4
$\begingroup$

Let $R$ be an integral domain and $S$ a multiplicative subset with $ 0\notin S$. I am attempting to prove the following statement: If $R$ is a euclidean domain then the localization of $R$ at $S$, $S^{-1}R$, is a euclidean domain.

I have looked at the discussion here and I can follow some of it but I am unfamiliar with saturation and so I do not understand why the norm is well defined. Is there another way to prove this without using saturation?

Thank you!

$\endgroup$
1
$\begingroup$

I know that one can also prove the claim without saturation but the computations would become messy. That's why I wanted to explain the saturation instead. I realize that this is not really an answer to your question.

For a multiplicative subset $D$ of $R$, the saturation of $D$ is defined as $D_{sat}=\{a\in A\mid \exists b\in A : ab\in D\}$. The claim is that the localization of $R$ at $D$ and the localization of $R$ at $D_{sat}$ are isomorphic.

I want to explain the saturation and the claim using an example : Say you want to localize $\mathbb{Z}$ at nonzero even integers and $1$ i.e. $E=\{x\in\mathbb{Z}-\{0\} \mid x\mbox{ is even } \}\cup\{1\}$ which is clearly multiplicative. Localizing $\mathbb{Z}$ at $E$ means that you invert every nonzero even integer. But observe that : $$\frac{3}{1}\cdot\frac{2}{6}=\frac{6}{6}=\frac{1}{1}$$ which means that you necessarily inverted $3$ too! That is because for every odd integer $n$, $2\cdot n$ is even. So when you invert $2\cdot n$, you necessarily invert $n$ since $\frac{1}{n}=\frac{2}{2\cdot n}$.

This is because of the saturation. If for an element $a\in A$ there exists an element $b\in A$ such that $a\cdot b\in D$, when you localize, you invert $a\cdot b$. But $\frac{1}{a} = \frac{b}{a\cdot b}$ so you also inverted $a$ automatically.

This is why $D^{-1}R$ is isomorphic to $D_{sat}^{-1}R$. That also explains why one can always define a multiplicative set as a set which contains $1$ and closed under multiplication. Because the saturation of a multiplicative set necessarily contains $1$.

About your second question. The well-definedness of the new defined norm function is explained in the second paragraph of the hint. If you still have some problems with that explanation, I can try to clarify.

$\endgroup$
  • $\begingroup$ Thank you for your response. This helps me see why $D^{-1}_{sat}R$ is isomorphic to $D^{-1}R$ but I am still struggling to see why the new norm is well-defined. For any $ \frac{a}{b} \in D^{-1}R$, may I assume a and b are coprime since R is a euclidean domain and therefore a unique factorization domain? $\endgroup$ – user509848 Dec 4 '17 at 13:03
  • 1
    $\begingroup$ Exactly. You can write $a=p_1\cdot ...\cdot p_n$ and $b=q_1\cdot ...\cdot q_m$ where $p_i$ and $q_i$ are the prime divisors of $a$ and $b$ respectively. Then for example if $p_1=q_1$, you can write $\frac{a}{b}=\frac{p_2\cdot...\cdot p_n}{q_2\cdot ...\cdot q_m}$. Note that by assuming $D$ is saturated, you can show that each $q_i$ must be in $D$. $\endgroup$ – Levent Dec 4 '17 at 13:14
  • $\begingroup$ This was very helpful. Thank you! $\endgroup$ – user509848 Dec 4 '17 at 14:04
  • $\begingroup$ You are welcome. $\endgroup$ – Levent Dec 4 '17 at 14:04
  • $\begingroup$ There is one little thing, in the other answer the OP claims that $s_1t_2=s_2t_1$ and that $a_1=a_2$, when it's only true that $s_1t_2\sim s_2t_1$ and $a_1\sim a_2$ (and that's enough). $\endgroup$ – Xam Dec 4 '17 at 20:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy