Modular Function of Weight k > 2: The Eisenstein Series

Let $k > 2$ be an integer, and consider the following function defined on the upper half plane of the complex numbers: $$G_k(τ) = \sum_{(c,d)∈ℤ^2-(0,0)} \dfrac{1}{(cτ+d)^k}$$ We shall prove that the series above converges absolutely for all $τ$ in the upper half plane, and as such the summands may be permuted freely. Then, we shall prove that the series converges uniformly on compact subsets, and thus defines a holomorphic function on the upper half plane. Finally, we shall prove that $G_k(τ)$ is weakly modular of weight $k$, and also $G_k(τ)$ is bounded as $\text{Im}(τ)→∞$, and so defines a modular function of weight $k$.

Proving that the series absolutely converges for any given $τ$ is more or less the same as showing the the same for the Weierstrass series with respect to the lattice $[τ,1]$; we outline the idea. One may contain the lattice points in disjoint balls of uniform size $δ > 0$. Since each annulus of radii $R-1, R+2$ has area $O(R)$, the convergence of $\sum \dfrac{1}{|cτ+d|^k}$ follows by comparison with the convergent series $\sum_{R=1} \dfrac{1}{R^{k-1}}$.

For positive real numbers $A,B$, let $Ω = \{τ~|~|\text{Re}(τ)| ≤ A, |\text{Im}(τ)| ≥ B\}$. We show $G_k(τ)$ converges uniformly on $Ω$, and hence on compact subsets in general, thereby proving $G_k(τ)$ defines a holomorphic function. We do this by showing that $\sum_{(c,d)∈ℤ^2-(0,0)} σ_{c,d}$ converges on $Ω$, where $$σ_{c,d} = \sup_{τ∈Ω} \dfrac{1}{|cτ+d|^k}$$ Note that this supremum is approached by bringing $cτ$ as close to $-d$ as possible; when $|d|$ is large enough, say $|d| > N$, this supremum is in fact achieved by letting $τ$ be one of $\pm A + Bi$. Thus, we know that $\sum σ_{c,d}$ converges when only large enough $d$ are summed, by considering $G_k(τ)$ converges absolutely for each $τ$. As for $\sum_c σ_{c,d}$ for each fixed $d$ with $|d| ≤ N$, we see $σ_{c,d} ≤ \dfrac{1}{|cB|^k} = \dfrac{1}{|B|^k} \dfrac{1}{|c|^k}$ generally when $c ≠ 0$, so the aforementioned sum converges. We conclude $G_k(τ)$ converges uniformly on $Ω$ and is holomorphic on the upper half plane. This calculation also shows that $G_k(τ)$ is bounded on $Ω$; the fact that $G_k(τ)$ is bounded as $\text{Im}(τ)→∞$ will therefore follow once we show it is weakly modular, and thereby $ℤ$-periodic.

Consider the action of a special fractional linear transformation on the lattice $L=[τ,1]$; for generic integers $α,β$ we see $$ατ+β↦α \dfrac{aτ+b}{cτ+d} + β = \dfrac{(aα+cβ)τ + (bα + dβ)}{cτ+d}$$ Therefore, noting that $\begin{bmatrix}a & b \\ c & d \end{bmatrix}^T$ permutes $ℤ^2$, we see the special linear transformation induces a bijection $L→\dfrac{L}{cτ+d}$. We conclude that $G_k(\dfrac{aτ+b}{cτ+d}) = (cτ+d)^k G_k(τ)$, so $G_k(τ)$ is weakly modular, hence fully modular by the above comment.$~\square$

Note: There is an error in the above proof, namely "when $|d|$ is large enough, say $|d| > N$, this supremum is in fact achieved by letting $τ$ be one of $\pm A + Bi$." One can instead show that there exists a constant $C > 0$ such that $|γτ+δ| > C \cdot \sup \{|γ|,|δ|\}$ for all $τ∈Ω$ and real $γ, δ$ not both zero, implying $$\sum σ_{c,d} ≤ \dfrac{1}{C} \sum \dfrac{1}{(\sup \{|c|,|d|\})^k}$$ is finite.

Galois Correspondence of Deck Transformations (5.2)

Let $p : Y → X$ be the universal covering, with $\text{Deck}(Y/X)$ its set of deck transformations. If $G ≤ \text{Deck}(Y/X)$ is a subgroup, $π : Y → Z$ is the appropriate quotient of $Y$ by equivalence modulo $G$, and $q : Z → X$ factors $p$ through the quotient, show that $q$ is a covering map that is Galois precisely when $G$ is a normal subgroup of $\text{Deck}(Y/X)$, in which case $$\text{Deck}(Z/X) ≅ \text{Deck}(Y/X)/G$$
Proof: We show $q$ is a covering map. Throughout let bars denote passage into the quotient. Let $x∈X$ be given. Since $p$ is a covering map, let $U$ be a neighborhood of $x$ with $$p^{-1}(U) = ∪_{j∈J} V_j$$ for disjoint open subsets $V_j$ of $Y$ with each $p|V_j → U$ a homeomomorphism. If $J$ is empty, then we are done. Otherwise, let $V$ be some given $V_j$, and define $V_g = g(V)$ for $g∈\text{Deck}(Y/X)$. By virtue of the universality of $p$, we have $$p^{-1}(U) = ∪_{g∈\text{Deck}(Y/X)} V_g$$ for disjoint open subsets $V_g$ of $Y$ with each $p|V_g → U$ a homeomomorphism. If $\{Gh_i\}_{i∈I}$ is a choice of representatives for the right cosets of $G$, and $W_i = \overline{∪_{g∈Gh_i}V_g}$, we claim $$q^{-1}(U) = ∪_{i∈I} W_i$$ for disjoint open subsets $W_i$ of $W$ with each $q|W_i → U$ a homeomomorphism. Indeed, each $W_i$ is open as it is the image of a $π$-saturated open set in $Y$, and are disjoint since the $V_g$ are disjoint. As well, it is clear that each $π|h_i(V)→W_i$ is a homeomorphism with inverse $φ_i$, hence each $q|W_i→U$ is a homeomorphism as $q = p∘φ_i$ on $W_i$. Thus $q$ is a covering map.

Suppose $G$ is a normal subgroup of deck transformations. Then for $\overline{y_0},\overline{y_1}∈Z$ lying in a common fiber over $q$, we see $y_0,y_1∈Y$ lie in a common fiber over $p$, so let $σ∈\text{Deck}(Y/X)$ be such that $σ(y_0) = y_1$. We define $\overline{σ}(\overline{y}) = \overline{σ(y)}$. This is seen to be independent of the choice of representative in $y$, since if $g∈G$, then $$\overline{σ(g(y))} = \overline{σ(g(σ^{-1}(σ(y))))} = \overline{g'(σ(y))} = \overline{σ(y)}$$ for some other $g'∈G$ since $G$ is normal. $\overline{σ}$ is seen to be a continuous fiber-preserving endomorphism of $Z$ with inverse $\overline{σ^{-1}}$, hence $\overline{σ}∈\text{Deck}(Z/X)$. Therefore $q : Z → X$ is Galois, and since $\overline{σ}\overline{ψ} = \overline{σψ}$ with $\overline{σ} = 1$ iff $σ∈G$, we have $$\text{Deck}(Z/X) ≅ \text{Deck}(Y/X)/G$$ Conversely, suppose $q : Z → X$ is Galois, with $σ∈\text{Deck}(Y/X)$, $g∈G$ given. We will show $h = σgσ^{-1}∈G$. Choose some $y_0∈Y$, and let $σ(y_0)=y_1$. Let $ψ∈\text{Deck}(Z/X)$ be such that $ψ(\overline{y_0}) = \overline{y_1}$. Then both $π∘σ$ and $ψ∘π$ are continuous fiber-preserving maps from $Y$ to $Z$ (relative to $p$ and $q$) sending $y_0$ to $\overline{y_1}$, thus they are equal on all of $Y$. Thus, $$\overline{σ(y)} = ψ(\overline{y})$$ Substituting $y$ with $g(y)$, we have $$\overline{σ(g(y))} = \overline{h(σ(y))} = ψ(\overline{g(y)}) = ψ(\overline{y}) = \overline{σ(y)}$$ so that, since $σ$ is a bijection and $y$ is arbitrary, $$\overline{y} = \overline{h(y)}$$ for all $y∈Y$. But this is to say $h$ preserves the fibers of $π$; necessarily $h∈G$, which concludes the proof that $G$ is normal.$~\square$

Algebraic Properties of Character Values (18.3.11-12,14,16-24)

Dummit and Foote Abstract Algebra, section 18, exercises 11, 12, 14-24:

Let $G$ be a finite group of order $n$.

11. Let $χ$ be an irreducible character of $G$. Prove that for every $z∈Z(G)$ we have $χ(z) = ζχ(1)$ for some root of unity $ζ$.

12. Let $ψ$ be the character of some representation $φ$ of $G$, and let $g∈G$. Prove (a) $ψ(g) = ψ(1)$ if and only if $g∈\text{ker }φ$, and (b) $|ψ(g)| = ψ(1)$ implies $g ∈ Z(G)$ if $φ$ is faithful.

In the exercises that follow, let $F = \overline{ℚ}$ be the field of algebraic numbers.

14. Prove that a representation $φ$ of $G$ over $F$ is irreducible over $F$ if and only if it is irreducible when considered as a representation over $ℂ$. Deduce that the group characters of $G$ when considered over $F$ are the same as when considered over $ℂ$, and that every complex representation of $G$ is equivalent to a representation over $F$.

16-17. Let $σ$ be an automorphism of the field extension $F/ℚ$. If $φ$ is a representation of $G$ over $F$ with character $ψ$, let $φ^σ$ be the mapping given by applying the automorphism $σ$ to each of the entries of $φ$. Prove that $φ^σ$ is a representation with character $ψ^σ = σ∘ψ$.

18. If $ψ$ is again a character of $G$, let $ℚ(ψ)$ be the (finite) extension of $ℚ$ by each value $ψ(g)∈F$. Prove that $ℚ(ψ)⊆ℚ(ζ_n)$, where $ζ_n$ is a primitive $n^\text{th}$ root of unity. Deduce that $ℚ(ψ)$ is a Galois extension of $ℚ$ with abelian Galois group.

19-24. For each $a$ relatively prime to $n$, let $σ_a ∈ \text{Gal}(ℚ(ζ)/ℚ) ≅ (ℤ/nℤ)^×$ be the automorphism given by $ζ_n ↦ ζ_n^a$. Show that $ψ^{σ_a}(g) = ψ(g^a)$. Prove that every character of a group has all integer values if and only if for all $g∈G$, one has $g$ is conjugate to $g^k$ for all $k$ relatively prime to $|g|$.

Proof: (11) Let $φ$ be the corresponding representation of $G$ on $V$. Since $z$ commutes with every element of $G$, it commutes with every element of $ℂG$. Therefore, multiplication by $z$ is a $ℂG$-module endomorphism on $V$. Diagonalize $V$ with respect to $z$, so that $φ(z)$ is given by a matrix with roots of unity on the diagonal and zeros elsewhere. Let $ζ$ be one such root. By Schur's lemma, $z - ζ$ is either an automorphism or an annihilator of $V$. Since $z - ζ$ annihilates at least one basis element of $V$ (that from which $ζ$ was chosen from $φ(z)$), we see it must be the latter, so that $φ(z) = ζI$ and $χ(z) = ζχ(1)$.

(12)(a-b) If $g ∈ \text{ker }φ$, then clearly $ψ(g) = ψ(1)$. Conversely, if $ψ(g) = ψ(1)$, then after diagonalization with respect to $g$, we may consider $g$ as a matrix with $ψ(1)$ roots of unity $ζ_i$ on the diagonal. If $ψ(g) = \sum ζ_i = ψ(1)$, then also $|ψ(g)| = |\sum ζ_i| = \sum |ζ_i| = ψ(1)$, so by the triangle inequality each $ζ_i$ is equal. From here we see (b), whereby $φ(g)$ commutes with every matrix in $φ(G)⊆\text{GL}(ℂ)$, then $φ(g)∈Z(φ(G))$ implying $g∈Z(G)$ if $φ$ is faithful. Given the stronger condition of (a) that $ψ(g) = ψ(1)$, we in fact see $ζ_i = 1$, so $g∈\text{ker }φ$.

(14) Note that the computation of a character norm is independent of the field over which the matrices are considered, and irreducibility of a character/representation is equivalent to the norm being $1$. Thus, a representation is irreducible over $F$ if and only if it is irreducible over $ℂ$. Furthermore, there are $r$ irreducible characters over $F$ that are equivalent to their corresponding complex characters, where $r$ is the number of conjugacy classes of $G$; this is also independent of the field considered, so this is the full set of irreducible complex characters of $G$. Thus, the $F$-characters and complex characters are the same. Finally, since every representation over $ℂ$ will entail a character identical to that of some representation over $F$, this implies every representation over $ℂ$ is equivalent to one over $F$.

(16-17) The action of $σ$ on $\text{GL}(F)$ is in fact an automorphism, so the composition $φ^σ$ is a homomorphism from $G$ into $\text{GL}(V)$ corresponding to which we have clearly have the character $ψ^σ$.

(18) Examining Proposition 14 more closely, we see in fact that $ψ(g)∈ℚ(ζ_k)$ for each $g∈G$ with order $|g|=k$. Since necessarily $k$ will divide $n$, and $ℚ(ζ_k)⊆ℚ(ζ_n)$, we have $ℚ(ψ)⊆ℚ(ζ_n)$, in fact $ℚ(ψ)=ℚ(ζ_e)$ for the exponent $e$ of $φ(G)$. Thus it is a Galois extension with abelian Galois group.

(19-24) If we diagonalize with respect to $g$, we see that $σ_a(ζ_n^k) = (ζ_n^k)^a$ for all $k$, hence $\text{tr }φ^{σ_a}(g) = \text{tr }φ(g)^a = \text{tr }φ(g^a) = ψ(g^a)$. If $g$ is conjugate to $g^a$ for all $a$ relatively prime to $|g|$, then $ψ^{σ_a} = ψ$ as characters for all $a$ relatively prime to $n$, so $ψ(g)$ is within the fixed field of the Galois extension $ℚ(ζ_n)/ℚ$, that is, $ℚ$. By Proposition 14, character values are algebraic integers, so $ψ$ actually takes values in $ℤ$.

Conversely, suppose $g$ isn't conjugate to $g^a$ for some $a$ relatively prime to $|g|$. Let $n = p_1^{α_1}p_2^{α_2}...p_k^{α_k}$ and $|g| = p_1^{β_1}p_2^{β_2}...p_k^{β_k}$ be factorizations, necessarily with $β_i ≤ α_i$. Reordering the primes if necessary, let $β_i = 0$ for $i > m$. By the Chinese Remainder Theorem, let $b$ be such that $$b≡a \mod p_i^{α_i}$$ for $i≤m$, and $$b≡1 \mod p_i^{α_i}$$ for $i > m$. Then $b ≡ a \mod |g|$ (so $g^a = g^b$), and $b$ is relatively prime to $n$. Then since the irreducible characters of $G$ form a basis for the space of class functions on $G$, there is some irreducible character $χ$ such that $χ(g)≠χ(g^b)=χ^{σ_b}(g)=σ_b(χ(g))$. Thus $χ(g)$ is not within the fixed field of $σ_b$, hence not an integer.$~\square$

Reorganization of Chapter 4.3

William Fulton Algebraic Curves, chapter 4, section 3

Throughout, let $φ=φ_{n+1}$, and let $τ : k[x_1,...,x_{n+1}] → k[x_1,...,x_n]$ be the surjective ring morphism given by evaluation $x_{n+1}↦1$. Given $V⊆\mathbb{A}^n$, the projective closure $V^*$ of $V$ is defined to be the smallest projective algebraic set containing $φ(V)⊆\mathbb{P}^n$ (that is, $V(I(φ(V)))$). Given $V⊆\mathbb{P}^n$, the dropdown $V_*$ of $V$ is defined to be $φ^{-1}(V)⊆\mathbb{A}^n$. Note that the dropdown of a projective algebraic set is an affine algebraic set, with defining ideal $τ(I(V))$.

Lemma: For finite unions, $(∪V_i)_* = ∪{V_i}_*$ and $(∪V_i)^* = ∪V_i^*$. Proof: The first equality is clear, as unions commute with function pullbacks. As for the second, note that $∪V_i^*$ is the smallest algebraic set containing each $V_i^*$ (as it is itself an algebraic set). Therefore, since any algebraic set containing $φ(V_i)$ contains $V_i^*$, it is the smallest algebraic set containing each $φ(V_i)$. That is, it is the smallest algebraic set containing $∪φ(V_i)=φ(∪V_i)$—and this is exactly the definition of $(∪V_i)^*$.

Proposition 3: Let $V, W$ denote algebraic sets throughout.

1. If $V⊆\mathbb{A}^n$, then $φ(V) = V^* ∩ U_{n+1}$ and $(V^*)_* = V$.
2. Projective closures and dropdowns are inclusion-preserving operations.
3. If $V⊆\mathbb{A}^n$ is a variety, then $V^*⊆\mathbb{P}^n$ is a variety.
4. If $V=∪V_i$ is the irreducible decomposition of $V$ in $\mathbb{A}^n$, then $V^* = ∪V_i^*$ is the irreducible decomposition of $V^*$ in $\mathbb{P}^n$.
5. If $V⊂\mathbb{A}^n$ is nonempty, then $V^*$ does not lie in or contain $H_∞$.
6. If $V⊆\mathbb{P}^n$ and no component of $V$ lies in or contains $H_∞$, then $V_*⊂\mathbb{A}^n$ and $(V_*)^* = V$.

Proof: (1) Suppose $[a,1]∈V^* \setminus φ(V)$. Then $a∉V$, so $f(a) = f^*([a,1]) ≠ 0$ for some $f∈I(V)$. Thus, since $f^*$ vanishes on $φ(V)$, we would have $V⊆V(f^*)∩V^*⊂V^*$, a contradiction. Now we compute $$(V^*)_* = φ^{-1}(V^*) = φ^{-1}(V^*∩U_{n+1}) = φ^{-1}(φ(V)) = V$$ (2) This is obvious.

(3) Suppose $V^* = W_1∪W_2$. Then since $$(W_1∪W_2)_* = {W_1}_*∪{W_2}_* = (V^*)_* = V$$ is a variety, we may suppose ${W_1}_* = V$ and $W_1$ contains $φ(V)$, hence contains $V^*$.

(4) By the Lemma and (3), it follows $∪V_i^*$ is an irreducible decomposition of $V^*$. To see that it is not redundant, suppose $V_i^* ⊆ V_j^*$: then by (1) and (2), we would see $V_i ⊆ V_j$.

(5) Note $V^* ⊈ H_∞$ by (1). Now, choose some nonzero $f∈I(V)$. Then we note $f^*∈I(V^*)$, the leading form $f'$ of $f$ is not zero at some point $a∈\mathbb{A}^n$, and $f^*([a,0]) = f'(a) ≠ 0$, so that $[a,0]∉V^*$ and $V^*$ does not contain $H_∞$.

(6) We may assume $V$ is nonempty. Then $V$ itself does not lie in or contain $H_∞$, for this latter is irreducible (with defining ideal $(x_{n+1})$). If we show $U_{n+1}^* = \mathbb{P}^n$, then it will follow $V_*$ must be properly contained in $\mathbb{A}^n$. As such, suppose $f$ is a form vanishing on $U_{n+1}$. Then $τ(f)$ is necessarily $0$, so write $f = (x_{n+1}-1)·h$. But the expression on the right is not a form unless $h=f=0$. Thus $I(U_{n+1}^*) = (0)$ and $U_{n+1}^* = \mathbb{P}^n$.

Note that $V_i = ({V_i}_*)^* ∪ (H_∞∩V_i)$ for each component $V_i$ of $V$, so since $V_i$ does not lie in $H_∞$, we must have $V_i = ({V_i}_*)^*$. Now we compute $$(V_*)^* = ∪({V_i}_*)^* = ∪V_i = V~~\square$$

Plane Curve Multiplicities Under Polynomial Mappings (3.1.8)

William Fulton Algebraic Curves, chapter 3, section 1, exercise 8:

Let $k$ be an algebraically closed field, $F∈k[x,y]$, and $T : \mathbb{A}^2 → \mathbb{A}^2$ be a polynomial map with $T(P)=Q$. Show that $m_P F^T ≥ m_Q F$, with equality achieved if the Jacobian of $T$, defined as the matrix $(dT_i / dx_j (Q))$ where $T = (T_1,T_2)$, is invertible. Show that the converse is false in considering $T = (x^2,y)$ and $F=y-x^2$, $P=Q=(0,0)$.

Proof: Suppose $P = Q = (0,0)$. Then necessarily $T_1$ and $T_2$ are of the form $$T_1 = ... + ax+by$$ $$T_2 = ... + cx+dy$$ with no constant terms. Write $F = F_m + ... + F_n$ in terms of its form decomposition; then we see $F^T = F_m^T + ... + F_n^T$, and for each $i$, $F_i^T$ contributes to the sum forms of degree $≥i$. This shows the nonzero forms in $F^T$ are all of degree $≥m$, so that $m_{(0,0)} F^T ≥ m_{(0,0)} F$. Now consider the general case for points $P,Q$. Let $S_1, S_2$ be the translations mapping the origin to $P,Q$ respectively. Then $$m_Q F^T = m_{(0,0)} F^{T∘S_2} = m_{(0,0)} F^{S_1∘S_1^{-1}∘T∘S_2}$$ $$= m_{(0,0)} (F^{S_1})^{S_1^{-1}∘T∘S_2} ≥ m_{(0,0)} F^{S_1} = m_P F$$ Now to consider the properties of the Jacobian: Again suppose $P = Q = (0,0)$, and factor $F_m$ into a product of linear factors. Since $F_i^T$ for $i > m$ only contribute forms of degree $> m$, it will suffice to show $F_m^T$ contributes a nonzero form of degree $m$ when the Jacobian is invertible. When the Jacobian is invertible, that is to say its rows are linearly independent, so that if $ex+fy$ is a nonzero linear term, then $T(ex+fy) = eT_1(x) + fT_2(y)$ will have a nonzero form of degree $1$ in its decomposition. Since the smallest nonzero form of a product is the product of the smallest nonzero forms, we see that $m_{(0,0)} F^T = m_{(0,0)} F$. The argument for the general case of $P, Q$ is exactly as before, together with the observations that (1) the Jacobian of a composition of polynomial maps is the same as the matrix product of the Jacobian of polynomial maps at the appropriate points, and (2) that the Jacobians of translations are the identity. As for the demonstration that the converse is not in general true, we note that in the example given, $m_{(0,0)} F = 1 = m_{(0,0)} y - x^4 = m_{(0,0)} F^T$ despite the fact that the Jacobian of $T$ at $(0,0)$ is the noninvertible $$J_{(0,0)}T = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}~~\square$$

Finite Dimensional Rings Containing Algebraically Closed Fields (2.9.47)

William Fulton Algebraic Curves, chapter 2, section 9, exercise 47:

Let $R$ be a commutative ring sharing an identity with $k⊆R$, an algebraically closed field. Show that $R$ is isomorphic to a finite direct product of local rings.

Proof: Let $r_1,...,r_n$ be a basis for $R$ over $k$. Then $k[x_1,...,x_n] / I ≅ R$ for some ideal $I$ via the natural identification $x_i ↦ r_i$. We see that distinct (surjective) morphisms $R → k$ preserving $k$ thus correspond to maximal ideals in $k[x_1,...,x_n]$ containing $I$, which themselves correspond to points in the locus of $I$. It follows that if we can prove there are only finitely many such morphisms, then Proposition 6 of this same section will imply the result. To this end, we show that the image of $r_i$ in $k$ under a morphism $φ$ may be among only finitely many elements; indeed, this follows by considering that $1,r_i,r_i^2,...$ are linearly dependent over $k$, and thus $φ(r_i)$ must be a root of the polynomial of $r_i$ over $k$ induced by this dependence.$~\square$

Multiple Representations are Required to Identify the Pole Set of a Rational Function (2.4.20)

William Fulton Algebraic Curves, chapter 2, section 4, exercise 20:

Let $k$ be algebraically closed, and consider the rational function $f = x/y = z/w$ over the algebraic variety $V = \mathcal{Z}(xw-yz)$ in $\mathbb{A}^4(k)$. Show that the pole set of $f$ is precisely $\{(t_1,0,t_2,0)~|~t_1,t_2∈k\}$, and that it is impossible to write $f = a/b$ for $b(p)≠0$ for all $p∈V$ outside the pole set of $f$.

Proof: First of all, $y$ and $w$ are nonzero on points of $V$, so they are nonzero in $\Gamma(V)$. As well, $\mathcal{I}(\mathcal{Z}(xw-yz)) = (xw-yz)$ since $xw-yz$ is irreducible. Let $J_f = \{G ∈ \Gamma(V)~|~Gf ∈ \Gamma(V)\}$. It is clear that $J_f$ is an ideal in $\Gamma(V)$, and that $y,w∈ J_f$. Suppose $gf = u ∈ \Gamma(V)$ for some $g∈k[x,z]$; then we may write $$gx - uy = 0 \mod (xw-yz)$$ $$gx = 0 \mod (xw,y)$$ from which we conclude $g = 0$, so that $J_f = (y,w)$. This allows us to conclude that the denominator of $f$ in any way that we write it will always have $\{(t_1,0,t_2,0)\}$ as zeros, so that this is precisely the pole set of $f$. Suppose we could write $f = a / b$ with $b$ having zeros precisely at the pole set of $f$; this would be to say $$\mathcal{Z}(b,xw-yz) = \mathcal{Z}(y,w,xw-yz) = \mathcal(Z)(y,w)$$ $$\mathcal{Z}(b,xw-yz) ∩ \mathcal{Z}(x,z) = \mathcal{Z}(x,y,w,z)$$ $$\mathcal{Z}(b,x,z) = \{(0,0,0,0)\}$$ Since $b∈(y,w)⊆\Gamma(V)$, it will thus suffice to show that given $b ∈ (y,w) ⊆ k[y,w]$, we may find a zero besides $(0,0)$. Indeed, this is equivalent to showing $$\mathcal{Z}(b) = \mathcal{Z}(y,w)$$ is impossible. Indeed, every nonconstant polynomial in multiple variables over an algebraically closed field has infinitely many zeros (1.2.14), while $\mathcal{Z}(y,w)$ is just a single point.$~\square$