# Prime ideals over the ring of power series over $\mathbb{Z}$

Let $$R\subset \mathbb{Z}[[T]]$$ be a subring of the the ring of power series over the integers. We define $$R$$ with the following property: $$\sum a_iT^i\in R$$ if and only if there exist a $$L>0$$ with $$\mid a_i\mid\leq L$$ for every coefficent $$a_i$$.

I choose this ring because for $$p,q\in \mathbb{Z}$$ the ideal generated by $$pT-q$$ does not contain the unity.

Let $$q>p>0$$ be prime integers. Let $$I_{\infty}$$ be the ideal of power series that vanish in $$p/q$$ with the real topology. Let $$I_{p}$$ be the ideal of power series that vanish in $$p/q$$ with the p-adic topology. And let $$I$$ be the ideal generated by $$pT-q$$.

The first question is if $$I$$ is a prime ideal.

The second question follows as a consequence of the first question. My question is if $$I=I_{\infty}= I_p$$.

I do this question because I want to understand if there is a relationship beetwen the convergence over the p-adic topology and the real topology.

• Are you sure $R$ is a subring? If $f$ is the power series with every coefficient equal to 1, then $f$ is in $R$, but $f^2$ has unbounded coefficients so is not in $R$. Am I misunderstanding something? – Matt Feller Aug 20 '19 at 2:07

Your $$R$$ is not a ring. I replace it with $$R^\prime$$, the ring of all power series, whose convergence radius is at least 1.
Then one of the ideals is distinct. Indeed, take $$f(T) = \sum a_k T^k \in I_p$$, plug in $$p/q$$ and examine $$p$$-adic convergence term by term. It follows that $$a_0$$ is divisible by $$p$$, which is not necessary for an element of the other two ideals.
Now $$I_\infty$$ is a prime ideal. You can find $$f(T)\in R^\prime$$ such that $$f(p/q)$$ converges to any real number. Hence, $$R^\prime / I_\infty$$ is isomorphic to the real field.
Finally, $$I= I_\infty$$. This follows from the fact that $$R^\prime$$ is zariskian and all ideals in a zariskian ring are closed. See "Zariskian Filtrations" by H. Li and F. Van Oystaeyen.
• Thanks I want to construct this ring. I have a doubt the element $pT-q$ belongs to $I_p$. So $I$ is a subset of $I_p$. With this equality we have a morphism between the real numbers and the p-adic numbers. It is posible?. – camilo Aug 20 '19 at 15:21
• @camilo $pT-q$ is your typo. You probably mean $qT-p$ that vanishes at $p/q$ and belongs to $I_p$. The ring $R^\prime /I_p$ is not $p$-adic numbers. I am not sure what this ring is. – Bugs Bunny Aug 20 '19 at 18:52