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Let $F$ be a field. Denote $R = \{ f/g \mid f, g \in F[x], g\neq 0 \}$, which is the fraction field of $F[x]$. Choose an element $a \in F$ and set $R_a = \{ f/g \in R\mid g(a) \neq 0 \}$.

Show that $R_a$ is a principal ideal domain, and describe all ideals of $R_a$.

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$R_a$ is the localization of $F[X]$ at the maximal ideal $(X-a)$. This is a DVR and all its ideals are powers of the maximal ideal. – user26857 Dec 15 '12 at 1:23

You have a natural injective ring map $i: F[x] \hookrightarrow R_a$, which sends $f(x) \mapsto \frac{f(x)}{1}$. Now given any ideal $I \subset R_a$, $i^{-1}(I)$ is principal. Try to use this to show that $I$ must be principal (Hint: if $\frac{f(x)}{g(x)} \in I$, then $f(x) \in i^{-1}(I)$). Why $R_a$ is a domain should be clear from the definition.

As a further exercise, can you show what the prime ideals of this ring are? There are only finitely many.

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The ideals of $R_a$ are of the form $IR_a$, with $I$ is an ideal of $R$ then $IR_a=(b)R.R_a=bR_a$ that is principal.

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