4

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.

kubo
  • 1,918
  • https://math.stackexchange.com/questions/3249312/how-do-p-adic-fields-degenerate-for-non-prime-p isn't about your problem directly, but maybe it can help you shed some light on what's going on. – Arthur Nov 27 '23 at 11:36
  • 1
    You wrote However, (2) and (3) are not closed.... But did you intend to write However, (2) and (3) are not open...? – Lee Mosher Nov 27 '23 at 12:24
  • 2
    We know that both $(6)$ and $3+(6)$ are open, so their union is open. – Andrew Hubery Nov 27 '23 at 12:39
  • @LeeMosher yes, thank you – kubo Nov 27 '23 at 16:58

1 Answers1

0

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.

tkr
  • 2,218