Algebraic Geometry, PS 3

Bryden Cais

October 16, 2002

Exercise 1.

Exercise 2. Let A = ℤ/(6). Consider the ideal 2A. If x ∈ A \ 2A then x is odd. It follows that the ideal (x, 2A) is all of A (one way to see this is that 2 ∈ 2A and since x is odd, we have integers a, b mod 6 such that 2a + x = 1). Thus 2A is maximal. For the ideal 3A, if y ∈ A \ 3A then y = 1, 2, 4 or 5. If y = 1 or 5, y is a unit in ℤ/(6). Otherwise, we have 3 − 2 = 4 − 3 = 1 so that in any case (3A, y) is the unit ideal. Hence 3A is maximal. Since if x ∉ 2A, 3A we have x = 1 or 5 (modulo 6) we see that any such x is a unit. It follows that 2A, 3A ore the only maximal ideals of A. Now consider the localization A_{2A}. This consists of all fractions a/b with a, b ∈ A and b ∉ 2A under the equivalence a/b ∼ c/d iff there exists s ∈ A \ 2A with s(ad − bc) = 0 in A. Set s = 3. Then we have a/b ∼ c/d iff ad ≡ bc mod 2. Since b, d are odd, we see that

0 ∼ \frac{2}{∗} ∼ \frac{4}{∗}

and

\frac{1}{∗} ∼ \frac{3}{∗} ∼ \frac{5}{∗},

where ∗ denotes any element of A \ 2A. Thus, A_{2A} has two elements and is clearly isomorphic to ℤ/(2) since the numerator of a/b + a/b is even and hence the sum is equivalent to 0. A similar analysis applies to the localization A_{3A}. In this case, we choose s = 2 and find that a/b ∼ c/d iff

ad ≡ cd mod (3).

Now observe that −a, b − a, 2b − a are three consecutive terms in arithmetic progression so that one of them is divisible by 3. It follows that a/b ∼ 0, 1 or 2. Since 3 ∼ 0 we see that A_{3A} ≃ ℤ/(3). Now let P be a projective A module of rank 1. Then we know that P_{2A}, P_{3A} are free A_{2A}, A_{3A} (resp.) modules of rank 1, so pick x ∈ P_{2A}, y ∈ P_{3A} such that

P_{2A} = A_{2A}x
P_{3A} = A_{3A}y.

By clearing denominators, we may assume that x, y ∈ P. We now claim that P = A(x + y) and that x + y is not a zero divisor. It then follows that P ≃ A. But how to show this claim??

Exercise 3. Set

A = ℂ[T_{1}, T_{2}]/(T_{1}^{2} − T_{2}(T_{2} − 1)(T_{2} − 2))

and let t_{1}, t_{2} be the images of T_{1}, T_{2} under the quotient map. Let I = (t_{1}, t_{2}). We show that for any maximal ideal 𝔐 ⊂ A the localization I_{𝔐} is a free A_{𝔐} module of rank 1. First, the maximal ideals of A are in bijective correspondence to the maximal ideals of ℂ[T_{1}, T_{2}] containing the principal ideal T_{1}^{2} − T_{2}(T_{2} − 1)(T_{2} − 2). Since every maximal ideal of ℂ[T_{1}, T_{2}] has the form (T_{1} − α, T_{2} − β), we see that the maximal ideals of A are precisely the ideals

𝔐_{α,β} = (t_{1} − α, t_{2} − β)

where (T_{1}, T_{2}) = (α, β) is a root of T_{1}^{2} − T_{2}(T_{2} − 1)(T_{2} − 2). Suppose (α, β) ≠ (0, 0). Then there exists some f ∈ I with f ∉ 𝔐_{α,β} (since I = 𝔐_{0,0}, 𝔐_{α,β} are distinct maximal ideals). It follows that I_{𝔐_{α,β}} contains 1 and is therefore equal to A_{𝔐_{α,β}}, that is, is free of rank 1. Now consider I_{I}. Since I is the ideal of all f ∈ A with no constant term, we find that any f ∈ A_{I} with a nonzero constant term is invertible. In particular, (t_{2} − 1)(t_{2} − 2) is invertible. It follows that

\frac{t_{1}}{(t_{2} − 1)(t_{2} − 2)}t_{1} = t_{2}

with \frac{t_{1}}{(t_{2}−1)(t_{2}−2)} ∈ A_{I}. From this, we see that I_{I} = (t_{1}, t_{2})I_{=}A_{I}t_{1} + A_{I}t_{2} = A_{I}t_{1} since A_{I}t_{2} ⊂ A_{I}t_{1}. Now the elliptic curve T_{1}^{2} − T_{2}(T_{2} − 1)(T_{2} − 2) has distinct roots and is therefore irreducible. It follows that A is an integral domain and hence that t_{1} in particular is not a zero divisor. Therefore, I_{I} = A_{I}t_{1} is free of rank one (sending t_{1} ↦ 1 gives a surjective homomorphism from I_{I} to A_{I}; injectivity follows from the fact that t_{1} is not a zero divisor). Therefore, since the localization of I at every maximal ideal of A is free of rank 1, we see that I is a projective A module of rank 1. Now we must show that I is not free, or equivalently that I is not a principal ideal.

Exercise 4. Suppose that I is a homogenous ideal of k[T] such that m_{+}^{s} ⊂ I for some s. Let F ∈ k[T]. Then for any G ∈ m_{+}^{s} we have GF ∈ m_{+}^{s} ⊂ I so that F ∈ I^{sat}. Thus, I^{sat} = k[T]. For the second part of the question, I am convinced that there is a typo and that it cannot be “Proposition 1.” But then what proposition is it supposed to be?

Exercise 5. Let I = (T_{0}^{2}, T_{0}T_{1}) ⊂ k[T_{0}, T_{1}]. Then since m_{+} = (T_{0}, T_{1}) we have T_{0}G ∈ I for any G ∈ m_{+}. Thus, T_{0} ∈ I^{sat} so that I ⊂ (T_{0}) ⊂ I^{sat}. We claim that I^{sat} = (T_{0}). For suppose that F G ∈ (T_{0}) for all G ∈ m_{+}^{s} for some s. In particular, T_{1}^{s}F ∈ (T_{0}) so that F is divisible by T_{0} and hence F ∈ (T_{0}). It follows that (T_{0}) is saturated and that I^{sat} = (T_{0}).

Exercise 6. We first homogenize the equations to find that the projective closure is given by the equations

Z_{2}Z_{0} − Z_{1}^{2}, Z_{3}Z_{0}^{2} − Z_{1}^{3}.

The solution set can be parameterized. Indeed, if Z_{0} = 0 then we must have Z_{1} = 0 also, whence we obtain the points

(0, 0, 1, t)

and

(0, 0, 0, 1)

for any t ∈ k. Otherwise, Z_{0} ≠ 0 whence we obtain the parameterization

(1, t, t^{2}, t^{3})

for any t ∈ k. Thus, the projective closure is

{(1, t, t^{2}, t^{3}) : t ∈ k} ∪ {(0, 0, 1, t) : t ∈ k} ∪ (0, 0, 0, 1).

Exercise 7. If F is a homogenous polynomail free of multiple factors then (F) = (F)^{sat}. We know that the set of solutions of F in ℙ_{k}^{n}(K) is irreducible if and only if the corresponding coordinate ring is an integral domain. But the coordinate ring is

A = k[T]/(F)^{sat} = k[T]/(F),

and is therefore an integral domain if and only f F is irreducible.

Exercise 8. Clearly, P has only simple roots in k̄ iff P, P′ have no common roots. By exercise 9 this is the case iff Res(P′, P) ≠ 0 (since the resultant is a product over all pairwise differences of the roots of P and P′). Let us now compute Res(P, P′) in the case where P is of degree less than 4.

1. deg(P) = 2 : Set P (Z) = a_{0}Z^{2} + a_{1}Z + a_{2} so that P′(Z) = 2a_{0}Z + a_{1}. We then have

Res(P, P′) = |\begin{matrix}a_{0} & a_{1} & a_{2} \\ 2a_{0} & a_{1} & 0 \\ 0 & 2a_{0} & a_{1} \end{matrix}|
= −a_{0}(a_{1}^{2} − 4a_{2}a_{0}).

2. deg(P) = 3 : Set P (Z) = a_{0}Z^{3} + a_{1}Z^{2} + a_{2}Z + a_{3} so that P′(Z) = 3a_{0}Z^{2} + 2a_{1}Z + a_{2}. We then have

Res(P, P′) = |\begin{matrix}a_{0} & a_{1} & a_{2} & a_{3} & 0 \\ 0 & a_{0} & a_{1} & a_{2} & a_{3} \\ 3a_{0} & 2a_{1} & a_{2} & 0 & 0 \\ 0 & 3a_{0} & 2a_{1} & a_{2} & 0 \\ 0 & 0 & 3a_{0} & 2a_{1} & a_{2} \end{matrix}|

= 27a_{3}^{2}a_{0}^{3} + (−18a_{3}a_{2}a_{1} + 4a_{2}^{3})a_{0}^{2} + (4a_{3}a_{1}^{3} − a_{2}^{2}a_{1}^{2})a_{0}.

3. deg(P) = 4 : Set P (Z) = a_{0}Z^{4} + a_{1}Z^{3} + a_{2}Z^{2} + a_{3}Z + a_{4} so that P′(Z) = 4a_{0}Z^{3} + 3a_{1}Z^{2} + 2a_{2}Z + a_{3}. Proceeding as above (and using PARI), we have

Res(P, P′) = |\begin{matrix}a_{0} & a_{1} & a_{2} & a_{3} & a_{4} & 0 & 0 \\ 0 & a_{0} & a_{1} & a_{2} & a_{3} & a_{4} & 0 \\ 0 & 0 & a_{0} & a_{1} & a_{2} & a_{3} & a_{4} \\ 4a_{0} & 3a_{1} & 2a_{2} & a_{3} & 0 & 0 & 0 \\ 0 & 4a_{0} & 3a_{1} & 2a_{2} & a_{3} & 0 & 0 \\ 0 & 0 & 4a_{0} & 3a_{1} & 2a_{2} & a_{3} & 0 \\ 0 & 0 & 0 & 4a_{0} & 3a_{1} & 2a_{2} & a_{3} \end{matrix}|

= 256a_{4}^{3}a_{0}^{4} + (−192a_{4}^{2}a_{3}a_{1} + (−128a_{4}^{2}a_{2}^{2} + 144a_{4}a_{3}^{2}a_{2} − 27a_{3}^{4}))a_{0}^{3} + ((144a_{4}^{2}a_{2} − 6a_{4}a_{3}^{2})a_{1}^{2}

+ (−80a_{4}a_{3}a_{2}^{2} + 18a_{3}^{3}a_{2})a_{1} + (16a_{4}a_{2}^{4} − 4a_{3}^{2}a_{2}^{3}))a_{0}^{2} + (−27a_{4}^{2}a_{1}^{4} + (18a_{4}a_{3}a_{2} − 4a_{3}^{3})a_{1}^{3}

+ (−4a_{4}a_{2}^{3} + a_{3}^{2}a_{2}^{2})a_{1}^{2})a_{0}.

Exercise 9. Let

P = a_{0}(Z − α_{1}) ⋯ (Z − α_{n}) = a_{0}Z^{n} + a_{1}Z^{n−1} + ⋯ + a_{n},
Q = b_{0}(Z − β_{1}) ⋯ (Z − β_{m})= b_{0}Z^{m} + b_{1}Z^{m−1} + ⋯ + b_{m}.

Let V (X_{1}, X_{2}, …, X_{r}) denote the matrix

(\begin{matrix}1 & X_{1} & ⋯ & X_{1}^{r} \\ 1 & X_{2} & ⋯ & X_{2}^{r} \\ ⋮ & ⋮ & ⋯ & ⋮ \\ 1 & X_{r} & ⋯ & X_{r}^{r} \end{matrix}) ,

set M = V (β_{m}, ⋯, β_{1}, α_{n}, ⋯, α_{1}) and

R = (\begin{matrix}a_{0} & & & b_{0} \\ a_{1} & ... & & ⋮ & ... \\ ⋮ & ... & a_{0} & b_{m−1} & ... & b_{0} \\ a_{n} & ... & a_{1} & b_{m} & ... & ⋮ \\ & ... & ⋮ & & ... & b_{m−1} \\ & & a_{n} & & & b_{m} \end{matrix}) .

We first show that det(V (X_{1}, …, X_{r})) = ∏_{1≤i<j≤r}(X_{i} − X_{j}). To see this, for 1 ≤ j ≤ r − 1, subtract X_{1} times the j^{th} column from the (j + 1)^{st} column of the matrix V (X_{1}, …, X_{r}). These operations do not alter the determinant, and we find that

det(V (X_{1}, …, X_{r})) = ∏_{2≤i≤r}(X_{i} − X_{1}) det(V (X_{2}, …, X_{r})).

Induction then gives the proposed formula. We now compute the determinant of MR in two different ways. It is not difficult to see that

(\begin{matrix}1 & β_{m} & ⋯ & β_{m}^{m+n−1} \\ ⋮ & ⋮ & ⋯ & ⋮ \\ 1 & β_{1} & ⋯ & β_{1}^{m+n−1} \\ 1 & α_{n} & ⋯ & α_{n}^{m+n−1} \\ ⋮ & ⋮ & ⋯ & ⋮ \\ 1 & α_{1} & ⋯ & α_{1}^{m+n−1} \end{matrix}) (\begin{matrix}a_{0} & & & b_{0} \\ a_{1} & ... & & ⋮ & ... \\ ⋮ & ... & a_{0} & b_{m−1} & ... & b_{0} \\ a_{n} & ... & a_{1} & b_{m} & ... & ⋮ \\ & ... & ⋮ & & ... & b_{m−1} \\ & & a_{n} & & & b_{m} \end{matrix})

= (\begin{matrix}P (β_{m}) & ⋯ & P (β_{m})β_{m}^{m−1} \\ ⋮ & ⋯ & ⋮ \\ P (β_{1}) & ⋯ & P (β_{1})β_{1}^{m−1} \\ & & & Q(α_{n}) & ⋯ & Q(α_{n})α_{n}^{n−1} \\ & & & ⋮ & ⋯ & ⋮ \\ & & & Q(α_{1}) & ⋯ & Q(α_{1})α_{1}^{n−1} \end{matrix}) ,

and hence that

det(MR) =

det(V (β_{m}, …, β_{1})V (α_{n}, …, α_{1})) ∏_{1≤j≤m} P (β_{j}) ∏_{1≤i≤n} Q(α_{i}).

On the other hand,

det(MR) = det(M) det(R^{T})
= det(V (β_{m}, …, β_{1}, α_{n}, …, α_{1})) Res(P, Q),

where R^{T} denotes the transpose of R. Now, using our formula for the determinant of V (X_{1}, …, X_{r}), we see that

det(V (β_{m}, …, β_{1}, α_{n}, …, α_{1}))
= det(V (β_{m}, …, β_{1})) det(V (α_{n}, …, α_{1})) ∏_{1≤i≤n} ∏_{1≤j≤m}(α_{i} − β_{j})

= det(V (β_{m}, …, β_{1})) det(V (α_{n}, …, α_{1}))\frac{1}{b_{0}^{n}} ∏_{1≤i≤n} Q(α_{i}).

It thus follows that

Res(P, Q) = b_{0}^{n} P (β_{1}) ⋯ P (β_{m}).

Notice that Res(P, Q) = (−1)^{mn} Res(Q, P) since to swap the roles of the a_{i}’s and the b_{j}’s in the defining determinant of the resultant requires a bubble sort involving mn flips of columns. Thus, we also have

Res(P, Q) = (−1)^{mn}a_{0}^{m} Q(α_{1}) ⋯ Q(α_{n}).

Exercise 10. Let X be the cubic curve defined projectively by T_{1}^{2}T_{0} −T_{2}^{2} −T_{0}^{3} = 0. Except for the point at infinity, (0, 1, 0), all points on X(ℂ) lie in 𝔸_{ℂ}^{1} (where we take the copy of 𝔸_{ℂ}^{1} ⊂ ℙ_{ℂ}^{1} given by T_{0} = 1. We thus proceed to determine explicit formulae for the group law on the affine curve X′ given by y^{2} = x^{3} + 1. Let P_{0} = (x_{0}, y_{0}), P_{1} = (x_{1}, y_{1}) be any two distinct points on X′. Then line through these points intersects X at a third point whose x coordinate is the third root of

(\frac{y_{1} − y_{0}}{x_{1} − x_{0}} (x − x_{0}) + y_{0})^{2} = x^{3} + 1.

Call this root x_{2}. Then since x_{0}, x_{1} are also roots, we know that

(\frac{y_{1} − y_{0}}{x_{1} − x_{0}} )^{2} = x_{0} + x_{1} + x_{2}.

Similarly, y_{2} is given by

y_{2} = \frac{y_{1} − y_{0}}{x_{1} − x_{0}} (x_{2} − x_{0}) + y_{0}.

Taking the point (0, 1, 0) as the zero element, we find that the sum P_{0} + P_{1} on X is the point P_{2} = (x_{2}, −y_{2}). Observe that

x_{2} = (\frac{y_{1} − y_{0}}{x_{1} − x_{0}} )^{2} − x_{0} − x_{1}

= (\frac{x_{1}^{3} − x_{0}^{3}}{(x_{1} − x_{0})(y_{1} + y_{0})})^{2} − x_{0} − x_{1}

= (\frac{x_{0}^{2} + x_{0}x_{1} + x_{1}^{2}}{y_{1} + y_{0}} )^{2} − x_{0} − x_{1},

so that if P_{0} = P_{1} we have 2P_{0} = (x_{2}, −y_{2}), where

x_{2} = (\frac{3x_{0}^{2}}{2y_{0}} )^{2} − 2x_{0},

and

y_{2} = (\frac{3x_{0}^{2}}{2y_{0}} ) (x_{2} − x_{0}) + y_{0}.

All of these formulae make sense except when x_{0} = x_{1} but P_{0} ≠ P_{1} (in which case the sum P_{0} + P_{1} is seen to be the point at infinity) and when y_{0} = 0, when x_{0} is a cube root of −1 whence the sum 2P_{0} is again the point at infinity.

Exercise 11. Using our formulae above, we can determine explicit formulae for 3P when P is a point on X. Let P = (x, y). It is not difficult to see from the formulae that the points of order 3 in X(C) correspond to those points P with 2P = −P, i.e. those points P with

(\frac{3x^{2}}{2y} )^{2} − 3x = 0.

This gives

x^{4} + 4x = 0,

or x = 0, −\sqrt[3]{4}, ρ\sqrt[3]{4}, −ρ\sqrt[3]{4}, where ρ = e^{πi/3} is a cube root of −1. For each of these four values of x there are two corresponding values of y, giving 8 points of order exactly 3. Since the point at infinity O also satisfies 3O = O, we see that there are nine points in X(C) satisfying 3P = O. They are

O = (0, 1, 0)
±P_{1} = (1, ±1, 0)
±P_{2} = (1, ±i\sqrt{3}, −\sqrt[3]{4})
±P_{3} = (1, ±i\sqrt{3}, ρ\sqrt[3]{4})
±P_{4} = (1, ±i\sqrt{3}, −ρ\sqrt[3]{4}).

Now let us determine the inflection points of X in ℂ. A point (T_{0}, T_{1}, T_{2}) is an inflection point of X if and only if

|\begin{matrix}−6T_{0} & 2T_{1} & 0 \\ 2T_{1} & 2T_{0} & 0 \\ 0 & 0 & 6T_{2} \end{matrix}| = −24T_{2}(3T_{0}^{2} + T_{1}^{2})= 0.

These are the points on X with T_{2} = 0 or T_{1} = ±i\sqrt{3}T_{0}. It is easy to see that these are precisely the 9 points listed above. Of course, a more intuitive way to think about this problem is that the points of order 3 are those points P such that the tangent line at P has third point of intersection P (this is just the requirement that 2P = −P). But this is equivalent to requiring that the tangent line at P intersect X with multiplicity 3 (the maximum multiplicity, according to Bezout’s theorem), i.e. that P be an inflection point of X.

Now it can be seen that any line containing two of the eight points of order 3 contains a third, and hence that this is an example of a configuration of 8 points not satisfying the assumption of Lemma 1.

Exercise 12. Let X be the curve given by T_{1}^{2}T_{0} − T_{2}^{3}. For any field K, define the map f : X(K) \ (1, 0, 0) → K by

(T_{0}, T_{1}, T_{2}) ↦ \frac{T_{2}}{T_{1}} .

Observe that this is a well defined map since T_{1} = 0 iff T_{2} = 0 and we are excluding the point (1, 0, 0); in addition, the value of f is independent of the homogenous coordinates chosen. We now show that f induces an isomorphism of the group structure on X(K) with the additive group K^{+}. Recall that X is rational, that is, we have the parameterization (1, t^{3}, t^{2}) (excluding the point at infinity). Thus, let P = (1, α^{3}, α^{2}), Q = (1, β^{3}, β^{2}) be any two points of X(K). Let us determine P +Q. Passing to affine coordinates by setting y = T_{1}/T_{0}, x = T_{2}/T_{0} and writing y = mx+b for the line (in affine coordinates) passing through P, Q, we see that α^{2}, β^{2} are both roots of (mx+b)^{2} = x^{3} and that the third root, γ is the x coordinate of the sum P + Q. Then we have α^{2} + β^{2} + γ = m^{2}. By now this process is standard, and we find that the x coordinate of P + Q is given by:

(\frac{α^{3} − β^{3}}{α^{2} − β^{2}} )^{2} − (α^{2} + β^{2}) = \frac{α^{6} + β^{6} − 2α^{3}β^{3} − (α^{2} − β^{2})(α^{4} − β^{4})}{(α^{2} − β^{2})^{2}}

= \frac{α^{2}β^{2}(α − β)^{2}}{(α^{2} − β^{2})^{2}}

= (\frac{αβ}{α + β} )^{2} .

It then follows that the y coordinate of P + Q is given by

(\frac{αβ}{α + β} )^{3} .

Therefore, in homogenous coordinates, we have

P + Q = (1, (\frac{αβ}{α + β} )^{3} , (\frac{αβ}{α + β} )^{2})

if α ≠ −β (that is, if P ≠ −Q) and P + Q = (0, 1, 0) if P = −Q. Now we have

f(0, 1, 0) = 0

and

f(P + Q) = f (1, (\frac{αβ}{α + β} )^{3} , (\frac{αβ}{α + β} )^{2})

= \begin{matrix}(\frac{αβ}{α+β} ) & 2 \\ (\frac{αβ}{α+β} ) & 3 \end{matrix}

= \frac{α + β}{αβ}

= f(P) + f(Q),

so that f is indeed a group homomorphism. We have already seen that f is bijective, so we are done.

Exercise 13. This is similar to the above. Let X be given in projective coordinates by T_{1}^{2}T_{0} − T_{2}^{2}(T_{2} + T_{0}). For any field K, define the map f : X(K) \ (1, 0, 0) → K by

(T_{0}, T_{1}, T_{2}) ↦ \frac{T_{1} − T_{2}}{T_{1} + T_{2}} .

As before, it is not difficult to see that this is a well defined map and that f(0, 1, 0) = 1 ∈ K^{∗}. We claim that f induces an isomorphism of the group structure on X(K) with the multiplicative group K^{∗}. As above, X is parameterized by (1, t(t^{2} − 1), t^{2} − 1) (excluding the point at infinity). Thus, let P = (1, α(α^{2} − 1), α^{2} − 1), Q = (1, β(β^{2} − 1), β^{2} − 1) be any two points of X(K). Observe first that

f(P) = \frac{α − 1}{α + 1}
f(Q) = \frac{β − 1}{β + 1}.

Let us determine P + Q. Proceeding as above, set y = T_{1}/T_{0}, x = T_{2}/T_{0} and write y = mx + b for the line (in affine coordinates) passing through P, Q. Then we see that α^{2} − 1, β^{2} − 1 are both roots of (mx + b)^{2} = x^{3} + x^{2} and that the third root, γ is the x coordinate of the sum P + Q. Then we have α^{2} + β^{2} + γ − 2 = m^{2} − 1. By now this process is standard, and we find that the x coordinate of P + Q is given by:

(\frac{α(α^{2} − 1) − β(β^{2} − 1)}{α^{2} − β^{2}})^{2} − (α^{2} + β^{2} − 1) = (\frac{α^{2} + β^{2} + αβ − 1}{α + β})^{2} − (α^{2} + β^{2} − 1)

= \frac{α^{2}β^{2} − α^{2} − β^{2} + 1}{(α + β)^{2}}

= (\frac{αβ + 1}{α + β})^{2} − 1.

We therefore see that

P + Q = (1, \frac{αβ + 1}{α + β} ((\frac{αβ + 1}{α + β})^{2} − 1) , (\frac{αβ + 1}{α + β})^{2} − 1) ,

so that

f(P + Q) = \frac{αβ − α − β + 1}{αβ + α + β + 1} = \frac{(α − 1)(β − 1)}{(α + 1)(β + 1)} = f(P)f(Q).

Thus f is a homomorphism. The fact that f is bijective follows from the rationality of our curve and the definition of f.