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Suppose we have the ring $\mathbb{Z}$ and the ideal $I=(6)$ on it. Thus we can talk about the $(6)$-adic topology. Supposedly, $\mathbb{Z}$ with the $(6)$-adic topology is a topological ring. For that, we have to check that the multiplication map $m:\mathbb{Z}\times \mathbb{Z}\rightarrow \mathbb{Z}$ is continuous.

The preimage of the open set $(6)$ under $m$ is $(\mathbb{Z}\times (6))\cup ((2)\times (3))\cup ((3)\times (2))\cup((6)\times \mathbb{Z})$, and it should be an open set of $\mathbb{Z}\times \mathbb{Z}$ with the product topology. However, $(2)$ and $(3)$ are not open in the $(6)$-adic topology, right? So what am I doing wrong?

Edit: This is a concrete example, but my question refers more generally to the $I$-adic topology when $I$ is not prime/primary.

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  • $\begingroup$ math.stackexchange.com/questions/3249312/… isn't about your problem directly, but maybe it can help you shed some light on what's going on. $\endgroup$
    – Arthur
    Nov 27, 2023 at 11:36
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    $\begingroup$ You wrote However, (2) and (3) are not closed.... But did you intend to write However, (2) and (3) are not open...? $\endgroup$
    – Lee Mosher
    Nov 27, 2023 at 12:24
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    $\begingroup$ We know that both $(6)$ and $3+(6)$ are open, so their union is open. $\endgroup$ Nov 27, 2023 at 12:39
  • $\begingroup$ @LeeMosher yes, thank you $\endgroup$
    – kubo
    Nov 27, 2023 at 16:58

1 Answer 1

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An answer in the comments illustrates you are wrong in assuming that $(2)$ and $(3)$ are not open.

To be an open ideal in the $(6)$-adic topology on $\mathbb{Z}$ is to contain $(6^k)$ for some integer $k \geq 0$. Since $6$ is a multiple of $2$ we have $(6) \subseteq (2)$, and so the ideal $(2)$ is open (take $k=1$). Similarly, $(6) \subseteq (3)$ and so $(3)$ is open.

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