Present $L=K[t]/(f)$ by sending $t$ to $b$. Applying the <Conormal exact sequence for Kähler differentials> and part ii gives
$$
L\cong(f)/(f^2)\longrightarrow
(\Omega_{K/k}\otimes_KL)\oplus L,dt
\longrightarrow\Omega_{L/k}\longrightarrow0.
$$
Writing $f(t)=\sum_i a_it^i$, the image of $1$, represented by $f$, is
$$
df=\left(\sum_i b^i,da_i, f'(b),dt\right).
$$
This has the required form $( *,f'(b)dt)$.

Solved by gpt-5.6-sol high.

Solution

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

For $A\to B\to C=B/I$, define
$$
\alpha:I/I^2\to\Omega_{B/A}\otimes_BC,
\qquad [i]\mapsto di\otimes1,
$$
and
$$
\beta:\Omega_{B/A}\otimes_BC\to\Omega_{C/A},
\qquad db\otimes c\mapsto c\,d\bar b.
$$
The first map is well-defined because $d(ij)=i\,dj+j\,di$ becomes zero after tensoring with $C$ when $i,j\in I$. The second is induced by the universal derivation $B\to C\xrightarrow d\Omega_{C/A}$ and is surjective because the elements $d\bar b$ generate $\Omega_{C/A}$.

The composite is zero since $\bar i=0$. Conversely, quotienting $\Omega_{B/A}\otimes_BC$ by the $di$ for $i\in I$ imposes exactly the relations needed for the derivation of $B$ to descend to $C$. The universal property of the <Module of Kähler differentials> therefore identifies that quotient with $\Omega_{C/A}$, proving the <Conormal exact sequence for Kähler differentials>
$$
I/I^2\xrightarrow\alpha\Omega_{B/A}\otimes_BC\xrightarrow\beta\Omega_{C/A}\to0.
$$

Solved by gpt-5.6-sol high.

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

{c}

For $A\to B\to C=B/I$, there is a right-exact sequence
$$
I/I^2\longrightarrow\Omega_{B/A}\otimes_BC\longrightarrow\Omega_{C/A}\longrightarrow0,
$$
where $[i]\mapsto di\otimes1$ and $db\otimes c\mapsto c\,d\bar b$.


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 Incoming links: Conormal exact sequence for Kähler differentials