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I found these notes which present a variation on how to prove that $R = \{m/n:n \text{ is odd and m/n is a reduced fraction},n,m \in \mathbb{Z}\}$ with respect to other posts in SE. I would like to ask you about the last step of the proof.

Basically they prove that $R$ is a integrity domain and they try to show that if we define $n = min \{p:p/q \in I, p,q > 0\}$ then $I = \langle n \rangle$. They first state that $n$ must be even and show that $\langle n \rangle \subseteq I$. To prove the converse they take $p/q \in I$ and distinguish cases where $n|p$ (no problem) and $n \nmid p$.

In this latter case they perform euclidean division $p = an + r$ and they state that by minimality, $a \ge 1,0 < r < n$. The conclusion reads $\frac{r}{q} = \frac{p}{q} - \frac{a}{q} n \in I$ which contradicts the fact that $n$ is minimal.

My problem

I don't see very well why $a \ge 1$. Even in that situation I don't see why $a/q$ should be in $I$ so that $r/q$ is in $I$.

Can you help me to finish this argument?

user1868607
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1 Answers1

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I wish to help you. I will prove that $R$ is a PID by using second way.

$R=\{\frac{m}{n}, m,n\in \mathbb Z, 2\nmid n \}=\mathbb Z_{(2)} $ is the ring of fractions $\mathbb Z_S$ where $S=\mathbb Z-(2)$. Every ideal in $\mathbb Z_{(2)}$ of the form $\{(2^n); n\in \mathbb N \}$

Mustafa
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