Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 20G / b / iii

Solution

T^{3} + 3 T + 3

is Eisenstein at

3

, so part (ii) gives

3 ∤ [O_{K} : Z [α]] .

For a cubic

T^{3} + a T + b

- 4 a^{3} - 27 b^{2}

. Here

disc (1, α, α^{2}) = - 4 \cdot 3^{3} - 27 \cdot 3^{2} = - 351 = - 3^{3} \cdot 13.

- 351 = [O_{K} : Z [α]]^{2} d_{K} .

Any prime dividing the index must therefore occur to at least the second power in

351

; the only possibility is

3

, which has already been excluded. The index is one, and hence

O_{K} = Z [α] .

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