$R = \mathbb{C}[t]_{(t)}[x]/(x^3+x^2+t)$ is a dedekind domain. Therefore every proper ideal $I$ can be written as a product of prime ideals.
I want to find the factorization of $I = (t+x^2-x)$ into prime ideals in $R$. I tried to describe the prime ideals of $S = \mathbb{C}[t]_{(t)}[x]$ the following way:
- zero ideal $(0)$
- $(f)$ where $f$ irreducible in $S$
- $(p)$ where $p$ prime in $\mathbb{C}[t]_{(t)}$
- $(p,f)$ where $p$ prime in $\mathbb{C}[t]_{(t)}$ and $f$ irreducible mod $p$
Then, the prime ideals in $R$ would correspond to those of $S$ that contain $(x^3+x^2+t)$.
Am I going in the right direction with this? Also I'm not able to find the factorization, how do I have to do this?
I think you are on the right track. In fact, $I$ is a prime ideal.
In $R$, one sees that $$ t + x^2 -x = (-x^3-x^2) + x^2 - x= -x^3-x = -x(x-i)(x+i). $$ Thus any prime ideal of $P$ containing $I$ must contain one of the ideals $(t,x)R, (t -1 -i,x-i)R$, or $(t-1+i,x+i)R$. Notice that the second and third cases cannot happen as $t-1-i, t-1+i$ are units in $R$. Therefore, since $R$ is a Dedekind domain, $I = P^a$ for some $a$, where $P = (t, x)$ (it is straightforward to check $P$ is prime).
One can determine $a$ locally. In $R_P$, $x$ is a uniformizing parameter, and the images of $x-i,x+i$ will be units. Thus $IR_P = xR_P$. Hence $I = P$.
Of course, we could've checked that $R/I$ is a field directly, but I believe this approach is more general.