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.

I need little help in proving the following result :

Consider the ring $R:=\mathbb{F}_q[X]/(X^n-1)$, where $\mathbb{F}_q$ is a finite field of cardinality $q$ and $n\in\mathbb{N}$. Then any ideal $I$ of $R$ is principle and can be written as $I=(g(X))$, such that $g(X)|(X^n-1)$.

share|improve this question
2  
Any ideas, thoughts, self work, insights...? –  DonAntonio Nov 5 '12 at 11:46

2 Answers 2

If $\varphi\colon S\to T$ is an epimorphism of rings with $1$, and if $I$ is an ideal of $T$, then $J=\varphi^{-1}(I)$ is an ideal of $S$.

Now if $S$ is PID, that is, if every ideal is generated by a single element, then $J=xS$, for some $x\in S$. So $I=\varphi(J)=\varphi(s)I$, is also principle.

Apply the above to $\varphi\colon \mathbb{F}_q[X]\to R$ and you'll get that every ideal of $R$ is principle.

share|improve this answer

Hint: Think about $R=\Bbb F_q[X]/(X^n-1)$ as the ring of polynomials of degree $<n$, and multiplication is 'modulo $(X^n-1)$', meaning that $X^n=1$ is the rule to use in $R$.

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.