# What is the universal property of the algebra of formal power series over a commutative ring?

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.

• 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? – tomasz Jun 9 at 16:21
• 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. – tomasz Jun 9 at 16:55
• 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. – tomasz Jun 9 at 19:57
• The question in the body of your post is at best tangentially related to the title of the post. – Eric Wofsey Jun 9 at 21:37
• @Falq: It's enough to check monomials, or equivalently, coefficient by coefficient. The coefficient at $X^{\sigma(\nu)}$ is by definition $a_{\sigma^{-1}\sigma(\nu)}=a_{\nu}$. – tomasz Jun 10 at 13:48

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 $$A[[\{T_i\}_{i \in I}]] \to A[[\{T_j\}_{j \in J}]]$$.

• 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 – Falq Jun 15 at 17:09
• Notice that $\mu_i \leq \nu_{\sigma(i)}$. From this finiteness easily follows. – Martin Brandenburg Jun 15 at 18:49

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.

• 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. – Falq Jun 14 at 16:14
• @Falq, glad it's helpful. I remember in my own trajectory, it took a while to appreciate the virtues of "characterization" rather than "construction"... :) – paul garrett Jun 14 at 17:07