Does $R[[x]] cong S[[x]]$ imply $Rcong S$












5














Let $R,S$ be commutative unitary rings. Is it true that




$$R[[x]] cong S[[x]] quad Rightarrow quad Rcong S.$$




Here by $R[[x]], S[[x]]$ I mean the ring of formal power series and the isomorphisms as isomorphisms of rings.



In fact, in this question Does $R[x] cong S[x]$ imply $R cong S$? the answer for the polynomial ring is negative and I became curious about the formal power series case.






abstract-algebra commutative-algebra formal-power-series







share|cite|improve this question




















  • 1




    not an answer, but here is a paper from around the same time as the cited one. The result was "author does not know a counterexample."
    – Andres Mejia
    Apr 21 at 22:49


















5














Let $R,S$ be commutative unitary rings. Is it true that




$$R[[x]] cong S[[x]] quad Rightarrow quad Rcong S.$$




Here by $R[[x]], S[[x]]$ I mean the ring of formal power series and the isomorphisms as isomorphisms of rings.



In fact, in this question Does $R[x] cong S[x]$ imply $R cong S$? the answer for the polynomial ring is negative and I became curious about the formal power series case.






abstract-algebra commutative-algebra formal-power-series







share|cite|improve this question




















  • 1




    not an answer, but here is a paper from around the same time as the cited one. The result was "author does not know a counterexample."
    – Andres Mejia
    Apr 21 at 22:49
















5












5








5


2





Let $R,S$ be commutative unitary rings. Is it true that




$$R[[x]] cong S[[x]] quad Rightarrow quad Rcong S.$$




Here by $R[[x]], S[[x]]$ I mean the ring of formal power series and the isomorphisms as isomorphisms of rings.



In fact, in this question Does $R[x] cong S[x]$ imply $R cong S$? the answer for the polynomial ring is negative and I became curious about the formal power series case.






abstract-algebra commutative-algebra formal-power-series







share|cite|improve this question















Let $R,S$ be commutative unitary rings. Is it true that




$$R[[x]] cong S[[x]] quad Rightarrow quad Rcong S.$$




Here by $R[[x]], S[[x]]$ I mean the ring of formal power series and the isomorphisms as isomorphisms of rings.



In fact, in this question Does $R[x] cong S[x]$ imply $R cong S$? the answer for the polynomial ring is negative and I became curious about the formal power series case.






abstract-algebra commutative-algebra formal-power-series




abstract-algebra commutative-algebra formal-power-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 12 at 10:25









Watson

15.8k92970




15.8k92970










asked Apr 21 at 22:37









Severin Schraven

5,6381832




5,6381832








  • 1




    not an answer, but here is a paper from around the same time as the cited one. The result was "author does not know a counterexample."
    – Andres Mejia
    Apr 21 at 22:49
















  • 1




    not an answer, but here is a paper from around the same time as the cited one. The result was "author does not know a counterexample."
    – Andres Mejia
    Apr 21 at 22:49










1




1




not an answer, but here is a paper from around the same time as the cited one. The result was "author does not know a counterexample."
– Andres Mejia
Apr 21 at 22:49






not an answer, but here is a paper from around the same time as the cited one. The result was "author does not know a counterexample."
– Andres Mejia
Apr 21 at 22:49












1 Answer
1






active

oldest

votes


















6














This is not true.



An explicit counterexample is given on p. 154 of the paper On power-invariance, by Eloise A. Hamann, Pacific J. Math., Volume 61, Number 1 (1975), 153-159.
The example was given by Andy Magid at the Commutative Algebra Conference in June of 1974, based on Hochster's counterexample for polynomial rings.




More precisely, Hochster's counterexample provides (using hairy ball theorem) a projective non-free module $P$ over the commutative ring $A = Bbb R[x,y,z]/(x^2+y^2+z^2-1)$ such that $A^3 cong A oplus P$ as $A$-modules.



Let $R$ be the complete symmetric algebra of the $A$-modules $P$, i.e. the completion of the symmetric algebra $S_A(P)$ with respect to the ideal generated by $P$.
Let $S := A[[X,Y]]$.
Then, one can show that
$$R[[T]] cong A[[X,Y,Z]] cong S[[T]]$$ as rings (not necessarily as topological rings), but
$$R not cong A[[X,Y]] = S.$$




However, theorem 4 in Hamann's paper states that if $R$ has finitely many maximal ideals, then
$R[![x]!] cong S[![x]!] implies R cong S$, for every commutative ring $S$.



Moreover, theorem 1.4 in Joong Ho Kim's 1974 paper Power invariant rings shows that the same property holds if $R$ has a nilpotent Jacobson radical (i.e. the intersection $J$ of all maximal ideals of $R$ satisfies $J^n=0$ for some $n>0$). For instance, this holds if $R$ is artinian (e.g. a field) or semiprimitive (e.g. $Bbb Z$).






share|cite|improve this answer























  • Wow, that is quite a complete answer. Thank you very much.
    – Severin Schraven
    May 11 at 19:15











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









6














This is not true.



An explicit counterexample is given on p. 154 of the paper On power-invariance, by Eloise A. Hamann, Pacific J. Math., Volume 61, Number 1 (1975), 153-159.
The example was given by Andy Magid at the Commutative Algebra Conference in June of 1974, based on Hochster's counterexample for polynomial rings.




More precisely, Hochster's counterexample provides (using hairy ball theorem) a projective non-free module $P$ over the commutative ring $A = Bbb R[x,y,z]/(x^2+y^2+z^2-1)$ such that $A^3 cong A oplus P$ as $A$-modules.



Let $R$ be the complete symmetric algebra of the $A$-modules $P$, i.e. the completion of the symmetric algebra $S_A(P)$ with respect to the ideal generated by $P$.
Let $S := A[[X,Y]]$.
Then, one can show that
$$R[[T]] cong A[[X,Y,Z]] cong S[[T]]$$ as rings (not necessarily as topological rings), but
$$R not cong A[[X,Y]] = S.$$




However, theorem 4 in Hamann's paper states that if $R$ has finitely many maximal ideals, then
$R[![x]!] cong S[![x]!] implies R cong S$, for every commutative ring $S$.



Moreover, theorem 1.4 in Joong Ho Kim's 1974 paper Power invariant rings shows that the same property holds if $R$ has a nilpotent Jacobson radical (i.e. the intersection $J$ of all maximal ideals of $R$ satisfies $J^n=0$ for some $n>0$). For instance, this holds if $R$ is artinian (e.g. a field) or semiprimitive (e.g. $Bbb Z$).






share|cite|improve this answer























  • Wow, that is quite a complete answer. Thank you very much.
    – Severin Schraven
    May 11 at 19:15
















6














This is not true.



An explicit counterexample is given on p. 154 of the paper On power-invariance, by Eloise A. Hamann, Pacific J. Math., Volume 61, Number 1 (1975), 153-159.
The example was given by Andy Magid at the Commutative Algebra Conference in June of 1974, based on Hochster's counterexample for polynomial rings.




More precisely, Hochster's counterexample provides (using hairy ball theorem) a projective non-free module $P$ over the commutative ring $A = Bbb R[x,y,z]/(x^2+y^2+z^2-1)$ such that $A^3 cong A oplus P$ as $A$-modules.



Let $R$ be the complete symmetric algebra of the $A$-modules $P$, i.e. the completion of the symmetric algebra $S_A(P)$ with respect to the ideal generated by $P$.
Let $S := A[[X,Y]]$.
Then, one can show that
$$R[[T]] cong A[[X,Y,Z]] cong S[[T]]$$ as rings (not necessarily as topological rings), but
$$R not cong A[[X,Y]] = S.$$




However, theorem 4 in Hamann's paper states that if $R$ has finitely many maximal ideals, then
$R[![x]!] cong S[![x]!] implies R cong S$, for every commutative ring $S$.



Moreover, theorem 1.4 in Joong Ho Kim's 1974 paper Power invariant rings shows that the same property holds if $R$ has a nilpotent Jacobson radical (i.e. the intersection $J$ of all maximal ideals of $R$ satisfies $J^n=0$ for some $n>0$). For instance, this holds if $R$ is artinian (e.g. a field) or semiprimitive (e.g. $Bbb Z$).






share|cite|improve this answer























  • Wow, that is quite a complete answer. Thank you very much.
    – Severin Schraven
    May 11 at 19:15














6












6








6






This is not true.



An explicit counterexample is given on p. 154 of the paper On power-invariance, by Eloise A. Hamann, Pacific J. Math., Volume 61, Number 1 (1975), 153-159.
The example was given by Andy Magid at the Commutative Algebra Conference in June of 1974, based on Hochster's counterexample for polynomial rings.




More precisely, Hochster's counterexample provides (using hairy ball theorem) a projective non-free module $P$ over the commutative ring $A = Bbb R[x,y,z]/(x^2+y^2+z^2-1)$ such that $A^3 cong A oplus P$ as $A$-modules.



Let $R$ be the complete symmetric algebra of the $A$-modules $P$, i.e. the completion of the symmetric algebra $S_A(P)$ with respect to the ideal generated by $P$.
Let $S := A[[X,Y]]$.
Then, one can show that
$$R[[T]] cong A[[X,Y,Z]] cong S[[T]]$$ as rings (not necessarily as topological rings), but
$$R not cong A[[X,Y]] = S.$$




However, theorem 4 in Hamann's paper states that if $R$ has finitely many maximal ideals, then
$R[![x]!] cong S[![x]!] implies R cong S$, for every commutative ring $S$.



Moreover, theorem 1.4 in Joong Ho Kim's 1974 paper Power invariant rings shows that the same property holds if $R$ has a nilpotent Jacobson radical (i.e. the intersection $J$ of all maximal ideals of $R$ satisfies $J^n=0$ for some $n>0$). For instance, this holds if $R$ is artinian (e.g. a field) or semiprimitive (e.g. $Bbb Z$).






share|cite|improve this answer














This is not true.



An explicit counterexample is given on p. 154 of the paper On power-invariance, by Eloise A. Hamann, Pacific J. Math., Volume 61, Number 1 (1975), 153-159.
The example was given by Andy Magid at the Commutative Algebra Conference in June of 1974, based on Hochster's counterexample for polynomial rings.




More precisely, Hochster's counterexample provides (using hairy ball theorem) a projective non-free module $P$ over the commutative ring $A = Bbb R[x,y,z]/(x^2+y^2+z^2-1)$ such that $A^3 cong A oplus P$ as $A$-modules.



Let $R$ be the complete symmetric algebra of the $A$-modules $P$, i.e. the completion of the symmetric algebra $S_A(P)$ with respect to the ideal generated by $P$.
Let $S := A[[X,Y]]$.
Then, one can show that
$$R[[T]] cong A[[X,Y,Z]] cong S[[T]]$$ as rings (not necessarily as topological rings), but
$$R not cong A[[X,Y]] = S.$$




However, theorem 4 in Hamann's paper states that if $R$ has finitely many maximal ideals, then
$R[![x]!] cong S[![x]!] implies R cong S$, for every commutative ring $S$.



Moreover, theorem 1.4 in Joong Ho Kim's 1974 paper Power invariant rings shows that the same property holds if $R$ has a nilpotent Jacobson radical (i.e. the intersection $J$ of all maximal ideals of $R$ satisfies $J^n=0$ for some $n>0$). For instance, this holds if $R$ is artinian (e.g. a field) or semiprimitive (e.g. $Bbb Z$).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 25 at 15:19

























answered May 11 at 12:17









Watson

15.8k92970




15.8k92970












  • Wow, that is quite a complete answer. Thank you very much.
    – Severin Schraven
    May 11 at 19:15


















  • Wow, that is quite a complete answer. Thank you very much.
    – Severin Schraven
    May 11 at 19:15
















Wow, that is quite a complete answer. Thank you very much.
– Severin Schraven
May 11 at 19:15




Wow, that is quite a complete answer. Thank you very much.
– Severin Schraven
May 11 at 19:15


















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