Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $R_a$.

share|improve this question
1  
What have you tried so far? –  Alex Youcis Dec 13 '12 at 8:31
    
$R_a$ 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

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.