I ntrnat. J. Mh. Math. S.
Vol. 4 No.
(1981)485-491
485
A NOTE ON POWER INVARIANT RINGS
JOONG HO KIM
Department
of-Mathematics
East Carolina University
Greenville, N.C. 27834
U.S.A.
(Received September i, 1980)
ABS.TRACT.
Let R be a commutative ring with identity and R ((n))
the power series ring in n independent indeterminates X
I,... ,X
called power invariant if whenever S is a ring such that
R
S.
a
R such that there is R- homomorphism o:
,Xn]
R[ [X I,.
over R.
R[[XI]
R is
S[[XI]],
then
R is said to be forever-power-invariant if S is a ring and n is any positive
((n))
((n)) then R S Let I
integer such that R
S
(R) denote the set of all
C
is an ideal of R.
R[[X]]
/
R with (X)
a.
Then I
c(R)
It is shown that if I C -(R) is nil, R is forever-power-invariant
KEY WORDS AND PHRASES.
Power seri ring, Power inviant rng, Forever-power-
invaria, Ideal-adic topology.
1980 Matemati Subject
I
Classification
Cod.
IF25
INTRODUCTION.
In this paper all rings are assumed to be commutative and to have identity
elements
Throughout this paper the symbol
and
a
I
a
n
be elements of R
((n)) with the
ring R
R ((n)) generated by
((n))
(al,...,an)
a l,...,an.
Let (R
adic
are used to denote the
Let R ((n))
positive and negative integers, respectively
formal power series ring in n indeterminates X
O
R[ [XI,... ,Xn]
be the
,Xn over a ring R and let
I
((n))
(al,...,an)) denote
topology where (al,... ,an)
It is well known that (R
(a I ,...,an) J
sets.of
the topological
is the ideal of
((n)), (a l,...,an)) is
(0). In this case, the topological
((n)) is
((n)
ring R
metrizable, and we say that (R
(al,...,an)) is complete if
Hausdorff if and only if
nj
5
486
J.H.
((n)) converges in R ((n)). Clearly, (R ((n))
(x,... ,Xn))
((n))
is a complete Hausdorff space. If R
is uniquely expressible in the
then
each Cauchy sequence of R
n]
,X for each ]0 such that
I
] is 0 or a homogenous polynomial (that is form) of degree ] in X1,...,Xn over R. We call
0 ]
form
Y’]--0
where
j’
R[X
j
r.]=
the homogenous decomposition of a, and for each
.
homogenous component of
J0’
aj
is called the j-th
Coleman and Enochs [3] raised the following question:
Can there be non-lso-
morphic rings R and S whose polynomial rings R[X] and S[X] are isomorphic?
Hochster [8] answered the question in the affirmative.
The analogous question
If R[[X]]
about formal power series rings was raised by O’Malley [13]:
B?
must A
Hermann [7] showed that there are non-isomorphic rings R and S whose
formal power series ring R[[X]] and S[[X]] are isomorphic.
S[[X]], then R
Then" what is necessary
a ring R in order that whenever S is a ring such that
and sufficient conditions on
R[[X]]
S[[X]],
S?
Several authors
[7,10,13] investigated
sufficient
conditions on R so that R should be power invariant, but we do not know the neces-
sary conditions on R.
The fact that rings with nilpotent Jacobson radical are
power invariant is known in [i0] and
Hamann [7] proved that a ring R is
variant, if J(R), the Jacobson radical of R, is nil.
ower in-
In this paper we impose more
relaxed condition on J(R) so that R should be power invariant and forever-powerinvariant.
phlsm o:
Let I (R) denote the set of all aR such that there is an R-homomor-
C
R[[X]]
R with o(X)
Then I (R) is an ideal of R contained in J(R)
a.
C
and contains the nil-radical of R (by Theorem E, [4]).
Then I (R) may be properly
C
contained in J(R) and it may properly contain the nil-radical of R.
if A
Z/ (4) [X]
(2,X) is
then M
a maximal ideal of A
the nil-radical of R is 2R and I (R)
C
to see that the nil-radical of A
(2 Y) and J(R)
M is (2) and
Ic(AM)
that J(R
((n)))
proved in [6]:
+
,n
((n))
R
AM[[Y]]
Let R
(2,X,Y)
any positive integer n, I (R
C
((n)))
can not be nil.
(2,X).
This
It is well known
Analogously, the following
iI X.I
n
((n))
I c (R) +
I c (R ((n)))
therefore, for
i=El Xi R
j(R)
then
Also it is easy
(2) and J(AM)
shows that for some ring R, I (R) is nil, but J(R) is not nil.
C
For example,
relation was
any ring R and
487
POWER INVARIANT RINGS
2.
SOME POWER INVARIANT RINGS.
Let a
R[[X]].
aiXi
i=0
complete with respect to the
If
(a0n)
nl
(a0)-adic
(0) (or
nOl
n
(0)) and R
(a
is
topology (or R[[X]] is complete with respect
to the (a)-adic topology), then there is an R-endomorphism
of
R[[X]] such that
a, ([14] and [15]).
(X)
The following theorem from [15] will be needed for our main results.
Let a
TEOREM i
i=10 al
of R[[X]] such that (X)
(i)
(R[[X]], (a))
(2)
a
X
i
R[[X]].
Then there exists an R-automorphism
a if and only if the following conditions are satisfied:
is a complete Hausdorff space;
is a unit of R.
I
The next theorem (Theorem 5.6, [5]) is the more generalized form of Theorem I.
THEOREM_ 2.
Let a
i
sitions of elements of
(X i)
a
(i)
R((n))
j__Z0 aj
R ((n)). There exists
for i--l,...,n, be homogeneous decompoan R-automorphism
of R
((n)) such that
for each i if and only if the following conditions are satisfied:
i
(i)
(R
(2)
R
((n)),
,an))
(a I
al(1) +
+ R
is a complete Hausdorff space;
al(n)
+ R Xn"
R XI +
Moreover, if such an automorphism
exists, then it is unique.
Also, we need the following proposition:
PROPOSITION 3.
Let M be a unitary free R-module of finite rank n and let
n
__{x
i}
be a free basis for M.
Let
i=l
Mn(R)
denote the ring of n x n matrices over R,
n
and let
where
z_,l "’’’Zn
aij
be elements of M such that
e R for each i. and j.
+
+R
R
z.m j=IZ
ai j xj
for each i--1,...,n
Then the following conditions are equivalent:
+
+ R
x
n
(i)
R
(2)
det (A), the determinant of A, is a unit of R where A
z_
1
z
n
x_
+/-
(aij)
is the
n x n matrix.
(3)
n
{zl} i=l
is a free basis for M.
The proof of the proposition is stralgMtforward so we omit its proof
Finally, we list the theorem from [4] which plays a particularly important
role in this paper.
488
J.H. KIM
THEOREM 4.
I
{a
I
nnd
R
there exists an R-automorphism o:
{a
12
Let
R
It(R)
/
a
+
f for some X
R[[XI,...,Xn] R[[YI,...,Ym]
m
Zj= 1 Yj R[[YI,...,Ym )"
and f
i
X + a
R[[X]] with o(x)
there exists an R-homomorphism u:
such that u(Xi)
Then
R[[X]]
/
11 12.
Now we are ready for our first result.
THEOREM 5.
PROOF.
S[ [X] ].
If R is a ring such that
Suppose that
It(R)
Then %(R) [%(X)]
therefore, in order to show power invariance
of R, it suffices to show that R[[X]]
terminate over a ring S.
+ YV where a0
R, b 0
is nil, then R is power invariant.
Let $ be an isomorphism of R[[X]] onto
is nil.
S[ [XJ
It(R)
S[[Y]] implies R
Let W
R[[X]]
S and U,V
W.
S, where Y is an inde-
S[[Y]] and let Y
a
0
+ XU
and X
Then by Theorem 4, a
i=Z0 ci
a0
yi
where c
It(R)
0
and so a
S for each i
i
and we have
y
m0’
0
0
then c
i=ZO ci yi + b0U
I
c
I +
J(W), b 0 + YV
element of J(S).
Recall that c
therefore,
i
v0u0
of W
S[[Y]].
ai+I
Y= i_E0
R for each i
a
i
I
b0uI
0’ 6hen
X where
b0u I + v0u0 where
0
and
S[[Y]], respectively
Since X is an
J(S[[Y]]) and so b 0 is an
a nilpotent element of S, then c +
boU1 J(S);
I
is an element of
is a unit of S.
This forces U and V to be units
the constant term a
I
is a unit of R.
Then by Theorem I, there exists an R-automorphlsm
R[[X]]/(X)
u
i_E0 ai+I
X
i
Then
the X coefficient, is a unit of R, and (W,(Y)) is a complete
al,
which maps X onto Y
m0
(i)
If we consider U as an element of R[[X]] and let U
i
Hausdorff Space.
Then R
c
I is
0
Let
+ YVU
is the Y coefficient of U considered as an element of S[ [Y]].
element of J(R[[X]])
a
is nilpotent for each i
i
are constant terms of U and V considered as elements of
and u
Y
is a nilpotent element of R.
The Y coefficients in both sides of (1) yields 1
v
0
Clearly (W,(Y)) is a complete Hausdorff
space; therefore, there is an R-endomorphism o of R[[X]] such that o(X)
+ XU.
b
a
0
+ XU
i__Z0
W/(a0 + XU)
a
i
X
of
R[[X]]
i.
W/(Y)
S.
This completes the proof.
Let R[t] be the polynomial ring in an indeterminate t over a ring R, then
J(R[t]) coincides with the nil-radlcal of R[t]
therefore
I C (R[t]) is a nil ideal
POWER IN-VARIANT RINGS
of
(R[t])
489
and by Theorem 5, R[t] is power invariant.
the polynomial ring in n Indeterminates
tl,... ,tn
Similarly, if R[t l,...,t
over R, then R[t I,...
,in]
n]
is
is
power invariant.
It is natural to raise the following question:
isomorphic to S whenever R[ [XI,... ,X
n]
positive integer n?
DEFINITION.
and S[
For what kind of ring R, is R
[Xl,... ,Xn]
are isomorphic for some
To wit, we give the following definition.
A ring R is said to be forever-power-invariant provided R is is.-
morphic to S whenever there is a ring S and a positive integer n such that
R[ [X
...,Xn]]
I
and S[[X
I ...,Xn]] are isomorphic where XI
,X are independent
indetermlnates over R and S.
EXAMPLE.
integer n.
R[[Xl,...,Xn]]for
If R is a quasi-local ring then so is
Since any quasi-local ring is power invariant
power invariant if R is a quasi-local ring.
any positive
[7], R[ [X I,... ,Xn]
is
Then clearly every quasi-local ring is
forever-power-invarlant.
THEOREM 6.
If R is a ring such that I c (R) is nil, then R is forever-power-
invar ant.
PROOF.
S[[Y
I
Suppose that R is a ring such that
,Yn]].
Let
Yi
Ic(R)
is nil.
To prove this theorem, it suffices to show that R and S are iso-
a0(i) + XIUI (1) + + XnUn(i) and
b0(i) + YIVI (i)- + + YnVn(i) for each i l,...,n where Uk(1)
morphic.
Xi
elements of W for each i
l,...,n.
i
R[[X l,...,xn]]
Let W
Since
l,...,n and k
(W,(Yi))
I
and
()
R, b 0
Vk(1) are
S for each
is a complete Hausdorff space, there is a R-homomorphism
R[[XI]] into R[[XI,...,Xn]] such that
I (R)
X U (i). Then by Theorem 4, a0(1)
c
n n
of
nilpotent for each i
()
,n and a0
l,...,n.
(X I)
(Yi)
for each i
a0(i) + XlUl(1) +
l,...,n and so a0(1)
The relation defined between
Yis
and
Xis
+
are
yields the
following:
n
n
(i)
Yi ao(i) + kl bo(k)Uk + kl Vl(k)Uk (i)) Y1 +
n
n
+ k=El Vn(k) Uk i))Yn.
+ k--E1 vi(k)Uk (+/-) Yi +
Let a0(i)
k0 Ck(i) be a homogenous decomposition S[[YI,...,Yn]].
in
(1)
Then since
490
J.H.
a0(1)
Q-(1)
is nilpotent
+
Cln
(1)
Y
(1) is
n’ then Clj
(+/-)
and V (i)_
Uk() _ Z- Ukj
Uk(f) and Vk(f)
j
+
VkllYl
is nllpotent for each k
+
(i)
+
Clj
n
b
(k)
0
Ukl j
(i)
J(S) if i # j.
e J(S)
Let A
n
then AB
(k)
k=Z1 Vj0
Y.
n
kEffil b0
k__E1 Vj 0
(k)
(i)
Uklj
(k)
Uk0
(i)
+
YnVn0
+ YnVn (i)
+
(i) for each i
UkllYl
Mn(S).
+
+
+UklnYn
Since
eli
S[ [YI
’Yn
(Uk0(i))ki be n x n
j and it is in
matrices over
Then S
S[[YI,
R[[XI,
#(Yi) b 0
Yn]]/(Y I ,Yn
Xn]]/(XI
Xn)
R
(i)
So AB is invertible in M (S): therefore,
n
Clearly, (W, (X 1,...
+
S,
every diagonal entry is a unit of S and
n and the n x n matrix A
such that
Vkl
(1) fs
nilpotent and
,Xn))
is a complete
..,Yn ]]
1
(Vji))j
i
YlVl
(i)
+
b
i
is
()
0
YlVl0
+
of
+y V (i) for each i
n n
W/((YI),...,(Yn))
(+/-)
is invertlble in
Then by Theorem 2 and Proposition 3, there is an S-automorphism
Mn(S).
and
k=l Vj 0 (k) Uk0 (f)
considered as an element of S[[Y
i,
YI +
Let
l,...,n.
Recall that the linear homogeneous component of X
Hausdorff space.
YlVl
(1)
(i) is a
unit of S if i
(Vj 0 (k)) jk and B
(i)
Uk0 )ji in which
both A and B are invertible in
(+/-)
coefficient of the right side of (i) is
the rest of entries are elements of J(S).
+
Ukl
f, otherwise, 0.
which fs equal to i if j
k=Z1
and let
Then the
VklnYn
J
ell
be homogeneous decompositions of elements
j__E 0 Vkj
,Yn
1
CI(1)
Let
,n
a nilpotent element of S for each
k
in S[[Y
I
i
W/(XI,...,Xn)
This completes the proof.
R[tl,... ,t n] is the polynomial ring in indetermlnates
tl,... ,tn over a riDg R, then. it is a forever-power-invariant.
It is easy to see that if R is a ring such that R[[XI,...,Xn]] i power invarCORROLLARY 7.
If
iant for any positive integer n.
Then R is forever-power-invariant.
This raises
the following open question:
If R is a ring such that I C (R) is nil then for any
positive integer, is R[ [X
,Xn]
I
power invariant?
n.
491
POWER INVARIANT RINGS
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