Algebraic Geometry, PS 2

Bryden Cais

October 2, 2002

Exercise 1. Let X be the subvariety of 𝔸_{k}^{2} defined by the equation T_{2}^{2}−T_{1}^{2}−T_{1}^{3} and let f : 𝔸_{k}^{1} → X be the morphism given by

T_{1} ↦ T^{2} − 1, T_{2} ↦ T (T^{2} − 1).

We then have a map f^{∗} : O(X) → k[T] defined as above on the generators T_{1}, T_{2} of O(X). Observe that

g(T) = f^{∗}(p(T_{1}, T_{2})) = p(T^{2} − 1, T (T^{2} − 1)) ∈ k[T]

satisfies g(1) = g(−1) = p(0, 0) ∈ k. Conversely, suppose that g(T) satisfies g(1) = g(−1) and char(k) ≠ 2. Then the polynomial G(T) = g(T) − g(1) is divisible by T ± 1 and hence by T^{2} − 1. Now observe that T^{2n} = ((T^{2} − 1) + 1)^{n} is a polynomial in (T^{2} − 1) and T^{2n+1} = T · T^{2n} is T times a polynomial in T^{2} − 1. Thus, any polynomial may be written as a k linear combination of (T^{2} − 1)^{i} and T (T^{2} − 1)^{j} with i, j ≥ 0. It thus follows that g(T) ∈ f^{∗}(O(X)). Thus f^{∗}(O(X)) is the subring of k[T] consisting of polynomials g(T) such that g(1) = g(−1).

Now suppose that char(k) = 2. Then T^{2} − 1 = (T − 1)^{2} and if g(T) is in the image of f^{∗} then g(T) − g(1) has a double root at T = 1 (since it is divisible by (T − 1)^{2} and therefore g′(1) = 0. Conversely, suppose that g′(1) = 0 for g ∈ k[T]. Then g′(T) is a polynomial in T − 1 with no constant term. It follows that g(T) is a polynomial in T − 1 with no linear term. We can therefore write g(T) = (T − 1)^{2}h(T) + g(0). Now observe that T^{2n} = ((T − 1)^{2} + 1)^{n} is a polynomial in (T − 1)^{2} and T^{2j+1} = T ((T − 1)^{2} + 1)^{j} is a k linear combination of terms of the form (T − 1)^{2i} and T (T − 1)^{2j} for i, j ≥ 0. Therefore, h is a k linear combination of such terms so that g(T) is in the image of O(X) under f^{∗}.

When char(K) = 2, the variety X is isomorphic to the variety Y given by the equation U_{1}^{2}−U_{2}^{3} = 0. To see this, define

f^{∗} : O(Y) ⟶ O(X)

by

f^{∗}(U_{1}) = T_{2} − T_{1}, f^{∗}(U_{2}) = T_{1}.

This is clearly well defined since f(U_{1}^{2} − U_{2}^{3}) = T_{2}^{2} − T_{1}^{2} − T_{1}^{3}. Moreover, the morphism g given by

g^{∗}(T_{1}) = U_{2}, g^{∗}(T_{2}) = U_{1} + U_{2}

is inverse to f^{∗} (i.e. f^{∗}g^{∗} : O(X) → O(X) = id_{O(X)} and g^{∗}f^{∗} : O(Y) → O(Y) = id_{O(Y)}). It therefore follows that O(X), O(Y) are isomorphic and hence that X, Y are isomorphic.

Exercise 2. Let Y be the variety given by T_{1}T_{2} − 1 = 0. If Y were isomorphic to 𝔸_{k}^{1} then we would have an isomorphism

f^{∗} : k[T_{1}, T_{2}]/(T_{1}, T_{2}) ⟶ k[U].

So suppose that f^{∗}(T_{1}) = g(U) and f^{∗}(T_{2}) = h(U). Then f^{∗}(T_{1}T_{2}) = gh = f^{∗}(1) = 1. Therefore g, h are invertible elements of k[U] and hence are both elements of k. It follows that the image of f^{∗} is contained in k and hence f^{∗} is not an isomorphism of O(Y) with O(𝔸_{k}^{1}).

Exercise 3. Let f : K^{2} → K^{2} be given by (x, y) ↦ (x, xy). Then the image of f is neither open nor closed and dense in K^{2}. We first show that it is dense. Clearly, the image is

U = {(a, b) ∈ K^{2} : a ≠ 0 or a = b = 0} .

Suppose now that f ∈ k[T_{1}, T_{2}] vanishes on U. Then in particular, f vanishes on all points of the line (t, b) for any fixed b and t ≠ 0. However, on this line, f is a polynomial of the variable t alone with infinitely many roots. It follows that f ≡ 0 and thus that I(Ū) = (0), where Ū denotes the closure of U in the Zariski topology. Therefore, the image of f is dense. Since the image is dense and not all of K^{2}, it cannot be closed. Now suppose that U is open. Then the complement is closed. Let f(T_{1}, T_{2}) be any polynomial vanishing on T_{1} = 0, T_{2} ≠ 0. Then by the same argument as before, f must also vanish at (0, 0), a contradiction. Hence the image is not open.

Exercise 4. Suppose that f : 𝔸_{k}^{1} → A_{k}^{1} is an isomorphism. then we have an isomorphism

f^{∗} : k[T] ⟶ k[U].

Let g^{∗} be inverse to f^{∗}. Suppose that f^{∗}(T) = P (U) and that g^{∗}(U) = Q(T). Then Q∘P = P ∘Q = id. It follows that P, Q have. Conversely, and morphism f^{∗} : k[T] → k[U] given by f^{∗}(T) = aU + b for a, b ∈ k and a ≠ 0 is an isomorphism, with inverse given by g^{∗}(U) = a^{−1}(U − b).

Exercise 5. Recall that we have the identification

X(K) = Hom_{k} (O(X), K) .

Therefore, we want to show that

Hom_{k} (O(X), K) × Hom_{k} (O(Y), K) = Hom_{k} (O(X) ⊗ O(Y), K)

(that is, we have a bijection between the two sets). Given any homomorphism

(α, β) ∈ Hom_{k} (O(X), K) × Hom_{k} (O(Y), K)

we get a bilinear map

αβ : O(X) × O(Y) ⟶ K

given by αβ(x, y) : α(x)β(y). By the universal property of the tensor product, we get a unique linear map L : O(X) ⊗ O(Y) → K such that the usual diagram commutes. To give the correspondence in the reverse direction, let L ∈ Hom_{k} (O(X) ⊗ O(Y), K). Then by composition with the natural map N : O(X) ⊗ O(Y) → O(X) ⊗ O(Y) we get a bilinear map B : O(X) × O(Y) → K. This gives elements α = B(∗, 1) ∈ Hom_{k} (O(X), K) and β = B(1, ∗) ∈ Hom_{k} (O(Y), K). It is not difficult to see that B(x, y) = α(x)β(y) and hence, (recalling how multiplication is defined in the tensor product) that this is the reverse correspondence required. Thus we have a bijection and we are done.

Exercise 6. To show that the correspondence k → O(n, K) is an abstract affine algebraic k variety, it is enough to construct a finitely generated k algebra A such that for each K there is a bijection X(K) ↔ Hom_{k} (A, K). We therefore let

A = k[T_{11}, …, T_{nn}]/I,

where I is the ideal generated by

∑_{k=1}^{n} T_{ik}T_{jk} − δ_{ij}

for 1 ≤ i, j ≤ n, where

δ_{ij} = \begin{cases}1 & if i = j \\ 0 & otherwise \end{cases} .

Any φ ∈ Hom_{k} (A, K) is given by its values on generators, so let φ(T_{ij}) = b_{ij} ∈ K. Then the required bijection is given by

φ ↦ (b_{ij}) = M ∈ O(n, K).

It is clear that this satisfies all the required properties of an abstract affine algebraic variety.

Exercise 7. Let S_{1} = ℤ ⊂ ℂ. Consider the map f : S_{1} → S_{1} given by

f(x) = \begin{cases}x & if x ≠ 0, 1 \\ 0 & if x = 1 \\ 1 & if x = 0 \end{cases} .

Now the Zariski topology on ℤ ⊂ ℂ is just that the closed sets are the finite sets. Clearly the preimage of any closed (=finite) set of S_{1} under f is again a (finite=) closed set, so that f is continuous in the Zariski topology. On the other hand, f is not regular: if it were given by a polynomial then f − x is a polynomial map from S_{1} ⊂ ℂ to itself with infinitely many zeroes, and is therefore identically zero, which is a contradiction.

Exercise 8. Suppose that k is a field with char(k) ≠ 2. Let X be the affine algebraic K set defined by the equations T_{1}^{2} + T_{2}^{2} + T_{3}^{2} = 0, T_{1}^{2} − T_{2}^{2} − t_{3}^{2} + 1 = 0. It is not difficult to see that this system of equations is equivalent to the system

T_{1}^{2} + 1/2, T_{2}^{2} + T_{3}^{2} − 1/2.

If k does not contain a square root of −1/2 then this system has no solutions in k and hence is empty. Otherwise,

X = U_{1} ∪ U_{2},

where U_{i} is the variety given by the system of equations

(T_{1} + (−1)^{i}\sqrt{−\frac{1}{2}}, T_{2}^{2} + T_{3}^{2} − \frac{1}{2}) ,

each of which is obviously irreducible.

Exercise 9. It is not difficult to see that the system of equations

T_{2}^{2} − T_{1}T_{3} = 0, T_{1}^{2} − T_{2}^{3}

defining a variety X has one solution set

T_{1} = T_{2} = 0, T_{3} ∈ 𝔸_{k}^{1}.

Let V_{1} be the corresponding variety. Then V_{1} is clearly one irreducible component of X. Now we may suppose that T_{1}, T_{2} ≠ 0 so that our system of equations is equivalent to

T_{1}T_{2}^{2} − T_{2}^{3}T_{3} = 0, T_{1}^{2} − T_{2}^{3} = 0,

which is equivalent to

T_{1} − T_{2}T_{3} = 0, T_{1}^{2} − T_{2}^{3} = 0.

Since we are working over a field, this is equivalent to

T_{2} = T_{3}^{2}, T_{1} = T_{3}^{3}.

Thus, if we let V_{2} be the variety corresponding to

T_{2} = T_{3}^{2}, T_{1} = T_{3}^{3}, T + 3 ∈ 𝔸_{k}^{1},

we see that V_{2} is irreducible and that

X = V_{1} ∪ V_{2}.

Moreover, it is easy to see that V_{1}, V_{2} are birationally isomorphic to 𝔸_{k}^{1}. One simply takes the maps

ϕ_{1} :R(V_{1}) ⟶ k(T)
ϕ_{2} :R(V_{2}) ⟶ k(T)

given by

ϕ_{1}(T_{3}) = T, ϕ_{1}(T_{2}) = ϕ_{1}(T_{1}) = 0,
ϕ_{2}(T_{3}) = T, ϕ_{2}(T_{2}) = T^{2}, ϕ_{2}(T_{1}) = T^{3}.

It is clear that these maps are isomorphisms.

Exercise 10. Let X be the variety defined by T_{2}^{2} − T_{1}^{2} − T_{1}^{3} = 0 and let f : A_{k}^{1} → X be the morphism given by T_{1} ↦ T^{2} − 1, T_{2} ↦ T (T^{2} − 1). We claim that f is birational. We must show that it induces an isomorphism of the fields R(O(X)) and k(T). So define g : k(T) → R(O(X)) by T ↦ T_{2}/T_{1}. Then we easily check

g ∘ f(T_{1}) = (T_{2}^{2} − T_{1}^{2})/T_{1}^{2} = T_{1}^{3}/T_{1}^{2} = T_{1}
g ∘ f(T_{2}) = T_{2}(T_{2}^{2} − T_{1}^{2})/T_{1}^{3} = T_{2}T_{1}^{3}/T_{1}^{3} = T_{2}
f ∘ g(T) = f(T_{2}/T_{1}) = T (T^{2} − 1)/(T^{2} − 1) = T,

which shows that f and g are inverses and hence that f is birational.

Exercise 11. Let F (T_{1}, …, T_{n}) = G(T_{1}, , …, T_{n}) + H(T_{1}, , …, T_{n}) be irreducible, where G is homogenous of degree d − 1 and H is homogenous of degree d. Let X be the algebraic set V (F) and let

φ :R(X) ⟶ k(U_{1}, …, U_{n−1})

be given by

φ(T_{1}) = −\frac{G(1, U_{1}, …, U_{n−1})}{H(1, U_{1}, …, U_{n−1})}

and

φ(T_{i}) = −U_{i−1} \frac{G(1, U_{1}, …, U_{n−1})}{H(1, U_{1}, …, U_{n−1})}

if i > 1.

Writing λ for −G(1, U_{1}, …, U_{n−1})/H(1, U_{1}, …, U_{n−1}), we find that

φ(F) = G(λ, λU_{1}, …, λU_{n−1}) + H(λ, λU_{1}, …, λU_{n−1})
= λ^{d−1} (G(1, U_{1}, …, U_{n−1}) + λH(1, U_{1}, …, U_{n−1}))
= λ^{d−1} (G(1, U_{1}, …, U_{n−1}) − G(1, U_{1}, …, U_{n−1}))
= 0,

so that φ is well defined. Now let ψ : k(U_{1}, …, U_{n−1}) ⟶ R(X) be given by

ψ(U_{i}) = \frac{T_{i+1}}{T_{1}} .

Then it is not difficult to check that ψ ∘ φ : R(X) → R(X) and φ ∘ ψ : k(U_{1}, …, U_{n−1}) are both the respective identity mappings. It follows that φ is an isomorphism and hence that X is rational.

Exercise 12. We now show that the affine algebraic sets given by T_{1}^{3} + T_{2}^{3} − 1 = 0 and Y^{2} − X^{3}/3 + 1/12 = 0 are birationally isomorphic. Explicitly, let

f(T_{1}) = \frac{Y + 1/2}{X}
f(T_{2}) = \frac{−Y + 1/2}{X}
g(X) = \frac{1}{T_{1} + T_{2}}
g(Y) = \frac{1}{2} \frac{T_{1} − T_{2}}{T_{1} + T_{2}} .

Then one readily checks that f, g give maps

f : k[T_{1}, T_{2}]/(T_{1}^{3} + T_{2}^{3} − 1) ⟶ k[X, Y]/(Y^{2} − X^{3}/3 + 1/12)
g : k[X, Y]/(Y^{2} − X^{3}/3 + 1/12) ⟶ k[T_{1}, T_{2}]/(T_{1}^{3} + T_{2}^{3} − 1)

that are inverse to eachother (this is just a computation: we have, for example, f(T_{1}^{3} + T_{2}^{3} − 1) = (3Y^{2} + 1/4)/X^{3} − 1 = 3/X^{3}(Y^{2} − X^{3}/3 + 1/12), as required) and hence f, g are isomorphisms.