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48 2. Plane Algebraic Curves

(2.2) Remark. The above discussion extends to hypersurfaces H f in P n defined
by homogeneous polynomials f (y 0 ... , y n ) of degree d in k[y 0 ,... , y n ]. For an
extension field K of k the set H f (K) consists of all y 0 : ··· : y n in P n (K) such that
f (y 0 ,... , y n ) = 0.
Observe that the hyperplane H 0 given by y 0 = 0, called the hyperplane at
infinity, is contained in H f if and only if y 0 divides f . This is equivalent to
f (0, y 1 ,... , y n ) = 0 or to the relation deg(g)< deg( f ) where g(x 1 ,... , x n ) =
f (1, x 1 ,... , x n ). The two polynomials f and g have the same factorization into ir-
a
reducible factors except for the factor y , where a = deg( f )− deg(g). The affine
0
n
hypersurface H g (K) aff is the set of all (x 1 ,... , x n ) in K with g(x 1 ,... , x n ) = 0.
n
Observe that H g (K) aff = H f (K) ∩ K . For some questions, as in the next theorem,
it is more convenient to use the affine picture.

Note that if f divides f in k[y 0 ,... , y n ], then H f (K) ⊂ H f (K) and H g (K) aff

⊂ H g (K) aff with the above notations for the relation between f and g and between

f and g .

(2.3) Theorem. Let H f and H f be two hypersurfaces defined over k in P n , where

f is irreducible. If for some algebraically closed extension field L of K we have the
inclusion H f (L) ⊂ H f (L), then f divides f , and so H f (K) ⊂ H f (K) for all

extension fields K of k.
Proof. We can consider the inclusion of hypersurfaces in some affine space where
aff
H g (L) aff ⊂ H g (L) . In affine coordinates we can write

d
g(x 1 ,... , x n−1 , x n ) = a 0 (x 1 ,... , x n−1 ) + ··· + a d (x 1 ,... , x n−1 )x ,
n
where d ≥ 0and a i (x 1 ,... , x n−1 ) ∈ k[x 1 ,... , x n−1 ].

If f = 0, then clearly f divides f . Now consider the case where g ∈ k[x 1 ,... ,

x n−1 ]with g nonzero. Since L is infinite, there exists a point (x 1 ,... , n n−1 ) in L n−1
with g (x)a 0 (x) nonzero for x = (x 1 ,... , x n−1 ). Since L is algebraically closed, the

polynomial equation g(x 1 ,... , x n−1 , t) has a root t = x n in L, and thus the point
aff
(x 1 ,... , x n−1 , x n ) lies in H g (L) aff − H f (L) , which is a contradiction. Therefore,

if g is in k[x 1 ,... , x n−1 ], then f is zero.
e
Now suppose that g = b 0 + b 1 x n +· · · + b e x , where e > 0and b i ∈ k[x 1 ,
n
... , x n−1 ]. By (4.2) in the Appendix, there is a relation R(x 1 ,... , x n−1 ) = ug+vg
n
in k[x 1 ,... , x n−1 ][x n ]. If g(x 1 ,... , x n ) = 0 for (x 1 ,... , x n ) ∈ L , then
g (x 1 ,... , x n ) = 0

by hypothesis and thus by the resultant formula, R(x 1 ,... , x n−1 ) = 0. In other
aff
words, as affine hypersurfaces in n-dimensional space H f (L) aff ⊂ H R (L) .Since
R is a polynomial in x 1 ,... , x n−1 , it follows that R = 0 by the special case treated
in the previous paragraph. By (4.2) in the Appendix again, g and g must have a

common prime factor which must be g itself since g is irreducible. Thus g divides g

and, hence, f divides f . This proves the theorem.