Solutions to Exercise Sheet 1
Algebraic Geometry I
Prof. Dr. Nikita Semenov
Winter Semester 2015-16
Anand Sawant, PhD
Exercise 1. Let f (x1 , . . . , xn ) ∈ C[x1 , . . . , xn ] and let M denote the largest
degree of the variables x1 , . . . , xn appearing in f (x1 , . . . , xn ). Suppose that f
vanishes on the set
ZnM +1 := {(a1 , . . . , an ) ∈ ZnM +1 | 0 ≤ ai ≤ M + 1, for every i}.
Using induction on n, show that f = 0.
Solution. Suppose n = 1. Then the hypothesis says that the polynomial f (x1 ) ∈
C[x1 ] of degree M has more than M roots. Therefore, f = 0. By induction,
assume the result for every k < n.
Now, let f (x1 , . . . , xn ) ∈ C[x1 , . . . , xn ] be a polynomial satisfying the hypotheses. Without loss of generality, assume that degree of xn in f (x1 , . . . , xn ) is M .
Suppose, if possible, that f 6= 0. We can write
f (x1 , . . . , xn ) = f0 (x1 , . . . , xn−1 ) + f1 (x1 , . . . , xn−1 )xn + · · · + fM (x1 , . . . , xn−1 )xM
n ,
where fM (x1 , . . . , xn−1 ) is a nonzero polynomial in which the degrees of variables
n−1
appearing are ≤ M . By induction, there exists (a1 , . . . , an−1 ) ∈ ZM
+1 such that
fM (a1 , . . . , an−1 ) 6= 0. Consider the polynomial in one variable xn defined by
g(xn ) := f (a1 , . . . , an−1 , xn ).
By assumption on f , we have g(an ) = 0, for 0 ≤ an ≤ M + 1. Therefore, g = 0
by the one-variable case. But this means fM (a1 , . . . , an−1 ) = 0, a contradiction.
Exercise 2. For a prime number p, let Fp denote the finite field with p elements.
It is well known (Fermat’s theorem) that the polynomial f (x) = xp − x ∈ Fp [x]
satisfies f (a) = 0, for all a ∈ Fp . Find a (nonzero) polynomial g(x, y, z) ∈
F2 [x, y, z], in which all the three variables x, y, z appear, such that g(a, b, c) = 0,
for all a, b, c ∈ F2 (that is, g(x, y, z) is a zero function F32 → F2 ).
Solution. One can take
g(x, y, z) := (x2 − x)(y 2 − y)(z 2 − z).
Another example can be found by following the hint to Exercise 3.
Exercise 3. Let R be an integral domain. Let
φ : R[x1 , . . . , xn ] → M aps(Rn → R)
be the evaluation homomorphism given by
F 7→ {(a1 , . . . , an ) 7→ F (a1 , . . . , an )} .
1
2
Show the following:
(a) if R is infinite, then φ is injective but not surjective;
(b) if R is finite, then φ is surjective but not injective.
Hint: If R = Fq (where q = pm , for a prime p) and f : Rn → R is a function,
then show that it is given by the polynomial
n
X
Y
Pf (x1 , . . . , xn ) :=
f (y1 , . . . , yn ) ·
1 − (xi − yi )q−1 .
y=(y1 ,...,yn )∈Rn
i=1
Solution. (a) We first observe that the case n = 1 is easy to see, since a polynomial with infinitely many roots has to be identically zero. Let F (x1 , . . . , xn ) ∈
R[x1 , . . . , xn ] be such that φ(F ) = 0. Let S = R[x1 , . . . , xn ]; clearly S is also an
infinite integral domain. We can view F (x1 , . . . , xn ) as a polynomial in S[xn ]:
F (x1 , . . . , xn ) = F0 (x1 , . . . , xn−1 ) + F1 (x1 , . . . , xn−1 )xn + · · · + Fr (x1 , . . . , xn−1 )xrn .
Since φ(F ) = 0 and by the one-variable case, it follows that F = 0. Thus, φ is
injective.
To see that φ is not surjective, we consider the function g : Rn → R defined by
(
1, if (x1 , . . . , xn ) = (0, . . . , 0);
g(x1 , . . . , xn ) =
0, otherwise.
Suppose, if possible, that g lies in the image of φ. Then the function h : R → R
defined by h(x) := g(x, 0, . . . , 0) is also represented by a polynomial, which has to
be nonzero. But since R is infinite, h has infinitely many roots, a contradiction.
Therefore, g cannot lie in the image of φ.
(b) If R is finite, then it has to be a field. Therefore, R = Fq , for some primepower q. The zero polynomial in R[x1 , . . . , xn ] and the polynomial
P (x1 , . . . , xn ) = (xq1 − x1 ) · (xq2 − x2 ) · · · (xqn − xn )
both map to the same element (the zero map Rn → R) under φ. Hence, φ is not
injective. To show that φ is surjective, given a map f : Rn → R, consider the
polynomial Pf (x1 , . . . , xn ) defined in the hint above. For every a ∈ Fq \ {0}, we
have aq−1 = 1. Therefore, it follows that
(
n
Y
1, if (a1 , . . . , an ) = (y1 , . . . , yn );
1 − (ai − yi )q−1 =
0, otherwise.
i=1
From this it follows that φ(Pf ) = f .
Exercise 4. For a prime number p and q = pr , let Fq denote the finite field with
q elements. The following exercise is a special case of the Chevalley-Warning
theorem.
3
Let P ∈ Fq [x1 , . . . , xn ] be such that n > deg P . Define
Z := (a1 , . . . , an ) ∈ Fnq | P (a1 , . . . , an ) = 0 .
Show that |Z| ≡ 0 mod p. In particular, if P has no constant term, then the
equation P = 0 has at least two distinct solutions.
Background. A monomial Cxα1 1 · · · xαnn ∈ Fq [x1 , . . . , xn ] is called reduced, if we
have αi < q, for every i. A polynomial is called reduced if all its nonzero monomials are reduced. (For example, the polynomial Pf in Exercise 3 is reduced.)
For every a ∈ Fq , we have
ar = aq · ar−q = a · ar−q = ar−(q−1) ,
if r ≥ q. Given a polynomial F ∈ Fq [x1 , . . . , xn ], we can associate to it a reduced
polynomial Fred by performing the above operation on variables (that is, replacing
r−(q−1)
each term of the form xri by xi
if r ≥ q) repeatedly. We will call Fred a
reduced polynomial representative of F .
The total number of reduced monomials in Fq [x1 , . . . , xn ] is q n . Hence, the
n
total number of reduced polynomials in Fq [x1 , . . . , xn ] is q q . The cardinality of
n
M aps(Fnq → Fq ) is also q q . Therefore, it follows that the restriction of the map
φ : Fq [x1 , . . . , xn ] → M aps(Fnq → Fq )
from Exercise 3 to the set of reduced polynomials is a bijection. As a consequence,
it follows that the reduced polynomial representative of a given polynomial is
uniquely determined.
Solution. Set Q(x1 , . . . , xn ) := 1 − P (x1 , . . . , xn )q−1 . Observe that
(
1, if (a1 , . . . , an ) ∈ Z;
Q(a1 , . . . , an ) =
0, otherwise.
By Exercise 3 and uniqueness of the reduced polynomial representative, the polynomial
n
X Y
QZ (x1 , . . . , xn ) :=
1 − (xi − ai )q−1
(a1 ,...,an )∈Z i=1
is the reduced polynomial representative of Q. Therefore,
deg QZ ≤ deg Q = (q − 1) deg P < (q − 1)n.
q−1
Now, the coefficient of the monomial xq−1
· · · xnq−1 in the polynomial QZ is
1 x2
n
(−1) |Z|. Suppose, if possible, that |Z| 6≡ 0 mod p. Then the term
q−1
(−1)n |Z| · xq−1
· · · xq−1
1 x2
n
in QZ is nonzero. Hence, deg QZ ≥ (q − 1)n, a contradiction.
4
Theorem (Chevalley-Warning). For a prime number p and q = pr , let Fq denote
the finite field with q elements. Let P1 , . . . , Pr ∈ Fq [x1 , . . . , xn ] be such that n >
r
P
deg Pi . Define
i=1
Z := (a1 , . . . , an ) ∈ Fnq | Pi (a1 , . . . , an ) = 0, for all 1 ≤ i ≤ r .
Then |Z| ≡ 0 mod p.
r
Q
(1 − Pi (x1 , . . . , xn )q−1 ). Observe that
i=1
(
1, if (a1 , . . . , an ) ∈ Z;
Q(a1 , . . . , an ) =
0, otherwise.
Proof. Consider Q(x1 , . . . , xn ) =
By Exercise 3 and uniqueness of the reduced polynomial representative, the polynomial
n
X Y
QZ (x1 , . . . , xn ) :=
1 − (xi − ai )q−1
(a1 ,...,an )∈Z i=1
is the reduced polynomial representative of Q. Therefore,
r
X
deg QZ ≤ deg Q = (q − 1)
deg Pi < (q − 1)n.
i=1
q−1
xq−1
1 x2
Now, the coefficient of the monomial
in the polynomial QZ is
· · · xq−1
n
n
(−1) |Z|. Suppose, if possible, that |Z| 6≡ 0 mod p. Then the term
(−1)n |Z| · x1q−1 xq−1
· · · xnq−1
2
in QZ is nonzero. Hence, deg QZ ≥ (q − 1)n, a contradiction.