Bourbaki

Example 4

Tag 06FN, permanent. Cite it as [Bourbaki, Tag 06FN].

Algebra, chapter IV, § 6. Symmetric functions

Consider the polynomial

F=A0X3+3A1X2+3A2X+A, F = A_0X^3 + 3A_1X^2 + 3A_2X + A,
with coefficients in Q[A0,A1,A2,A3]Q[A_0, A_1, A_2, A_3]. We have
DF=3(A0X2+2A1X+A2), DF = 3(A_0X^2 + 2A_1X + A_2) ,
F−1/3X⋅DF=A1X2+2A2X+A3. F - 1/3X \cdot DF = A_1X^2 + 2A_2X + A_3 .
By the formulae (54) (IV, p. 84) and (29) (IV, p. 77) we have
A0⋅dis3(F)=−res2,3(DF,F)=−res2,3(DF,F−1/3X⋅DF) A_0 \cdot \mathrm{dis}_3(F) = - \mathrm{res}_{2,3}(DF, F) = - \mathrm{res}_{2,3}(DF, F - 1/3X \cdot DF)
Applying Cor. 2 of IV, p. 80 we finally obtain
dis3(F)=−27res2,2(A0X2+2A1X+A2, A1X2+2A2X+A3). \mathrm{dis}_3(F) = - 27 \mathrm{res}_{2,2}(A_0X^2 + 2A_1X + A_2,\ A_1X^2 + 2A_2X + A_3) .
By Example 3 of IV, p. 80 we thus have
Δ~(A0,3A1,3A2,A3)=−27(A0A3−A1A2)2−108(A1A3−A22)(A12−A0A2). \tilde{\Delta}(A_0, 3A_1, 3A_2, A_3) = - 27 (A_0A_3 - A_1A_2)^2 - 108 (A_1A_3 - A_2^2)(A_1^2 - A_0A_2) .
After some calculations we find, that if f=a0X3+a1X2+a2X+a3f = a_0X^3 + a_1X^2 + a_2X + a_3, we have
dis3(f)=a12a22+18a0a1a2a3−4a13a3−4a0a23−27a02a32. \mathrm{dis}_3(f) = a_1^2a_2^2 + 18a_0a_1a_2a_3 - 4a_1^3a_3 - 4a_0a_2^3 - 27a_0^2a_3^2 .
In particular, we have
dis(X3+pX+q)=−(4p3+27q2). \mathrm{dis}(X^3 + pX + q) = - (4p^3 + 27q^2) .

Exercises

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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Transcribed from the printed edition, 2003, Springer. Not translated.