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What are the prime ideals in $R=\mathbb{Z}[\frac{1}{10}]$?

An element of this ring is of the form $a_0+\sum_{i=1}^{n}\frac{a_i}{10^i}$ where $a_i\in\mathbb{Z}$ for all $0\leq i\leq n$, so I presume that the ideals $(0)$ and $(p)$ where $p$ is a prime except $2$ and $5$ would be prime ideals in this ring. But are there more since the prime ideals of $\mathbb{Z}[X]$ are $(0)$, $(p)$, $(f)$ and $(p,f(x))$ where $p$ is a prime in $\mathbb{Z}$ and $f$ is an irreducible polynomial in $\mathbb{Z}[X]$ and in the last ideal $f$ is irreducible modulo $p$? Also, is it correct to think about the prime ideals of $R$ by considering the case of the polynomial ring?

Anish Ray
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  • @AnneBauval not completely because the thread you provided doesn't mention what are the prime ideals in $\mathbb{Z}[1/10]$? I need that information to solve the actual problem that I am working on. But I think the prime ideals $(0)$ & $(p)$ for $p$ prime except $2,3$ that I have mentioned are the only prime ideals, right? – Anish Ray Feb 09 '23 at 11:44
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    The question linked by Anne does answer your question, althuogh not completely obvious: In fact, the ring $\mathbb{Z}[\frac1{10}]$ from your question is equal to the ring $\mathbb{Z}_{10}$ from the question Anne linked. – student91 Feb 09 '23 at 12:03
  • @student91 Yes, I do understand that. But what are the prime ideals in this ring? – Anish Ray Feb 09 '23 at 12:14

1 Answers1

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@student91 Yes, I do understand that. But what are the prime ideals in this ring?

Okay I will answer your question.

I presume that the ideals $(0)$ and $(p)$ where $p$ is a prime except 2 and 5 would be prime ideals in this ring.

Indeed, correct! :D

But are there more since the prime ideals of ℤ[] are (0), (), () and (,()) where is a prime in ℤ and is an irreducible polynomial in ℤ[] and in the last ideal $f$ is irreducible modulo $p$?

Also, is it correct to think about the prime ideals of by considering the case of the polynomial ring?

I don't think this is particularly helpful, although you could think of $\mathbb{Z}[\frac1{10}]$ as $\mathbb{Z}[X]/(10X-1)$. Now if you would like to do it this way, the prime ideals of $\mathbb{Z}[X]/(10X-1)$ are exactly the image of the prime ideals of $\mathbb{Z}[X]$ under the quotient map.

However, some of these prime ideals have the same image under the quotient map. In particular:

  • The ideals $(2)$ and $(5)$ reduce to $(1)$, because $1=2\cdot(5X)=5\cdot(2X)$.
  • Any ideal of a polynomial $f=\sum_{i=0}^na_iX^i$ reduces to the ideal generated by $X^n\left(\sum_{i=0}^n a_i(10)^{n-i}\right)$. But $X^n$ is a unit (Because $10^nX^n=1)$, so this is the ideal generated by $\sum_{i=0}^n a_i(10)^{n-i}$, which is just a number. So we did not find any new prime ideals here.
  • The ideal $(p,f(x))$ is a combination of the situations above, so we did not find anything new here either.
student91
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