# $(2)$ is not open in the $(6)$-adic topology

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.

• 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. Nov 27, 2023 at 11:36
• You wrote However, (2) and (3) are not closed.... But did you intend to write However, (2) and (3) are not open...? Nov 27, 2023 at 12:24
• We know that both $(6)$ and $3+(6)$ are open, so their union is open. Nov 27, 2023 at 12:39
• @LeeMosher yes, thank you
– kubo
Nov 27, 2023 at 16:58

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.