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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

Curious student
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2 Answers2

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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 $q(x)\mid p(x)$.

SMM
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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?

Ehsaan
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