# Ideals of the ring $\mathbb{F}_q[X]/(X^n-1)$

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 $g(X)|(X^n-1)$.

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Any ideas, thoughts, self work, insights...? – DonAntonio Nov 5 '12 at 11:46

## 2 Answers

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.

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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 $R$.

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