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 $pT-q$.

I have two question about this ring:

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.

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    $\begingroup$ 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? $\endgroup$ – 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.

  • $\begingroup$ 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?. $\endgroup$ – camilo Aug 20 '19 at 15:21
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    $\begingroup$ @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. $\endgroup$ – Bugs Bunny Aug 20 '19 at 18:52

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