Given the polynomial ring over a field $\mathbb{K}[x]$ and $f\in \mathbb{K}[x],$ one may define the residual polynomial ring $\mathbb{K}[x]/(f).$ I am wondering how does an ideal of the residual ring look like ? Can you give an example? While I know how the elements of $\mathbb{K}[x]/(f)$ can be described by considering it as a vector space, I am having trouble in figuring out how an ideal in that ring can be formally described. I have the same question in the case of the residual polynomial ring $\mathbb{K}[x_1,..,x_n]/(f), \,f\in \mathbb{K}[x_1,..,x_n] $. It will be great if you provide an example in each case. Many thanks.
-
3Are you familiar with the correspondence theorem? – Mark Mar 19 '23 at 19:38
-
Thanks. I know that it provides a correspondence between the ideals in $\mathbb{K}[x]$ which contain the ideal $(f)$ and the ideals of $\mathbb{K}[x]/(f).$ But I have not used it much. So, I am not familiar. Can you maybe provide an example an thus an answer to my post ? – user996159 Mar 19 '23 at 19:45
-
See for example this post. – Dietrich Burde Mar 19 '23 at 19:55
1 Answers
In general, if $R$ is a ring and $I \subset R$, the ideals of $R/I$ are in one-to-one correspondence with the ideals of $R$ that contain $I$. Every ideal of $R/I$ is of the form $I' / I$, where $I'$ is an ideal of $R$ that contains $I$.
In your example, $R = k[X]$ and $I = (f)$. $k[X]$ is a principal ideal domain. An ideal $I' = (g)$ contains the ideal $I = (f)$ if and only if $g$ divides $f$. Therefore, the ideals of $k[X]/(f)$ are of the form $(g)/(f)$, where $g$ divides $f$.
Let's make this more concrete. Suppose $R = \mathbb Q[X]$ and $I = (X^2 - 1)$. Then $(X + 1)/(X^2 - 1)$ is an ideal in $R/I$.
It might be worth examining some elements of $(X + 1)/(X^2 - 1)$ to get a feel for what this thing is. Remember: every element in this ideal is really an equivalence class modulo $(X^2 - 1)$. For example:
- $[X + 1]_{\text{mod } (X^2 - 1)}$ is an element in this ideal.
- $[(X + 1)(X + 14)]_{\text{mod } (X^2 - 1)}$ is an element in this ideal, since $(X + 1)(X + 14)$ is divisible by $(X + 1)$.
- $[(X + 1)(3X + 12)]_{\text{mod }(X^2 - 1)}$ is an element in this ideal. In fact, it's the same element as the previous one, since $(X + 1)(3X + 12)$ and $(X + 1)(X + 14)$ represent the same equivalence class mod $X^2 - 1$ (since $(X + 1)(3X + 12) = (X + 1)(X + 14) + 2(X^2 - 1)$).
I hope this gives you some intuition for what these ideals are.
- 33,403
-
Thanks. Great answer. Does this play out in a similar way in $\mathbb{K}[x_1,..,x_n]$? Also, in case we consider $\mathbb{K}[x]/(x^2-1)$ as a vector space, whereby it's elements have the form $ax+b,$ is there any difference between the two spaces, i.e. at one hand seen as vector space and at the other hand as a residual ring ? Actually, when one considers it as a vector space, one usually does not consider it's elements as equivalence classes. – user996159 Mar 19 '23 at 21:17
-
As a vector space, $K[x]/(x^2 - 1)$ has dimension two. A basis is $[x]{mod (x^2 - 1)}$ and $[1]{mod (x^2 - 1)}$. – Kenny Wong Mar 19 '23 at 21:34
-
$K[x_1, ... , x_n]$ is similar, expect that it is not a principal ideal domain. So you may have to work slightly harder to verify whether some ideal contains some other ideal. – Kenny Wong Mar 19 '23 at 21:35