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62 Appendix to Chapter 2

(c 0 , c 1 ,... , c m+n−1 ) = (α 0 ,... ,α n−1 ,β 0 ,... ,β m−1 )[R( f, g)],

where a(x) f (x)+b(x)g(x) = c(x) = c 0 +c 1 x +···+c m+n−1 x m+n−1 . Now (2) and
(3) are equivalent because there exists a nonzero (α 0 ,... , α n−1 ,β 0 ,... ,β m−1 ) for
c 0 ,... , c m+n−1 ) = (0,... , 0) if and only if R( f, g) = det[R( f, g)] = 0. Moreover,
the second statement holds by choosing (c 0 ,... , c m+n−1 ) = (R( f, g), 0,... , 0),
and using the cofactor formulas for the inverse of the matrix [R( f, g)]. This proves
the theorem.

(4.3) Proposition. Let f = a m + a m−1 x +· · · + a 0 x m and g = b n + b n−1 x +
n
··· + b 0 x be polynomials over the ring R[y 0 ,... , y r ], where R is factorial. If a k
and b k are homogeneous of degree k, then R( f, g) ∈ R[y 0 ,... , y r ] is homogeneous
of degree mn.

Proof. We calculate R( f, g)(ty) using the homogeneous character of a 1 (y 0 ,... , y r )
and b j (y 0 ,... , y r ) and at the time multiplying each row by a suitable power of t so
that each column contains a constant power of t.

n m n m−1 n
t t a m t t a m−1 ··· t a 0 0 ··· 0
n−1 m n−1 n−1
0 t t a m ··· t t
ta 1 a 0 ··· 0

................................................
m m−1
0 0 tt a m tt
u a m−1 ··· ta 0
t R( f, g)(ty) =
m n m n−1 b n−1 ··· t b 0 0 ··· 0
m
t t b n t t
m−1 n m−1 m−1
0 t t b n ··· t tb 1 t b 0 ··· 0

................................................

n n−1
0 0 ··· tt b m tt b ··· tb 0

v
Since each column contains a fixed power of t, this determinant equals t R( f, g)(y)
where
m(m + 1) n(n + 1)
u = +
2 2
and
(m + n)(m + n + 1)
v = .
2
Since v − u = mn, we deduce that R( f, g)(ty) = t mn R( f, g)(y). Thus R( f, g)(y)
is homogeneous of degree mn and this proves the proposition.
Exercises to Appendix
1. Verify the elementary properties of valuations given in definition (1.6), and verify that
ord p is a valuation on the field F of fractions of R where p is an irreducible in a factorial
ring R.
2. In §3 carry out the details to show P → v P defining P 1 (k) → V is a bijection.