[proofplan]
We first verify that the $I$-adically open sets form a topology on $R$ by checking the three axioms (empty set and full space, closure under arbitrary unions, closure under finite intersections). We then show that addition and multiplication are continuous with respect to this topology by verifying that preimages of basic open sets under the ring operations are open.
[/proofplan]
[step:Verify the topology axioms for the $I$-adic open sets]
Recall that a set $U \subseteq R$ is $I$-adically open if for every $x \in U$, there exists $n \geq 0$ such that the coset $x + I^n \subseteq U$.
**Empty set and full space:** The empty set $\varnothing$ is vacuously open. For $R$ itself, every $x \in R$ satisfies $x + I^0 = x + R = R \subseteq R$, so $R$ is open.
**Arbitrary unions:** Let $\{U_\lambda\}_{\lambda \in \Lambda}$ be a collection of $I$-adically open sets, and let $U = \bigcup_\lambda U_\lambda$. If $x \in U$, then $x \in U_\lambda$ for some $\lambda$, so there exists $n$ with $x + I^n \subseteq U_\lambda \subseteq U$. Hence $U$ is open.
**Finite intersections:** Let $U$ and $V$ be $I$-adically open, and let $x \in U \cap V$. There exist $n, m \geq 0$ with $x + I^n \subseteq U$ and $x + I^m \subseteq V$. Set $N = \max(n, m)$. Since $I^N \subseteq I^n$ and $I^N \subseteq I^m$ (as $N \geq n, m$ and $I^{k+1} = I \cdot I^k \subseteq I^k$), we have
\begin{align*}
x + I^N \subseteq x + I^n \subseteq U \quad \text{and} \quad x + I^N \subseteq x + I^m \subseteq V.
\end{align*}
Therefore $x + I^N \subseteq U \cap V$, so $U \cap V$ is open. By induction, any finite intersection of open sets is open.
[guided]
The $I$-adic topology on $R$ is defined by declaring that a set $U \subseteq R$ is open if for every $x \in U$, there exists $n \geq 0$ such that the coset $x + I^n$ is contained in $U$. The cosets $x + I^n$ play the role of "basic open neighbourhoods" of $x$ — they are the non-archimedean analogue of open balls in a metric space.
The key property used for finite intersections is that the ideals $I^n$ form a decreasing filtration: $R = I^0 \supseteq I^1 \supseteq I^2 \supseteq \cdots$. This means $I^N \subseteq I^n$ whenever $N \geq n$. So if $x + I^n \subseteq U$ and $x + I^m \subseteq V$, then taking $N = \max(n, m)$ gives $x + I^N \subseteq (x + I^n) \cap (x + I^m) \subseteq U \cap V$.
Why does this fail for infinite intersections? Consider $R = \mathbb{Z}$ and $I = (p)$. The sets $U_n = \{x : x + p^n\mathbb{Z} \subseteq U_n\}$ get finer and finer, and the infinite intersection $\bigcap_n (x + p^n\mathbb{Z}) = \{x\}$ is a single point, which need not be open (and is not, unless $\bigcap_n I^n$ is itself an ideal making $\{x\}$ a coset).
[/guided]
[/step]
[step:Show addition is continuous with respect to the $I$-adic topology]
We must show that the addition map
\begin{align*}
+: R \times R &\to R \\
(x, y) &\mapsto x + y
\end{align*}
is continuous, where $R \times R$ carries the product topology. It suffices to check that for every $I$-adically open set $U \subseteq R$, the preimage $+^{-1}(U) = \{(x, y) \in R \times R : x + y \in U\}$ is open in $R \times R$.
Let $(x, y) \in +^{-1}(U)$. Since $x + y \in U$ and $U$ is open, there exists $n \geq 0$ with $(x + y) + I^n \subseteq U$. We claim $(x + I^n) \times (y + I^n) \subseteq +^{-1}(U)$. Indeed, for any $(x + a, y + b)$ with $a, b \in I^n$:
\begin{align*}
(x + a) + (y + b) = (x + y) + (a + b) \in (x + y) + I^n \subseteq U,
\end{align*}
since $a + b \in I^n$ (as $I^n$ is an ideal, hence an additive subgroup). Therefore $(x + a, y + b) \in +^{-1}(U)$.
Since $(x + I^n) \times (y + I^n)$ is a basic open set in $R \times R$ containing $(x, y)$ and contained in $+^{-1}(U)$, the preimage $+^{-1}(U)$ is open.
[/step]
[step:Show multiplication is continuous with respect to the $I$-adic topology]
We must show that the multiplication map
\begin{align*}
\cdot: R \times R &\to R \\
(x, y) &\mapsto xy
\end{align*}
is continuous. Let $(x, y) \in \cdot^{-1}(U)$ with $xy \in U$, so there exists $n \geq 0$ with $xy + I^n \subseteq U$.
We claim $(x + I^n) \times (y + I^n) \subseteq \cdot^{-1}(U)$. For any $a, b \in I^n$:
\begin{align*}
(x + a)(y + b) = xy + xb + ay + ab.
\end{align*}
Since $b \in I^n$, we have $xb \in I^n$ (as $I^n$ is an ideal of $R$ and $x \in R$). Similarly $ay \in I^n$. And $ab \in I^n \cdot I^n = I^{2n} \subseteq I^n$. Therefore
\begin{align*}
xb + ay + ab \in I^n,
\end{align*}
so $(x + a)(y + b) = xy + (xb + ay + ab) \in xy + I^n \subseteq U$.
This shows $(x + I^n) \times (y + I^n) \subseteq \cdot^{-1}(U)$, so $\cdot^{-1}(U)$ is open, and multiplication is continuous.
[guided]
The key computation is that the "error term" in the product $(x + a)(y + b) - xy = xb + ay + ab$ belongs to $I^n$ whenever $a, b \in I^n$. Let us verify each summand:
- $xb \in I^n$: since $I^n$ is an ideal of $R$ and $x \in R$, we have $xb \in R \cdot I^n = I^n$.
- $ay \in I^n$: same reasoning, since $a \in I^n$ and $y \in R$.
- $ab \in I^{2n} \subseteq I^n$: since $a \in I^n$ and $b \in I^n$, the product $ab \in I^n \cdot I^n = I^{2n}$. And $I^{2n} \subseteq I^n$ because $2n \geq n$.
So $xb + ay + ab \in I^n + I^n + I^n = I^n$ (since $I^n$ is closed under addition as an ideal).
Notice that unlike addition (where we needed only the additive subgroup structure of $I^n$), continuity of multiplication uses the full ideal structure: the fact that $R \cdot I^n \subseteq I^n$.
This proof works uniformly for all $n$ and does not require adjusting $n$ based on $x$ and $y$. In an archimedean metric topology, one would typically need to choose the neighbourhood size depending on the magnitudes of $x$ and $y$, but here the ideal structure of $I^n$ handles this automatically.
[/guided]
[/step]
