On the Amitsur property of radicals
The Amitsur property of a radical says that the radical of a polynomial ring is again a polynomial ring. A hereditary radical γ has the Amitsur property if and only if its semisimple class is polynomially extensible and satisfies: f(x) ∈ γ(A[x]) implies f(0) ∈ γ(A[x]). Applying this criterion, i...
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Інститут прикладної математики і механіки НАН України
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| Цитувати: | On the Amitsur property of radicals / N.V. Loi, R. Wiegandt // Algebra and Discrete Mathematics. — 2006. — Vol. 5, № 3. — С. 92–100. — Бібліогр.: 9 назв. — англ. |
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irk-123456789-1573772019-06-21T01:28:15Z On the Amitsur property of radicals Loi, N.V. Wiegandt, R. The Amitsur property of a radical says that the radical of a polynomial ring is again a polynomial ring. A hereditary radical γ has the Amitsur property if and only if its semisimple class is polynomially extensible and satisfies: f(x) ∈ γ(A[x]) implies f(0) ∈ γ(A[x]). Applying this criterion, it is proved that the generalized nil radical has the Amitsur property. In this way the Amitsur property of a not necessarily hereditary normal radical can be checked. 2006 Article On the Amitsur property of radicals / N.V. Loi, R. Wiegandt // Algebra and Discrete Mathematics. — 2006. — Vol. 5, № 3. — С. 92–100. — Бібліогр.: 9 назв. — англ. 1726-3255 2000 Mathematics Subject Classification: 16N60. http://dspace.nbuv.gov.ua/handle/123456789/157377 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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The Amitsur property of a radical says that
the radical of a polynomial ring is again a polynomial ring. A
hereditary radical γ has the Amitsur property if and only if its
semisimple class is polynomially extensible and satisfies: f(x) ∈
γ(A[x]) implies f(0) ∈ γ(A[x]). Applying this criterion, it is proved
that the generalized nil radical has the Amitsur property. In this
way the Amitsur property of a not necessarily hereditary normal
radical can be checked. |
| format |
Article |
| author |
Loi, N.V. Wiegandt, R. |
| spellingShingle |
Loi, N.V. Wiegandt, R. On the Amitsur property of radicals Algebra and Discrete Mathematics |
| author_facet |
Loi, N.V. Wiegandt, R. |
| author_sort |
Loi, N.V. |
| title |
On the Amitsur property of radicals |
| title_short |
On the Amitsur property of radicals |
| title_full |
On the Amitsur property of radicals |
| title_fullStr |
On the Amitsur property of radicals |
| title_full_unstemmed |
On the Amitsur property of radicals |
| title_sort |
on the amitsur property of radicals |
| publisher |
Інститут прикладної математики і механіки НАН України |
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2006 |
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http://dspace.nbuv.gov.ua/handle/123456789/157377 |
| citation_txt |
On the Amitsur property of radicals / N.V. Loi, R. Wiegandt // Algebra and Discrete Mathematics. — 2006. — Vol. 5, № 3. — С. 92–100. — Бібліогр.: 9 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
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AT loinv ontheamitsurpropertyofradicals AT wiegandtr ontheamitsurpropertyofradicals |
| first_indexed |
2025-07-14T09:48:59Z |
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2025-07-14T09:48:59Z |
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1837615325723492352 |
| fulltext |
Algebra and Discrete Mathematics RESEARCH ARTICLE
Number 3. (2006). pp. 92 – 100
c© Journal “Algebra and Discrete Mathematics”
On the Amitsur property of radicals
N. V. Loi and R. Wiegandt
Communicated by L. Márki
Abstract. The Amitsur property of a radical says that
the radical of a polynomial ring is again a polynomial ring. A
hereditary radical γ has the Amitsur property if and only if its
semisimple class is polynomially extensible and satisfies: f(x) ∈
γ(A[x]) implies f(0) ∈ γ(A[x]). Applying this criterion, it is proved
that the generalized nil radical has the Amitsur property. In this
way the Amitsur property of a not necessarily hereditary normal
radical can be checked.
1. Introduction
All rings considered are associative, not necessarily with unity element.
Radicals are meant in the sense of Kurosh and Amitsur. A radical γ is
hereditary, if I ⊳ A ∈ γ implies I ∈ γ. For details of radical theory the
readers are referred to [3].
Many classical radicals, for instance, the Baer, Levitzki, Köthe, Ja-
cobson, and Brown–McCoy radicals, enjoy an important property con-
cerning polynomial rings, called the Amitsur property: the radical of a
polynomial ring is again a polynomial ring.
In several cases it is not so easy to decide that a given radical has
the Amitsur property. So it seems to be desirable to have equivalent
conditions (as Krempa’s condition [5]) for testing the Amitsur property
of radicals. We are going to prove such a criterion for hereditary radicals
in Theorem 2.4.
Research supported by the Hungarian OTKA Grant # T043034
2000 Mathematics Subject Classification: 16N60.
Key words and phrases: Amitsur property, hereditary, normal and generalized
nil radical.
N. V. Loi, R. Wiegandt 93
A radical γ has the Amitsur property, if for every polynomial ring
A[x] it holds
γ(A[x]) = (γ(A[x]) ∩ A)[x]. (A)
The Amitsur property of a radical states that the radical of a polynomial
ring is again a polynomial ring. It seems to be folklore that also the
converse is true.
Proposition 1.1. A radical γ has the Amitsur property if and only if
γ(A[x]) is a polynomial ring in x.
Proof. If γ(A[x]) is a polynomial ring B[x], then the constant polynomials
on both sides are equal. Hence γ(A[x])∩A = B, and so γ has the Amitsur
property.
A useful criterion for the Amitsur property of a radical was given by
Krempa [5].
Proposition 1.2. For a radical γ to have the Amitsur property a neces-
sary and sufficient condition is
γ(A[x]) ∩ A = 0 implies γ(A[x]) = 0 (K)
for all rings A.
Let Z(A1) denote the center of the Dorroh extension A1 of a ring A.
We say that a radical γ is closed under linear substitutions, if f(x) ∈
γ(A[x]) implies f(ax + b) ∈ γ(A[x]) for all rings A and all a, b ∈ Z(A1).
Proposition 1.3. If a radical γ has the Amitsur property, then γ is
closed under linear substitutions. If a radical γ is closed under linear
substitutions, then γ satisfies condition
f(x) ∈ γ(A[x]) implies f(0) ∈ γ(A[x]) (T)
for all rings A.
Proof. Suppose that γ has the Amitsur property and let
f(x) =
n∑
i=0
cix
i ∈ γ(A[x]) = (γ(A[x]) ∩ A)[x].
Then for any a, b ∈ Z(A1) we have
f(ax + b) =
n∑
i=0
ci(ax + b)i = g(x).
94 On the Amitsur property of radicals
Since each ci ∈ γ(A[x]) ∩ A and a, b ∈ Z(A1), all the coefficients of g(x)
are in γ(A[x]) ∩ A. Hence
f(ax + b) = g(x) ∈ (γ(A[x]) ∩ A)[x] = γ(A[x]).
If a radical γ is closed under linear substitutions then γ satisfies triv-
ially condition (T).
We say that the semisimple class Sγ of a radical γ is polynomially
extensible if A ∈ Sγ implies A[x] ∈ Sγ. This notion was introduced and
studied in connection with the Amitsur property in [9].
Proposition 1.4. If a radical γ has the Amitsur property, then its semisim-
ple class
Sγ is polynomially extensible.
Proof. The statement is a special case of [9, Proposition 3.4].
Let us observe that the Amitsur property of a radical γ is independent
from the polynomial extensibility of γ (that is A ∈ γ implies A[x] ∈ γ),
as proved in [9, Corollary 3.8 (iii)].
2. Hereditary radicals and the Amitsur property
We shall denote by (f(x))A[x] the principal ideal of the polynomial ring
A[x] generated by the polynomial f(x) ∈ A[x].
Proposition 2.1. For a hereditary radical γ condition (T) is equivalent
to
(f(x))A[x] ∈ γ implies (f(0))A[x] ∈ γ. (S)
Proof. Straightforward.
The following Lemma may be useful also in other contexts.
Lemma 2.2. Let γ be a hereditary radical. If A ∈ γ and γ(A[x]) ⊆ xA[x],
then γ(A[x]) = 0.
Proof. Let us consider the set
K = {f ∈ xA[x] | xf ∈ γ(A[x])}.
Clearly γ(A[x]) ⊆ K ⊳ A[x].
For arbitrary polynomials f, g ∈ K we have xfg ∈ γ(A[x]) and g = xh
with a suitable polynomial h ∈ A[x]. Hence fh ∈ K, so fg = xfh ∈
γ(A[x]). Thus K2 ⊆ γ(A[x]), that is, (K/γ(A[x]))2 = 0.
N. V. Loi, R. Wiegandt 95
We define a mapping ϕ : K → γ(A[x])/xγ(A[x]) by the rule
ϕ(f) = xf + xγ(A[x]) ∀f ∈ K.
Obviously this mapping preserves addition. Further,
ker ϕ = {f ∈ K | xf ∈ xγ(A[x])},
so to each f ∈ ker ϕ there exists a g ∈ γ(A[x]) such that xf = xg, that
is, x(f − g) = 0. Since x is an indeterminate, f = g follows. Hence
ker ϕ ⊆ γ(A[x]). The inclusion γ(A[x]) ⊆ ker ϕ is obvious, therefore
ker ϕ = γ(A[x]). Taking into account that
im ϕ ∼= K/ ker ϕ = K/γ(A[x]),
by (K/γ(A[x]))2 = 0 we conclude that ϕ is a ring homomorphism. Since
γ is hereditary, from
K/γ(A[x]) ∼= im ϕ ⊳ γ(A[x])/xγ(A[x]) ∈ γ
it follows that K/γ(A[x]) ∈ γ. We have also
K/γ(A[x]) ⊳ A[x]/γ(A[x]) ∈ Sγ,
and therefore K/γ(A[x]) ∈ γ ∩ Sγ = 0. Thus K = γ(A[x]).
Let us define the ideal
M = {f ∈ A[x] | xf ∈ γ(A[x])}
of A[x]. Obviously M ∩ xA[x] = K = γ(A[x]). Then M/γ(A[x]) ⊳
A[x]/γ(A[x]) ∈ Sγ implies M/γ(A[x]) ∈ Sγ. Further,
M/γ(A[x]) = M/(M ∩ xA[x]) ∼= (M + xA[x])/xA[x] ⊳ A[x]/xA[x] ∼= A.
Since A ∈ γ, by the hereditariness of γ we have
(M + xA[x])/xA[x] ∈ γ ∩ Sγ = 0,
and so M ⊆ xA[x]. This implies
γ(A[x]) = M ∩ xA[x] = M.
Suppose that γ(A[x]) 6= 0 and p =
t∑
i=1
aix
i ∈ γ(A[x]) is a polynomial
of minimal degree. Taking into consideration that γ(A[x]) ⊆ xA[x], we
have a0 = 0 and p = xq with an appropriate polynomial q ∈ A[x]. By
the definition of M we have that q ∈ M = γ(A[x]). But the degree of q
is t − 1 < t, a contradiction. This proves γ(A[x]) = 0.
96 On the Amitsur property of radicals
The next statement is crucial in proving Theorem 2.4.
Lemma 2.3. Let γ be a hereditary radical. If γ satisfies condition (T)
and the semisimple class Sγ is polynomially extensible, then γ satisfies
Krempa’s condition (K).
Proof. For proving the validity of Krempa’s condition (K), we suppose
that γ(A[x]) ∩ A = 0, and have to show that γ(A[x]) = 0.
Let us consider an arbitrary polynomial f(x) ∈ γ(A[x]). By the
assumption condition (T) implies that f(0) ∈ γ(A[x])∩A = 0. Hence we
have got that γ(A[x]) ⊆ xA[x].
If A ∈ γ then an application of Lemma 2.2 yields that γ(A[x]) = 0.
If γ(A) = 0, then by the polynomial extensibility of Sγ it follows that
γ(A[x]) = 0, and Krempa’s condition is trivially fulfilled.
Hence we may confine ourselves to the case 0 6= γ(A) 6= A. We have
to prove that γ(A[x]) = 0. Since the semisimple class Sγ is polynomially
extensible, A/γ(A) ∈ Sγ implies that
A[x]/γ(A)[x] ∼= (A/γ(A))[x] ∈ Sγ.
Hence γ(A[x]) ⊆ γ(A)[x]. For the radical B = γ(A) of A, the hereditari-
ness of γ yields
γ(B[x]) = γ(A[x]) ∩ B[x] ⊆ γ(A[x]),
and so
γ(B[x]) ∩ B ⊆ γ(A[x]) ∩ A = 0
follows. Hence applying Lemma 2.2 to B = γ(A) ∈ γ, we get that
γ(B[x]) = 0. Thus, we arrive at
γ(A[x]) = γ(γ(A[x])) ⊆ γ(γ(A)[x]) = γ(B[x]) = 0.
From Propositions 1.2, 1.3, 1.4, Lemmas 2.2 and 2.3 we get immedi-
ately
Theorem 2.4. A hereditary radical γ has the Amitsur property if and
only if γ satisfies condition (T) and its semisimple class Sγ is polynomi-
ally extensible.
N. V. Loi, R. Wiegandt 97
3. Strict and special radicals
In this section we shall look at the Amitsur property of strict special
radicals.
A radical γ is strict if S ⊆ A and S ∈ γ imply S ⊆ γ(A) for every
subring S of every ring A.
Proposition 3.1. If γ is a strict radical, then γ satisfies condition (T).
Proof. The mapping ϕ : A[x] → A defined by ϕ(f(x)) = f(0) for all
f(x) ∈ A[x], is obviously a homomorphism onto A. Since γ is strict, we
have
ϕ(γ(A[x])) ⊆ γ(A) ⊆ γ(A[x]).
Hence f(x) ∈ γ(A[x]) implies that f(0) ∈ γ(A[x]).
An ideal I of A is said to be essential in A if I ∩ K 6= 0 for every
nonzero ideal K of A, and we denote this fact by I ⊳ ·A. A hereditary
class ̺ of prime rings is called a special class if I ⊳ ·A and I ∈ ̺ imply
A ∈ ̺. The upper radical
γ = U̺ = {A | A −→ f(A) ∈ ̺ ⇒ f(A) = 0}
is called a special radical. As is well known, every special radical is
hereditary and every γ-semisimple ring A ∈ Sγ is a subdirect sum of
rings in ̺, that is, Sγ is the subdirect closure ̺ of the class ̺ (see, for
instance [3, Theorem 3.7.12 and Corollary 3.8.5]).
Proposition 3.2. For a special class ̺ and special radical γ = U̺ the
following conditions are equivalent.
(i) A ∈ ̺ implies A[x] ∈ ̺,
(ii) the semisimple class ̺ = Sγ is polynomially extensible.
Proof. The implication (ii)⇒(i) is trivial.
Assume the validity of (i), and let A ∈ ̺. Then A is a subdirect
sum of rings A/Iλ ∈ ̺, λ ∈ Λ and ∩Iλ = 0. By condition (i) we have
(A/Iλ)[x] ∈ ̺ for every λ ∈ Λ. Since
A[x]/Iλ[x] ∼= (A/Iλ)[x]
and ∩Iλ[x] = 0, the ring A[x] is a subdirect sum of (A/Iλ)[x] ∈ ̺. Hence
A[x] ∈ ̺ holds.
Example 3.3. The generalized nil radical Ng is the upper radical of
all domains, that is, of all rings without zero-divisors. It is well known
that Ng is a strict special radical and the semisimple class SN g is the
98 On the Amitsur property of radicals
class of all reduced rings, that is, of all rings which do not possess nonzero
nilpotent elements (see, for instance, [3, Theorem 4.11.11 and Proposition
4.11.12]). Hence by Proposition 3.1 the radical Ng satisfies condition (T)
and a moment’s reflection shows – without making use of Proposition 3.2
– that the semisimple class SN g is polynomially extensible. Thus by
Theorem 2.4 the generalized nil radical Ng has the Amitsur property.
Let us mention that by Puczy lowski [6] the generalized nil radical Ng
is the smallest strict special radical.
4. Subidempotent, normal and A-radicals
A hereditary radical γ is called subidempotent, if the radical class γ con-
sists of idempotent rings, or equivalently, the semisimple class Sγ contains
all nilpotent rings.
Proposition 4.1. γ(A[x]) = 0 for every subidempotent radical γ and
every ring A, and every subidempotent radical γ has the Amitsur property.
Proof. If γ(A[x]) 6= 0 for a ring A, then by the hereditariness of γ we
have that xγ(A[x]) ∈ γ. Hence
0 6= xγ(A[x])/(xγ(A[x]))2 ∈ γ,
and so the subidempotent radical γ contains a non-zero ring with zero
multiplication, a contradiction. Thus γ(A[x]) = 0 follows. This means
that Krempa’s condition (K) in Proposition 1.2 is trivially fulfilled, and
therefore γ has the Amitsur property.
A radical γ is said to be an A-radical, if the radicality depends only
on the additive group of the ring; this may be defined as follows: A ∈ γ
if and only if the zero-ring A0 ∈ γ.
Proposition 4.2. Every A-radical γ has the Amitsur property.
Proof. Gardner’s [2, Proposition 1.5 (ii)] states that γ(A[x]) = γ(A)[x].
Hence by Proposition 1.1 the assertion follows.
Next, we shall focus our attention to normal radicals which are defined
via Morita contexts and characterized as left strong and principally left
hereditary radicals. A radical γ is said to be left strong, if L ⊳ℓ A and
L ∈ γ imply L ⊆ γ(A), and principally left hereditary if A ∈ γ implies
Aa ∈ γ for every a ∈ A. Jaegermann and Sands [4] proved the following
result. Let
γ0 = {A | A0 ∈ γ}
N. V. Loi, R. Wiegandt 99
be the A-radical determined by a radical γ, β the Baer (prime) radical
and L(γ ∪ β) the lower radical generated by γ and β, that is, L(γ ∪ β) is
the union γ ∨ β in the lattice of all radicals.
Proposition 4.3. Every normal radical γ is the intersection
γ = γ0 ∩ L(γ ∪ β).
Notice that the normal radical γ as well as the A-radical γ0 need
not be hereditary, but L(γ ∪ β), as a supernilpotent normal radical is
hereditary (cf. [3, Theorem 3.18.12]).
Puczy lowski [7] and Tumurbat [8] kindly informed us about
Proposition 4.4. The radicals with Amitsur property form a sublattice
in the lattice of all radicals.
Proof. Let γ, δ be radicals with Amitsur property. The union γ∨δ in the
lattice of all radicals is the lower radical ϑ = L(γ∪ δ) generated by γ and
δ. By Krempa’s criterion (K) it suffices to show that ϑ(A[x]) 6= 0 implies
ϑ(A[x]) ∩ A 6= 0. If ϑ(A[x]) 6= 0, then either γ(A[x]) 6= 0 or δ(A[x]) 6= 0.
Thus by (K) one of them has nonzero intersection with A. Since both of
them are contained in ϑ(A[x]), necessarily also ϑ(A[x]) 6= 0.
The meet τ = γ ∧ δ is just the intersection τ = γ ∩ δ of the radical
classes. For a given ring A, let I be the smallest ideal of A such that
τ(A[x]) ⊆ I([x]). Such an ideal I exists, it is the intersection of all ideals
containing τ(A[x]). We have
τ(A[x]) = τ(I[x]) ⊆ γ(A[x]),
and by the Amitsur property of γ it holds γ(A[x]) = J [x] with some
ideal J of I. Moreover, J [x] = γ(I[x]) ⊳ A[x], therfore J ⊳ A. Hence by
the minimality of I we conclude that I = J . Thus I[x] ∈ γ. By the
same token also I[x] ∈ δ. Consequently, τ(A[x]) = I[x] which means by
Proposition 1.1 that τ has the Amitsur property.
The next result shows that for the Amitsur property of a normal
radical γ it is enough to check the hereditary radical L(γ ∪ β).
Proposition 4.5. A normal radical γ has the Amitsur property if and
only if the hereditary normal radical L(γ ∪ β) has the Amitsur property.
Proof. Suppose that γ has the Amitsur property. Since β has the Amitsur
property, by Proposition 4.4 also L(γ ∪ β) has it.
Assume that L(γ∪β) has the Amitsur property. Then by Propositions
4.2, 4.3 and 4.4 also γ = γ0 ∩ L(γ ∪ β) has the Amitsur property.
100 On the Amitsur property of radicals
Corollary 4.6. A normal radical γ has the Amitsur property if and only
if the hereditary radical L(γ∪β) satisfies condition (T) and its semisimple
class is polynomially extensible.
Proof. Apply Theorem 2.4 and Proposition 4.5.
The authors are indebted to the referee for many improvements in
the manuscript.
References
[1] S. A. Amitsur, Radicals of polynomial rings, Canad. J. Math. 8 (1956), 355–361.
[2] B. J. Gardner, Radicals of abelian groups and associative rings, Acta Math. Acad.
Sci. Hungar. 24 (1973), 259–268.
[3] B. J. Gardner and R. Wiegandt, Radical theory of rings, Marcel Dekker, 2004.
[4] M. Jaegermann and A. D. Sands, On normal radicals, N -radicals and A-radicals,
J. Algebra 50 (1978), 337–349.
[5] J. Krempa, On radical properties of rings, Bull. Acad. Polon. Sci. 20 (1972), 545–
548.
[6] E. R. Puczy lowski, Remarks on stable radicals, Bull. Acad. Polon. Sci. 28 (1980),
11–16.
[7] E. R. Puczy lowski, private communication, 2005.
[8] S. Tumurbat, private communication, 2005.
[9] S. Tumurbat and R. Wiegandt, Radicals of polynomial rings, Soochow J. Math. 29
(2003), 425–434.
Contact information
N. V. Loi,
R. Wiegandt
A. Rényi Institute of Mathematics
P. O. Box 127
H–1364 Budapest
Hungary
E-Mail: nvloi@hotmail.com,
wiegandt@renyi.hu
Received by the editors: 04.04.2005
and in final form 28.09.2005.
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