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In $\mathbb{F}[x]$ the ideals are all principal . So the ideals are generated by monic polynomials. Now if $g(x)$ is a monic polynomial ,not contained in $(f(x))$ and which belongs to $\mathbb{F}[x]$ then $(g(x)/f(x))$ is an ideal in $\mathbb{F}[x]/(f(x))$ by the third isomorphism theorem.

This has been my attempt.Iam not sure whether the idea is clear and also I couldn't understand how to describe it in terms of factorization of $f(x)$.

Antimony
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  • For most $g$ you'll have $(g) = (1)$ in $F[x]/(f(x))$ – reuns May 11 '20 at 18:06
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    There's the correspondence: ideals in $R$ containing $I$ correspond to ideals in $R/I$ i.e. the ideals in this case will correspond to the ones generated by the divisors of $f(x)$ i.e. if $d(x)$ is a divisor of $f(x)$, the ideals are $(d(x)I)$ in $R/I$, $I = (f(x))$ – Naweed G. Seldon May 11 '20 at 18:16
  • Consider the almost equivalent but simpler example: What are the ideals of $\mathbb Z / f \mathbb Z$, for $f$ an integer? – lhf May 11 '20 at 19:06

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