Example 50.19.4. Let $A$ be a ring. Let $f = x^ d + \sum _{0 \leq i < d} a_{d - i} x^ i \in A[x]$. Let $B = A[x]/(f)$. By Proposition 50.19.3 we have a morphism of complexes

$\Theta _{B/A} : \Omega ^\bullet _ B \longrightarrow \Omega ^\bullet _ A$

In particular, if $t \in B$ denotes the image of $x \in A[x]$ we can consider the elements

$\Theta _{B/A}(t^ i\text{d}t) \in \Omega ^1_ A,\quad i = 0, \ldots , d - 1$

What are these elements? By the same principle as used in the proof of Proposition 50.19.3 it suffices to compute this in the universal case, i.e., when $A = \mathbf{Z}[a_1, \ldots , a_ d]$ or even when $A$ is replaced by the fraction field $\mathbf{Q}(a_1, \ldots , a_ d)$. Writing symbolically

$f = \prod \nolimits _{i = 1, \ldots , d} (x - \alpha _ i)$

we see that over $\mathbf{Q}(\alpha _1, \ldots , \alpha _ d)$ the algebra $B$ becomes split:

$\mathbf{Q}(a_0, \ldots , a_{d - 1})[x]/(f) \longrightarrow \prod \nolimits _{i = 1, \ldots , d} \mathbf{Q}(\alpha _1, \ldots , \alpha _ d), \quad t \longmapsto (\alpha _1, \ldots , \alpha _ d)$

Thus for example

$\Theta (\text{d}t) = \sum \text{d} \alpha _ i = - \text{d}a_1$

Next, we have

$\Theta (t\text{d}t) = \sum \alpha _ i \text{d}\alpha _ i = a_1 \text{d} a_1 - \text{d}a_2$

Next, we have

$\Theta (t^2\text{d}t) = \sum \alpha _ i^2 \text{d}\alpha _ i = - a_1^2 \text{d} a_1 + a_1 \text{d}a_2 + a_2 \text{d}a_1 - \text{d}a_3$

(modulo calculation error), and so on. This suggests that if $f(x) = x^ d - a$ then

$\Theta _{B/A}(t^ i\text{d}t) = \left\{ \begin{matrix} 0 & \text{if} & i = 0, \ldots , d - 2 \\ \text{d}a & \text{if} & i = d - 1 \end{matrix} \right.$

in $\Omega _ A$. This is true for in this particular case one can do the calculation for the extension $\mathbf{Q}(a)[x]/(x^ d - a)$ to verify this directly.

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