Bourbaki

Example 2

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

Algebra, chapter IV, § 6. Symmetric functions

When p=q=1p = q = 1, we have

res⁡1,1(t1X+t0,u1X+u0)=t1u0−t0u1 \operatorname{res}_{1,1}(t_1 X + t_0, u_1 X + u_0) = t_1 u_0 - t_0 u_1

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

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