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I started studying about localization of rings and i am stuck at a question about $Spec(R_{S})$ where $R$ is a commutative ring and $S$ a multiplicatively closed set, so $R_{S}$ is also known as localization.

So, have to specify how the prime ideals of a certain $R_{S}$ look like. Given is the ring $R = \mathbb{Z}$, some $k\in \mathbb{Z}$, the localization is $\mathbb{Z}_{k}$. Our multiplicatively closed set is $S=\left \{ 1, k, k^{2}, \dots \right \}$ and $\mathbb{Z}_{k} = \left \{ \frac{r}{k^{n}} : r\in \mathbb{Z}, n\in \mathbb{N} \right \}$.

How to find the prime ideals of $\mathbb{Z}_{k}$? Can anybody help me, please? Thank you very much!

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2 Answers 2

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The prime ideals of $\mathbf Z_k$ correspond bijectively to the primes $p$ such that $k\not\in p\mathbf Z$ i.e. to the primes $p$ that do not divide $k$.

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The prime ideals in a ring of fractions $S^{-1}R$ correspond bijectively to the prime ideals $\mathfrak q$ of $R$ such that $\mathfrak q\cap S=\varnothing$. In your specification, with $R=\mathbb{Z}$, we would like that $k=p$ is prime, so that $S=R\setminus (p)$ is multiplicatively closed and $\mathfrak{p}=(p)$ is a prime ideal in $\mathbb{Z}$, with localisation $S^{-1}R=R_{\mathfrak{p}}=\mathbb{Z}_{(p)}$. Then the prime ideals in the localisation correspond to the prime ideals in $R=\mathbb{Z}$ which do not meet $S$.

Edit: The case $k=p$ seems for me the "natural" one to consider, with $S$ being the complement to $(p)$, but it is also possible to consider the multiplication set $S$ generated by any integer $k$ - see the comments.

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  • $\begingroup$ I dont think that we need k=p! $\endgroup$
    – math635
    Commented Dec 9, 2015 at 23:06
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    $\begingroup$ In the OP's example $S$ is not the complement of a prime, it's simply the multiplicative set generated by $k$. $\endgroup$
    – user26857
    Commented Dec 9, 2015 at 23:42

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