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

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

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

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

Thus for example

Next, we have

Next, we have

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

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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