elements	Nondegenerate trace form	$\dim_{k} R < \infty$
$(5.14) (1)$ contraction	yes	no	no	no
$(5.14) (2)$ algebraic case	yes	yes	no	no
$(5.14) (2)$ finite-dimensional case	yes	no	no	yes
$(5.14) (3)$ exponent bound	yes	yes	no	no
$(5.15)$ nilness	yes	no	no	no
$(5.16)$ , $(5.17)$	yes	yes	yes	no

Which property of $K / k$ each conclusion consumes

Relationship Map

All field extensions $K / k$

R \cap rad R^{K} \subseteq rad R

Algebraic

R \cap rad R^{K} = rad R

Finite of degree $n$

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

Separable

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

Galoistrace-form argument available directly

(5.14)

contraction

(5.16)

semiprimitivity ascends

(5.17)

rad

commutes with

\otimes K

splitting field theory

The downstream consumer is Splitting Fields for Algebras: to split a finite-dimensional algebra one enlarges $k$ until the semisimple quotient becomes a product of matrix rings, and $(5.17)$ is what guarantees the radical does not misbehave during the enlargement, provided the enlargement is separable. In the other direction, $(5.15)$ supplies the nilness assertion required by Amitsur's rational function field theorem $(5.13)$ .

Applications and Industry Use

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

Representation theory

Splitting fields and Schur indices

Enlarging scalars to split a group algebra is standard practice; $(5.17)$ guarantees that over a separable extension the radical, and hence the modular structure, extends without surprises.

Computational algebra

Wedderburn decomposition over extensions

Systems compute $rad A$ over the base field and then decompose $A / rad A$ after extending scalars. The correctness of that order of operations over finite and characteristic-zero fields is exactly $(5.17)$ , since those fields are perfect.

Coding theory

Codes over field extensions

Finite fields are perfect, so extending the alphabet from $𝔽_{q}$ to $𝔽_{q^{m}}$ never creates radical in an algebra used to build codes; subfield subcodes and trace codes rely on this stability.

Arithmetic of algebras

Brauer groups and central simple algebras

Central simple algebras remain simple under separable extension and split over a separable splitting field; the inseparable counterexample here explains why separability is built into the definition of the Brauer group.

The honest summary: these results are hygiene theorems. They are invoked to justify a base change that a practitioner would otherwise perform without comment, and their real content is the list of situations where that base change would be unjustified.

Design Considerations

Design considerations here means the choices made when modelling a problem with these algebraic structures.

Choose a separable splitting field. When you may choose the extension, choose a separable one: $(5.17)$ then makes every radical computation base-change stable, and descent back to $k$ is available.
Prefer perfect base fields. Over $ℚ$ , $ℝ$ , $ℂ$ or any finite field, all algebraic extensions are separable and the inseparable pathology cannot occur.
Extend scalars before or after quotienting? For separable extensions the two orders agree, by $(5.17 a)$ . For inseparable ones they do not, and quotienting by the radical first loses information that the extension would have exposed.
Finite-dimensional or not? If $\dim_{k} R < \infty$ the radical is nilpotent, which makes the extension inclusion automatic. Infinite-dimensional algebras with non-nil radical — power series rings are the standard model — are the ones where transcendental extensions bite.
Record the degree. For $[K : k] = n$ , the exponent bound $(rad R^{K})^{n} \subseteq (rad R)^{K}$ is often all that is needed and costs nothing to record.

Failure Modes and Common Mistakes

Do not confuse $(rad R)^{K}$ — the extension of the radical — with $rad (R^{K})$ ; the theorems are precisely about when these differ.
Do not assume $(5.14) (3)$ gives equality: it gives an exponent bound, and the bound is attained in the inseparable degree- $p$ example.
Do not treat $R (T) = R \otimes_{k} k (T)$ as a special case of the algebraic theory; $k (T) / k$ is transcendental and needs $(5.13)$ and $(5.15)$ .
Do not assume the trace argument works without normality: the proof passes to the normal hull first, and $Gal (E / k)$ is used, not merely the embeddings of $L$ .
Do not forget that $rad R^{K}$ is invariant under all ring automorphisms, including the ones coming from $Gal (K / k)$ — that invariance is the whole basis of the averaging step.

Historical Notes and Lessons Learned

1907–1908WedderburnThe structure theory of finite-dimensional algebras is developed over a general field, and the failure of semisimplicity to survive scalar extension in characteristic $p$ is noticed.
1930sSeparable algebrasNoether, Deuring and Albert isolate separability as the condition making an algebra insensitive to base change, and the theory of central simple algebras and splitting fields is built on it.
1945Jacobson's radicalWith the radical defined for arbitrary rings, the base-change question can be asked outside the finite-dimensional setting.
1956AmitsurAmitsur proves that the contraction of the radical along a transcendental extension is nil, and settles the radical of the rational function field extension $R (T)$ .
1964ConsolidationThe results appear in systematic form in Jacobson's Structure of Rings and later texts, in the shape presented here.

The lesson is that base change is a hypothesis-consuming operation. Each strengthening of the conclusion — from inclusion, to equality of contractions, to equality of ideals — costs a further property of $K / k$ : finiteness, then algebraicity, then separability.

Quick Reference

$(5.14) (1)$

R \cap rad R^{K} \subseteq rad R

, always

$(5.14) (2)$ Equality if

K / k

algebraic or

\dim_{k} R < \infty

$(5.14) (3)$

[K : k] = n \Rightarrow (rad R^{K})^{n} \subseteq (rad R)^{K}

$(5.15)$

K / k

transcendental

\Rightarrow

R \cap rad R^{K}

nil

$(5.16)$

K / k

separable algebraic,

rad R = 0

\Rightarrow

rad R^{K} = 0

$(5.17)$

K / k

separable algebraic

\Rightarrow

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

Counterexample

K \otimes_{k} K

for

K = k (a^{1 / p})

: radical generated by

1 \otimes α - α \otimes 1

Key toolNondegeneracy of the trace form of a separable extension

Worked values
$k$ , $K$ , $R$	$rad R$	$rad R^{K}$
$ℝ$ , $ℂ$ , $ℍ$	$0$	$0$ (and $R^{K} ≅ M_{2} (ℂ)$ )
$𝔽_{2} (u)$ , $𝔽_{2} (u^{1 / 2})$ , $R = K$	$0$	$K (x - α)$ , square zero
$k$ , $k (t)$ , $k [[x]]$	$(x)$	contraction is $0$
$k$ , any separable $K$ , $T_{2} (k)$	strictly upper	strictly upper over $K$

Frequently Asked Questions

Why can scalar extension create radical at all?

Because $R^{K}$ has more elements than $R$ does, and new nilpotents can appear that were invisible over $k$ . The mechanism is concrete: if $K = k (α)$ with $α^{p} = a \in k$ , then $1 \otimes α - α \otimes 1$ is a nonzero element of $K \otimes_{k} K$ whose $p$ -th power is $1 \otimes a - a \otimes 1 = 0$ . Nothing like it exists over $k$ .

rad R \subseteq rad R^{K}

always true?

No. It holds for algebraic $K / k$ and for finite-dimensional $R$ , both by $(5.14) (2)$ . It fails for $R = k [[x]]$ and $K = k (t)$ , where $rad R = (x)$ meets $rad R^{K}$ in $0$ by $(5.15)$ .

How does the finite-degree bound interact with

(5.17)

When $K / k$ is separable of degree $n$ , $(5.17)$ gives equality, so the bound in $(5.14) (3)$ is not needed. The bound earns its keep in the inseparable case, where it says the discrepancy between $rad R^{K}$ and $(rad R)^{K}$ is nilpotent of index at most $n$ — and the degree- $p$ example shows the index $n$ is attained.

Does

(5.16)

need the extension to be Galois?

The trace argument is run over a finite Galois extension, but the statement only assumes separable algebraic. The proof reduces first to a finite subextension, then to its normal hull, which is Galois because a normal separable extension is Galois. Separability is thus used twice: to make the hull Galois and to make the trace form nondegenerate.

What does this say about semisimple algebras?

If $R$ is finite-dimensional and semisimple and $K / k$ is separable algebraic, then $R^{K}$ is again semisimple, since $rad R^{K} = (rad R)^{K} = 0$ and $\dim_{K} R^{K} = \dim_{k} R < \infty$ . Over a perfect field this extends to arbitrary extensions; over an imperfect field the inseparable counterexample shows it can fail.

Where is

(5.15)

used?

It supplies the missing half of Amitsur's rational function field theorem $(5.13)$ : the assertion that $N' = R \cap rad R (T)$ is nil. Since $k (T) / k$ is transcendental, $(5.15)$ applies verbatim.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §5, results (5.14)–(5.17) (pp. 79–81).
S. A. Amitsur, “Algebras over infinite fields”, Proceedings of the American Mathematical Society 7 (1956), 35–48.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964, Chapters V and X.
C. W. Curtis and I. Reiner, Representation Theory of Finite Groups and Associative Algebras, Interscience, 1962, Chapters IV and VII.
R. S. Pierce, Associative Algebras, Graduate Texts in Mathematics 88, Springer-Verlag, 1982, Chapters 10 and 11.

AI Suggested Questions

Prove that the trace form of a finite extension is nondegenerate if and only if the extension is separable.
Compute $rad (A \otimes_{k} K)$ for $A = M_{2} (k)$ and $K / k$ purely inseparable of degree $p$ .
Show that a finite-dimensional semisimple algebra over a perfect field stays semisimple under every scalar extension.
Give an example of a $k$ -algebra $R$ and a transcendental extension $K / k$ where $rad R^{K}$ is strictly larger than $(rad R)^{K}$ .
How do these results interact with the Brauer group and the choice of a separable splitting field?
What is the analogue of $(5.17)$ for scalar extension along a separable algebra rather than a field extension?
Explain how the composition length argument in $(5.14) (3)$ would change if $V$ were taken to be a left module.

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Two module structures, two theorems

Why separability is the dividing line

06Key Results

07Proof Techniques and Method

Read the tensor product as a module

Bound composition length

Average with the Galois group and use the trace form

08Worked Example

A separable extension: quaternions complexified

An inseparable extension in characteristic 2

A transcendental extension where the contraction shrinks

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

Splitting fields and Schur indices

Wedderburn decomposition over extensions

Codes over field extensions

Brauer groups and central simple algebras

12Design Considerations

13Failure Modes and Common Mistakes

14Historical Notes and Lessons Learned

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

An inseparable extension in characteristic $2$

Comparison and Classification

Relationship Map

Applications and Industry Use

Design Considerations

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions