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Let $A$ be a commutative ring. For every set $I$, $A^{\mathbb{N}^{(I)}}$ is the algebra of formal power series.

Suppose $\sigma:I\rightarrow J$ is a bijection. Ignoring topology, what is the canonical algebra isomorphism between $A^{\mathbb{N}^{(I)}}$ and $A^{\mathbb{N}^{(J)}}$?

For example, let $(\alpha_\nu)_{\nu\in\mathbb{N}^{(I)}}$ be an element of $A^{\mathbb{N}^{(I)}}$, what should be the image of this element in $A^{\mathbb{N}^{(J)}}$?

Edit:

Since $\sigma$ is a bijection, there exists a monoid isomorphism $f:\mathbb{N}^{(I)}\rightarrow\mathbb{N}^{(J)}$ such that $$f\circ\delta=\delta'\circ\sigma,$$ where $\delta,\delta'$ are the canonical injections of $I\rightarrow\mathbb{N}^{(I)}$ and $J\rightarrow\mathbb{N}^{(J)}$, respectively. Now, from set theory, we know that the mapping $$g:A^{\mathbb{N}^{(I)}}\rightarrow A^{\mathbb{N}^{(J)}},\,(\alpha_\nu)_{\nu\in\mathbb{N}^{(I)}}\mapsto(\alpha_{f^{-1}(\mu)})_{\mu\in\mathbb{N}^{(J)}}$$ is a bijection.

Definition: By $\mathbf{N}^{(I)}$ denote the subset of $\mathbf{N}^{I}$ consisting of sequences with finite support.

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    $\begingroup$ Why would you ignore the topology? The topology (or equivalently, the filtration given by the gradation) is essential. The image should be: replace each $i\in I$ (in the index) by its image in $J$. What else would it be? $\endgroup$ – tomasz Jun 9 at 16:21
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    $\begingroup$ It would be helpful if you wrote down the definitions of these objects. I'm kind of guessing here, since I'm not familiar with this notation, so I may be misunderstanding something. $\endgroup$ – tomasz Jun 9 at 16:55
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    $\begingroup$ Yes, that works. I was thinking about the question in the title, though. I think the formal power series ring is a free object in an appropriate category (something like the category of complete filtered commutative $A$-algebras), similarly to the polynomial ring, but I can't quite work it out now. $\endgroup$ – tomasz Jun 9 at 19:57
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    $\begingroup$ The question in the body of your post is at best tangentially related to the title of the post. $\endgroup$ – Eric Wofsey Jun 9 at 21:37
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    $\begingroup$ @Falq: It's enough to check monomials, or equivalently, coefficient by coefficient. The coefficient at $X^{\sigma(\nu)}$ is by definition $a_{\sigma^{-1}\sigma(\nu)}=a_{\nu}$. $\endgroup$ – tomasz Jun 10 at 13:48
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A more common notation is $A[[\{T_i\}_{i \in I}]]$. If $\sigma : I \to J$ is a map, it induces a unique continuous ring homomorphism $A[[\{T_i\}_{i \in I}]] \to A[[\{T_j\}_{j \in J}]]$ which extends the identity on $A$ and maps $T_i \mapsto T_{\sigma(i)}$. Thus, it maps a general power series $$\sum_{\mu \in \mathbb{N}^{(I)}} p_{\mu} \cdot \prod_{i \in I} (T_i)^{\mu_i}$$ to the power series $$\sum_{\mu \in \mathbb{N}^{(I)}} p_{\mu} \cdot \prod_{i \in I} (T_{\sigma(i)})^{\mu_i} = \sum_{\nu \in \mathbb{N}^{(J)}} \left(\sum_{\Large \mu \in \mathbb{N}^{(I)}, \, \nu_j = \sum_{\sigma(i)=j} \mu_i} p_{\mu}\right)\cdot \prod_{j \in J} T_j^{\nu_j}.$$ When you ignore the topology, uniqueness fails, and you need to verify that the above formula indeed defines a ring homomorphism.

The whole construction is compatible with composition (it defines a functor), so a bijection $I \to J$ gets mapped to an isomorphism $A[[\{T_i\}_{i \in I}]] \to A[[\{T_j\}_{j \in J}]]$.

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  • $\begingroup$ Martin, sorry to bother you again..In the right hand side of the last equation, is the index set $\{\mu\in\mathbb{N}^{(I)}\ |\ \nu_j=\sum_{\sigma(i)=j}\mu_j,\,\forall j\in J\}$ supposed to be finite? I am not able to see this $\endgroup$ – Falq Jun 15 at 17:09
  • $\begingroup$ Notice that $\mu_i \leq \nu_{\sigma(i)}$. From this finiteness easily follows. $\endgroup$ – Martin Brandenburg Jun 15 at 18:49
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I think @tomasz' deleted answer goes in a direction most appealing to me (at least by this point...), namely, to say that a "formal power series ring" $R[[\{x_i\}]]$ over a commutative ring $R$ (probably with identity) in variables $\{x_i:i\in I\}$ is a/the projective limit of quotients $R[\{x_i\}]/I_d$ of polynomial rings $R[\{x_i\}]$ by the ideals $I_d$ consisting of polynomials of total degree $>d$.

The projective limit characterization is that a map to the proj lim is given by a compatible family of maps to the limitands. A bijection of the index sets of the variables identifies the ideals $I_d$, etc.

Really, this amounts to an assertion that an isomorphism of index sets in the category of sets induces an isomorphism of corresponding formal power series rings (with fixed ring $R$).

In particular, and maybe this is an implicit question, the postulated uniqueness of the map to the proj lim, induced by a compatible family of maps to the limitands, shows that there is a unique self-map of that proj lim that respects all those maps. So, "unique up to unique isomorphism", with that latter qualification having some significance.

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  • $\begingroup$ Thank you Paul! I went for Martin's because I am still not clear about projective limit side of the story. But this is a good answer because it's something I can try to understand next. $\endgroup$ – Falq Jun 14 at 16:14
  • $\begingroup$ @Falq, glad it's helpful. I remember in my own trajectory, it took a while to appreciate the virtues of "characterization" rather than "construction"... :) $\endgroup$ – paul garrett Jun 14 at 17:07

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