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$\renewcommand{\power}{[\![X]\!]}$ Let $R$ be a commutative (unital) ring and $a \in R$. What does the quotient ring $R[[X]] / (X-a)$ look like? This might be too general, so we may assume that $R$ is a PID.

Known results:

  • It should hold that $R\power / (X) \cong R$ (the evaluation at $0$ being well-defined).

  • When $a \in R^{\times}$, then $X-a$ is a a unit in $R\power$, so the quotient is just $0$. For instance, $R \power / (X-1) \cong \{0\}$.

It is easy to show that $R[X]/(X-a)$ is isomorphic to the ring $R$. But for power series, I'm unsure what happens.

When $a=p$ is prime and $R=\Bbb Z$, we get the $p$-adic integers $\Bbb Z_p$. If $R=\Bbb Z$, maybe we can use the prime decomposition of $a \in \Bbb Z \setminus \{0\}$ (this is where the assumption of PID or UFD might be useful) to get a description of $R\power / (X-a)$.

Is $R\power / (X-a)$ related to the product ring $R^{\Bbb N}$ somehow (similarly to $\Bbb Z_p$ being a subring of $\prod_n \Bbb Z/p^n \Bbb Z$) ? At least if $a$ is a prime element in $R$ ?

(Notice that studying $R\power/(bX-a)$ might be even more difficult, see the analogue question for polynomial rings : What is $\Bbb Z[X]/(aX+b)$ isomorphic to?).

Alphonse
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1 Answers1

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If $R$ is noetherian, this is the $a$-adic completion of $R$, see Theorem 8.12 in Matsumura's Commutative Ring Theory.

In particular, by the explicit construction of the inverse limit, this is indeed a subring of $\prod_{n} R/a^n$.

MooS
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