# Thread: Ideals in a UFD and PID

1. ## Ideals in a UFD and PID

This problem is from Dummit and Foote Abstract Algebra

Exhibit all of the ideals in the ring

$\displaystyle F[x]/(p(x))$ , where F is a field and $\displaystyle p(x)$ is a polynomial in $\displaystyle F[x]$ (describe them in terms of the factorization of p(x)))

I know that since F is a field that $\displaystyle F[x]$ is a PID and a UFD.

So since $\displaystyle p(x) \in F[x]$ $\displaystyle p(x)=p_1(x)p_2(x) \cdots p_n(x)$ and where each $\displaystyle p_i(x)$ is irreduceable and this is unique upto associates. Also since each $\displaystyle p_i(x)|p(x)$ the Ideal of $\displaystyle (p(x)) \subseteq p_i(x)$

First I am not sure if the above observations help me, If they do I don't see where to go from here. Second I am having a hard time visualzing what elements of this ring would look like. I think it would eliminate all elements of F[x] that have the same degree as any factor of $\displaystyle P_i(x)$.

Thanks

TES

2. Originally Posted by TheEmptySet
This problem is from Dummit and Foote Abstract Algebra

Exhibit all of the ideals in the ring

$\displaystyle F[x]/(p(x))$ , where F is a field and $\displaystyle p(x)$ is a polynomial in $\displaystyle F[x]$ (describe them in terms of the factorization of p(x)))

I know that since F is a field that $\displaystyle F[x]$ is a PID and a UFD.

So since $\displaystyle p(x) \in F[x]$ $\displaystyle p(x)=p_1(x)p_2(x) \cdots p_n(x)$ and where each $\displaystyle p_i(x)$ is irreduceable and this is unique upto associates. Also since each $\displaystyle p_i(x)|p(x)$ the Ideal of $\displaystyle (p(x)) \subseteq p_i(x)$

First I am not sure if the above observations help me, If they do I don't see where to go from here. Second I am having a hard time visualzing what elements of this ring would look like. I think it would eliminate all elements of F[x] that have the same degree as any factor of $\displaystyle P_i(x)$.

Thanks

TES
suppose $\displaystyle p(x)=\prod_{j=1}^n (p_j(x))^{r_j}, \ r_j \geq 1,$ be the factorization of $\displaystyle p(x)$ into irreducibles. an ideal of $\displaystyle F[x]/(p(x))$ is in the form $\displaystyle I/(p(x)),$ where $\displaystyle I$ is an ideal of $\displaystyle F[x]$ which contains $\displaystyle (p(x)).$

we have $\displaystyle I=(f(x)),$ for some $\displaystyle f(x) \in F[x],$ since $\displaystyle F[x]$ is a PID. now $\displaystyle (p(x)) \subseteq I$ is equivalent to $\displaystyle f(x) \mid p(x),$ i.e. $\displaystyle f(x)=\prod_{j=1}^n (p_j(x))^{s_j}, \ 0 \leq s_j \leq r_j.$

3. Thank you very much .