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

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

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.

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    $\begingroup$ 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." $\endgroup$ – Andres Mejia Apr 21 '18 at 22:49
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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. $\Bbb Z$).

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  • $\begingroup$ Wow, that is quite a complete answer. Thank you very much. $\endgroup$ – Severin Schraven May 11 '18 at 19:15

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