Every $k$-derivation $D:K[t]\to M$ is uniquely determined by its restriction to $K$ and by $D(t)$. Conversely, a $k$-derivation $K\to M$ and an arbitrary element $m\in M$ extend uniquely by
$$
D\left(\sum_i a_it^i\right)=\sum_iD(a_i)t^i+\sum_iia_it^{i-1}m.
$$
Representing this natural decomposition of derivations gives
$$
\Omega_{K[t]/k}\cong(\Omega_{K/k}\otimes_KK[t])\oplus K[t],dt.
$$

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 126 / 1 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-126/1/ii/solution