Ideals in a UFD and PID

**TheEmptySet** · April 12th 2009, 05:00 PM

This problem is from Dummit and Foote Abstract Algebra

Exhibit all of 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 since F is a field that $F[x]$ is a PID and a UFD.

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

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 $P_i(x)$ .

Thanks

TES

**NonCommAlg** · April 12th 2009, 05:13 PM

Originally Posted by TheEmptySet

This problem is from Dummit and Foote Abstract Algebra

Exhibit all of 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 since F is a field that $F[x]$ is a PID and a UFD.

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

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 $P_i(x)$ .

Thanks

TES

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

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

**TheEmptySet** · April 12th 2009, 05:39 PM

Thank you very much

.

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