This is Exercise 9.2.5 in Dummit and Foote's Abstract Algebra
Exhibit all the ideals in the ring $F[x]/(p(x))$, where $F$ is a field and $p(x)$ is a polynomial in $F[x]$ (describe them in terms of the factorization of $p(x)$).
I know that similar questions was considered here Listing all Possible ideals of ring $F[x]/(p(x))$ where F is field and p(x) is polynomial in $F[x]$, here ideals in the ring $\frac{F[x]}{(p(x))}$ and here All ideals of $\mathbb{F}[x]/(f(x))$ in terms of factorization of $f(x)$, but these considerations seem to be incomplete because the Exercise explicitly requires to "describe the ideals in terms of the factorization of $p(x)$".
I came up with the following solution:
According to the Lattice Isomorphism Theorem for Rings, every ideal in the ring $F[x]/(p(x))$ has a preimage-ideal $I$ in the ring $F[x]$ containing the ideal $(p(x))$, i.e., $(p(x))⊆I$ (*). Since $F$ is a field, $F[x]$ is a Euclidean Domain, hence a Principal Ideal Domain, P.I.D., therefore, $I=(q(x))=(q)$ [denoting $q(x)=q$] for some polynomial $q∈F[x]$.
- If $q$ is a unit, then $(q)=F[x]$.
- If not, $(q)$ is a proper ideal. Since $F[x]$ is a Principal Ideal Domain, P.I.D., $F[x]$ is a Unique Factorization Domain, U.F.D., therefore, $p(x)=p_1 p_2…p_n$, where $p_i=p_i (x)$-s are polynomials-irreducibles in $F[x]$ and this factorization of $p(x)$ is unique up to associates. So, from (*) we get: $(p_1 p_2…p_n )⊆(q)$, hence $p_1 p_2…p_n∈(q)$, hence $p_1 p_2…p_n=qv$ for some polynomial $v∈F[x]$. Then, since $F[x]$ is a U.F.D., $q$ is necessarily a product $q=q_1 q_2…q_k$ with $1≤k≤n$ of associates to $p_i$-s (possibly renumbered) and $v=v_{k+1} v_{k+2}…v_n$ is a product of the associates to the remaining $n-k$ factors $p_i$ (possibly renumbered). So we have: $q_1 q_2…q_k=(u_1 p_1 )(u_2 p_2 )…(u_k p_k )$, where $u_i$-s are units, $⇔ q_1 q_2…q_k=up_1 p_2…p_k$, where $u=u_1 u_2…u_k$ is a unit, therefore, $(q_1 q_2…q_k )=(p_1 p_2…p_k) ⇔ (q) =(p_1 p_2…p_k )$ with $1≤k≤n$. So, every proper ideal $I=(q)$ in $F[x]$ containing $(p(x))$ is a principal ideal generated by some subproduct taken from the factorization of $p(x)$ into irreducibles in $F[x]$. The number of distinct subproducts and hence the number of ideals in $F[x]$ containing $p(x)$ is $≤\binom{n}{1}+\binom{n}{2}+⋯+\binom{n}{n}=\sum_{k=1}^n\binom{n}{k}$, where $n$ is the number of irreducible factors of $p(x)$ in $F[x]$ and the equality is obtained when all the irreducibles $p_i$ are distinct.
Therefore, the image of $I=(q)$ containing $(p(x))$ in $F[x]$ in the quotient ring $F[x]/(p(x))$ is $\overline{I}=\overline{(q)}=(\overline{q})=(\overline{p_1 p_2…p_k})$ with $1≤k≤n$. So (by the Lattice Isomorphism Theorem), every proper ideal in the ring $F[x]/(p(x))$ is a principal ideal generated by an image of some subproduct taken from the factorization of $p(x)$ into irreducibles in $F[x]$. And the number of proper ideals of $F[x]/(p(x))$ is equal to the number of the proper ideals of $F[x]$ containing $(p(x))$, hence is $≤\sum_{k=1}^n\binom{n}{k}$.
Thus, all the ideals in the ring $F[x]/(p(x))$, where $F$ is a field and $p(x)$ is a polynomial in $F[x]$, when described in terms of the factorization of $p(x)$, are:
- the ring $F[x]/(p(x))$ itself;
- the set of proper ideals, each of which is a principal ideal generated by an image of some subproduct taken from the factorization of $p(x)$ into irreducibles in $F[x]$. The number of proper ideals is $≤\sum_{k=1}^n\binom{n}{k}$, where $n$ is the number of irreducible factors of $p(x)$ in $F[x]$. If this subproduct is the entire factorization itself (i.e., when $k=n$), then the generated ideal is the zero ideal in $F[x]/(p(x))$.
⦁ What I would like to know, is the estimation $≤\sum_{k=1}^n\binom{n}{k}$ of the number of proper ideals in $F[x]/(p(x))$ correct, i.e., may there be more proper ideals in $F[x]/(p(x))$ than $\sum_{k=1}^n\binom{n}{k}$, where $n$ is the number of irreducible factors of $p(x)$ in $F[x]$?