— 1) Take K=R, A=C. Taking into account Alg., chap. VIII, §12, n° 2, prop. 4, we obtain modC(z)=∣z∣2 for all z∈C. (2)
2) Take K=R, and for A the quaternion field H (GT, VIII, §1, No. 4). Consider the following elements of M2(C):
X1=(0ii0)X2=(01−10)X3=(i00−i)
which, together with I2, form a basis of M2(C) over C. One verifies easily that
X12=X22=X32=−I2,X1X2=−X2X1=X3,
X2X3=−X3X2=X1,X3X1=−X1X3=X2.
The mapping a+bi+cj+dk↦aI2+bX1+cX2+dX3 may therefore be extended to a C-isomorphism of the algebra C⊗RH onto the algebra M2(C). Since [H:R]=4, C is a neutralizing field of H (Alg., chap. VIII, §10, n° 5), and the reduced norm of q=a+bi+cj+dk∈H is
Nrd(q)=det(aI2+bX1+cX2+dX3)
=det(a+idc+ib−c+iba−id)=a2+b2+c2+d2=∥q∥2.
By Alg., chap. VIII, §12, n° 3, prop. 8, one has
NH/R(q)=(NrdH/R(q))2=∥q∥4.
This stated, Prop. 17 shows that
modH(q)=∥q∥4.
A deeper study of the structure of locally compact fields will be made in CA, VI, §9.
(2) This also follows from Cor. 1 of Prop. 15 and the fact that left-multiplication by z=a+ib has matrix (ab−ba) with respect to the canonical basis 1,i of C over R.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.