CHAPTER 1 BASIC ALGEBRAIC GEOMETRY
1.6 Rational Points
In the previous section Proposition 1.5.7 states that an irreducible algebraic set V ⫅ An
K
defines a prime ideal IK(V) ≤K[x]. Conversely, if IK(V) ≤K[x] is prime for some algebraic set V ⫅An
K, then V is irreducible. We now examine a dual nature of Proposition 1.5.7.
Question 1.6.1.
(i) If ZK(P) is irreducible for some ideal P ≤K[x], may we conclude P is prime in K[x]? (ii) For a prime ideal P ≤K[x], may we concludeZK(P)is irreducible?
It turns out Question 1.6.1(i) and (ii) both have counter-examples. This section is dedicated to showing such examples. We first remind the reader of useful results and definitions.
Definition 1.6.2. A Pythagorean triple is a set of three integers x, y, z such that x2+y2 =z2; the triple is said to be primitive if gcd(x, y, z) =1.
Lemma 1.6.3. [ [1], Chapter 11, Lemma 1] Ifx, y, z is a primitive Pythagorean triple, then one of the integers xand y is even, while the other is odd.
Lemma 1.6.4. [ [1], Chapter 11, Lemma 2] If ab=cn, where gcd(a, b) = 1, then a and b are nth
powers; that is, there exist positive integers a1, b1 for which a=an1, b=bn1.
Theorem 1.6.5. [[1], Theorem 11.1] All the solutions of the Pythagorean equation
x2+y2 =z2
satisfying the conditions
gcd(x, y, z) =1 2∣x x>0 y>0 z>0
are given by the formulas
x=2st y=s2−t2 z=s2+t2 for integers s>t>0 such that gcd(s, t) =1 and s/≡t (mod 2).
We now give a counter-example to Question 1.6.1(ii). Consider the curve
f(x, y) =x2y+y2−1∈Q[x, y].
We first show f(x, y) is irreducible. Hence, f(x, y) generates a prime ideal in Q[x, y]. However, we then showZQ(x2y+y2−1) = {(0,−1),(0,1)}is a union of two points, which is a reducible affine algebraic set.
Proposition 1.6.6. The polynomial f(x, y) = (x2y+y2−1) is irreducible over
Q[x, y] and
ZQ(x
2y+y2−1) = {(0,−1),(0,1)}.
Proof. Suppose f(x, y) is reducible. That is f(x, y) = h(x, y)g(x, y) for some non-units h(x, y), g(x, y) ∈ Q[x, y]. Since f(x, y) is quadratic in the variable y, we only need consider two cases: either both h(x, y) and g(x, y) are linear iny or one is quadratic in y and the other is a polynomial in only x. For the first case we have,
f(x, y) =h(x, y)g(x, y) = (yh0(x) +h1(x))(yg0(x) +g1(x))
After distributing and equating degrees, we must have
h0(x)g0(x) =1, h1(x)g0(x) +g1(x)h0(x) =x2, and h1(x)g1(x) = −1.
The first and the last relation implyh0(x), h1(x), g0(x), g1(x) ∈Qbut this contradictsh1(x)g0(x)+ g1(x)h0(x) =x2 since deg(h1(x)g0(x) +g1(x)h0(x)) =0 /=deg(x2) =2. As for the second case and without loss of generality, let h(x)be quadratic in y. Then,
(y2h
0(x) +yh1(x) +h2(x))g(x) =f(x, y)
However, this impliesg(x) ∈Q, provingf(x, y)is irreducible by definition. Thus we may conclude f(x, y) is irreducible and generates a prime ideal in Q[x, y].
We now find the zeros for f(x, y) over Q. After solving f(x, y) =x2y+y2−1=0 for y, one gets y= ± √ x4+4−x2 2 .
We can see that y is rational if and only if √x4+4 is rational. Thus, let a b =
√
x4+4 and x= c d,
where gcd(a, b) =1 and gcd(c, d) =1. After squaring both sides, a2
b2 = c4 d4 +4,
where gcd(a, b) =1 and gcd(c, d) =1. Cross multiplying gives us
a2⋅d4= (c4+4d4) ⋅b2⇒b2∣d4⇒b∣d2.
Thus, b⋅k=d2 for some k∈
Z. We then substitute and factor:
a2⋅k2=c4+4b2⋅k2⇒k2(a2−4b2) =c4⇒k2 ∣c4⇒k ∣c2.
We have that k divides both c2 and d2; however, gcd(c, d) =1⇒gcd(c2, d2) =1. We may conclude that k =1. We now have a2 =c4+4d4 with gcd(a, b) =1, gcd(c, d) =1, and d2=b. Let
gcd(a, c) =w⇒a1⋅w=a, c1⋅w=c.
After substitution,
a21⋅w2 =c41⋅w4+4d4⇒w2(a21−c41⋅w2) =4d4⇒w2 ∣4d4.
Notice that gcd(c, d) = 1 ⇒ gcd(w, d) = 1, hence w2 ∣ 4 ⇒ w =1 or w = 2. Let us consider first the case when gcd(a, c) =1. Assume an integer solution exists for a2 =c4+4d4. Of all the integer solutions, we consider the solution with∣a∣to be the smallest. Notice thata, cmust both be odd or must both be even. In this case, a, cmust both be odd. Therefore, gcd(a,2d) =1 and gcd(a, d) =1 because gcd(a, b2) =1. We also have that gcd(c,2d) =1 since cis odd and gcd(c, d) =1. We may
now apply Theorem 1.6.5,
∃m, n∈Z, where a=m2+n2, c2=m2−n2,2d2=2m⋅n.
The last expression gives d2 =m⋅n; however, gcd(m, n) =1. Then we must have thatd2 =m2 1⋅n21, where m2
1 =m and n21=n. We also have thatc2+n2=m2.We apply Theorem 1.6.5 again because gcd(d, c) =1⇒ (c, n, m) =1. Thus,
∃s, t∈Z where s>t>0 and gcd(s, t) =1 such that m=s2+t2, n=2s⋅t, c=s2−t2.
We also know that s /≡t (mod 2). Without loss of generality, assume s is odd andt is even. Since t=2t1, n=2s⋅t=2s⋅2t1⇒ n 4 =t1⋅s. However since gcd(t1, s) =1, n=n21 ⇒ (n1 2 ) 2=t 1⋅s⇒t1=t22, s=s 2 1.
Notice now that
m=s2+t2⇒m2
1 =s41+4t21 ⇒m21 =s41+4t42. Therefore, (m1, s1, t2)is another integral solution such that
m1≤m21 =m≤m2<m2+n2 =a.
This is a contradiction to our assumption that awas the smallest solution. In the second case we have gcd(a, c) =2. This implies that
We then have
4a21=16c41+4d4⇒a21=4c41+d4.
Notice that gcd(a1, c1) = 1,gcd(c1, d) = 1, and (a1, d2) = 1 since gcd(a, b4) = 1. Once again, a1 and d must be of the same parity, and in this case they are both odd. Therefore (a1,2c2) = 1 and (d2,2c2
1) =1, which brings us back to the first case. Hence, the proof is exactly the same as before, showing that a2 =c4+4d4 has no non-trivial solutions. Therefore,√x4+4 is never rational for x /=0, so the only rational solutions to f(x, y) =x2y+y2−1 are (0,±1).
As for a counterexample over R, consider the curve h(x, y) = (x2 −1)2+y2 ∈
R[x, y]. The
irreducibility of h(x, y) is similar to that of the argument given above. As for the solutions of h(x, y) overR, we have ZR((x 2−1)2+y2) = Z R(x 2−1, y) = Z R(x 2−1) ∩ Z R(y) = {(±1,0)}
Hence, ZR((x2 −1)2 +y2) is once again reducible since it is a union of two points, although (x2−1)2+y2 generates a prime ideal.
We now consider Question 1.6.1(i). We will see in Chapter 2 that with an extra hypothesis on the prime ideal P the statement becomes valid. For the moment, we give a counter example:
Example 1.6.7. Consider the set ZR((x2+1)x,(x2+1)y) ⫅
AR. We have that ZR((x 2+1)x,(x2+1)y) = Z R(x 2+1) ∪ Z R(x, y) = ∅ ∪ ZR(y) = ZR(x, y),
which is the origin of the coordinate plane over R. The origin is a single point of the affine plane, so is irreducible. One easily notices however, that the ideal ⟨(x2+1)x,(x2+1)y)⟩is not prime in