# Listing all Possible ideals of ring $F[x]/(p(x))$ where F is field and p(x) is polynomial in $F[x]$

I wanted to list all Possible ideals of ring $$F[x]/(p(x))$$ where F is field and p(x) is polynomial in $$F[x]$$

I can list ideal but I do not know my list contain all possible .

My List of ideal

$$F[x]/(n,p(x))$$ where $$n\neq 0, n\in F$$ but as F is field it n becomes unit and $$F[x]/(n,p(x))=0$$

So $$F[x]/(g(x),p(x))$$ where 0

AS for degree greater than equal to p(x) we have same ideal that of original

Is am right ?

Please Help me

ANy help will be appreciated

## 2 Answers

All ideals in $$F[x]/\langle p(x)\rangle$$ are in bijections with the ideals of $$F[x]$$ containing $$\langle p(x)\rangle$$ (by correspondence theorem), i.e. containing $$p(x)$$. Since $$F$$ is a field, $$F[x]$$ is a PID, so any ideal is of the form $$\langle q(x)\rangle$$. Also, $$p(x)\in \langle q(x)\rangle$$ iff $$q(x)\mid p(x)$$. To summarize: all ideals of $$F[x]/\langle p(x)\rangle$$ are given by $$\langle q(x)\rangle/\langle p(x)\rangle$$ for $$q(x)\mid p(x)$$.

Let's do an example to see what happens.

Let's take $$F=\mathbf{R}$$ and $$p(x)=x^2+1$$. Then $$\mathbf{R}[x]/(x^2+1)\simeq \mathbf{C}$$ --- do you see why?

What happens if we change to $$F=\mathbf{C}$$? Then $$\mathbf{C}[x]/(x^2+1)\simeq \mathbf{C}\times \mathbf{C}$$. Try to work out these isomorphisms and you will get a good idea of what happens when you take a quotient of a polynomial ring.

It will also help to look up the "correspondence theorem": it says that the ideals of $$R/I$$ are in bijection with the ideals of $$R$$ that contain $$I$$. How does this apply to your particular case?