Example 3

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

Algebra, chapter IV, § 6. Symmetric functions

Let us prove the formula

\operatorname{res}_{2,2}(aX^2 + bX + c, a'X^2 + b'X + c') = (ac' - ca')^2 + (bc' - cb')(ba' - ab')

Arguing as for Prop. 9 (IV, p. 79) we see that it is enough to prove the formula when $a$ is invertible. There exists then a decomposition of the form

aX^2 + bX + c = a(X - x)(X - y)

in $B[X]$ , where $B$ is an appropriate ring containing $A$ as subring. By Cor. 1 above, the required resultant is equal to

R = a^2(a'x^2 + b'x + c')(a'y^2 + b'y + c') .

Now we have

ax + ay = -b , \quad axy = c

by (43), whence $(ax)^2 + (ay)^2 = b^2 - 2ac$ .

By (44) we have

R = {a'}^2(axy)^2 + a{b'}^2(axy) + a^2{c'}^2 + a'b'(axy)(ax + ay) +

+ a'c'((ax)^2 + (ay)^2) + ab'c'(ax + ay)

= {a'}^2c^2 + a{b'}^2c + a^2{c'}^2 - a'b'cb + a'c'(b^2 - 2ac) - ab'c'b

= (ac' - ca')^2 + (ab' - a'b)(b'c - c'b) ,

from which the desired result follows.

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

Cited by

Nothing.

Cites

Cor. 1