Let $k$ be a field. Then for the extension $R = k[t^2, t^3, s] \rightarrow S = k[t, s]$, I want to figure out the conductor ideal $ I $of $R $ which is defined to be the largest ideal of S contained in R which is $ I = \{x\in R : xS \subset R\}$ in order to define a Milnor square, so that:
I can explicitly compute $Pic(R[s])$.
But I am unable to construct the square and to conclude what the picard group is.
I was trying to construct the Milnor square via ideals $(t^2, t^3) $ that maps isomorphically onto the ideal $(t^2) $. But I am sure I am making some mistake. Please guide me with hints and suggestions. Thank you.
So I am putting up my entire computation for possible mistakes that I might be making: Let $R = k[t^2,t^3,s]$, and $R' = k[t^2,t^3]$ $\require{AMScd}$ \begin{CD} k[t^2,t^3,s] @>{f}>> k[t,s]\\ @VVV @VVV\\ k[t^2,t^3,s]/(t^2,t^3) = k[s] @>{\bar{f}}>> S/I =k[t,s]/t^2 \end{CD}
For a Milnor square like this the unit Pic sequence will be of the form $1 \rightarrow k^{\times} \rightarrow k^{\times} \times k^{\times} \rightarrow k^{\times} \times k \rightarrow Pic (R) \rightarrow 1.$ Thus $Pic(R) $ is a quotient of $k^{\times} \times k$ but this is the same quotient that I have as $Pic(k[t^2,t^3])$ as the Milnor square then is of the form \begin{CD} k[t^2,t^3] @>{f}>> k[t]\\ @VVV @VVV\\ k[t^2,t^3]/(t^2,t^3) = k @>{\bar{f}}>> k[t]/(t^2) \end{CD} and picard group of an UFD is always trivial. So what I am getting is that both $Pic(k[t^2,t^3])$ and $Pic(k[t^2,t^3,s])$ are isomorphic which should not be the case since $k[t^2,t^3]$ is not semi normal.
Why do you think you are making a mistake?
You can certainly check that the conductor ideal of the finite extension $R \subseteq S$ is $I = (t^2, t^3)R$, just as you wrote.
Indeed, $I = (t^2, t^3)R$ clearly satisfies $IS \subseteq R$. On the other hand, if $r \in R$ is such that $rS \subseteq R$, then in particular $rt \in R$. Note that every element $r \in R$ can be written uniquely as $f_0 + \sum_{i > 1} f_i t^i$ with $f_i \in k[s]$. If $rt = f_0 + \sum_{i > 1} f_i t^i$ for $r \in R$, this shows that $f_0$ is divisible by $t$, hence $f_0 = 0$. In turn this shows that $r \in I$.
The canonical Milnor square associated to the finite extension $R \subseteq S$ by way of its conductor ideal is then:
$\require{AMScd}$ \begin{CD} R @>{f}>> S\\ @VVV @VVV\\ R/I = k[s] @>{\bar{f}}>> S/I =k[t,s]/t^2 \end{CD}