The fourth power mean of Dirichlet $L$-functions in $\mathbb{F}_q [T]$

We prove results on moments of $L$-functions in the function field setting, where the moment averages are taken over primitive characters of modulus $R$, where $R$ is a polynomial in $\mathbb{F}_q [T]$. We consider the behaviour as $\textrm{deg} R \rightarrow \infty$ and the cardinality of the finite field is fixed. Specifically, we obtain an exact formula for the second moment provided that $R$ is square-full, and an asymptotic formula for the fourth moment for any $R$. The fourth moment result is a function field analogue of Heath-Brown's result in the number field setting, which was subsequently improved by Soundararajan. Both the second and fourth moment results extend work done by Tamam in the function field setting who focused on the case where $R$ is prime.

The study of moments of families L-functions is a central theme in analytic number theory. These moments are connected to the famous Lindelöf hypothesis for such L-functions and have many applications in analytic number theory. It is a very challenging problem to establish asymptotic formulas for higher moments of families of L-functions and until now we only have asymptotic formulas for the first few moments of any given family of L-functions. However, we do have precise conjectures for higher moments of families of L-functions due to the work of many mathematicians (see for example [CFK + 05] and [DGH03]). In this paper the focus is on the moments of Dirichlet L-functions associated to primitive Dirichlet characters.
In 1981, Heath-Brown [HB81] proved that * χ(mod q) where, for all positive integers q, * χ(mod q) represents a summation over all primitive Dirichlet characters of modulus q, φ * (q) is the number of primitive characters of modulus q, and ω(q) is the number of distinct prime divisors of q and L(s, χ) is the associated Dirichlet L-function.
In the equation above (1), in order to ensure that the error term is of lower order than the main term, we must restrict q to ω(q) ≤ log log q − 7 log log log q log 2 .
Soundararajan [Sou07] addressed this by proving that * χ(mod q) Here, the error terms are of lower order than the main term without the need to have any restriction on q.
In a breakthrough paper, Young [You11] obtained explicit lower order terms for the case where q is an odd prime and was able to establish the full polynomial expansion for the fourth moment of the associated Dirichlet L-functions. In other words, he proved that where the constants c i are computable. The error term was subsequently improved by Blomer et al. [BFK + 17] who proved that In the function field setting Tamam [Tam14] Here, Q is an irreducible, monic polynomial in F q [T ] with F q a finite field with q elements; χ 0 is the trivial character (in this case, of modulus Q); and, for non-trivial characters of modulus Q, where M is the set of monic polynomials F q [T ].
In this paper we prove the function field analogue of Soundararajan's result, which is also an extension of Tamam's fourth moment result. In order to accomplish this we prove, along the way, a function field analogue of a special case of Shiu's Brun-Titchmarsh theorem for multiplicative functions [Shi80]. We also obtain an exact formula for the second moment, similar to Tamam's second moment result; our result holds for square-full moduli, R, whereas Tamam's result holds for irreducible moduli, Q.
Acknowledgment: The first author is grateful to the Leverhulme Trust (RPG-2017-320) for the support through the research project grant "Moments of L-functions in Function Fields and Random Matrix Theory". The second author is grateful for an EPSRC Standard Research Studentship (DTP).

Statement of Results
Let q ∈ N be a prime-power. We denote the finite field of order q by F q . We denote the ring of polynomials over the finite field F q by A := F q [T ]. Unless otherwise stated, for a subset S ⊂ A we define S n := {A ∈ S : deg A = n}. We identify A 0 with F q . Also, if we have some non-negative real number x, then range deg A ≤ x is not taken to include the polynomial A = 0.
The norm of A ∈ A\{0} is defined by |A| := q deg A , and for the zero polynomial we define |0| := 0.
We denote the set of monic polynomials in A by M. For a ∈ (F q ) * we denote the set of polynomials, with leading coefficient equal to a, by aM. Because A is an integral domain, an element is prime if and only if it is irreducible. We denote the set of prime monic polynomials in A by P, and all references to primes (or irreducibles) in the function field setting are taken as being monic primes. Also, when indexing, the upper-case letter P always refers to a monic prime. Furthermore, if we range over polynomials E that divide some polynomial F , then these E are taken to be the monic divisors only. Due to point 2, we can view a character χ of modulus R as a function on A\RA. This makes expressions such as χ(A −1 ) well-defined for A ∈ A\RA * .
We can deduce that χ(1) = 1 and |χ(A)| = 1 when (A, R) = 1. We say that χ is the trivial character of modulus R if χ(A) = 1 when (A, R) = 1, and this is denoted by χ 0 . Otherwise, we say that χ is non-trivial. Also, there is only one character of modulus 1 and it simply maps all A ∈ A to 1.
It can easily be seen that the set of characters of a fixed modulus R forms an abelian group under multiplication. The identity element is χ 0 . The inverse of χ is χ, which is defined by χ(A) = χ(A) for all A ∈ A. It can be shown that the number of characters of modulus R is φ(R).
A character χ is said to be even if χ(a) = 1 for all a ∈ F q . Otherwise, we say that it is odd. The set of even characters of modulus R is a subgroup of the set of all characters of modulus R. It can be shown that there are 1 q−1 φ(R) elements in this group.
Definition 2.2 (Primitive Character). Let R ∈ M, S | R and χ be a character of modulus R. We say that S is an induced modulus of χ if there exists a character χ 1 of modulus S such that χ is said to be primitive if there is no induced modulus of strictly smaller norm than R. Otherwise, χ is said to be non-primitive. φ * (R) denotes the number of primitive characters of modulus R.
We note that all trivial characters of some modulus R = 1 are non-primitive as they are induced by the character of modulus 1. We also note that if R is prime, then the only non-primitive character of modulus R is the trivial character of modulus R. We denote a sum over primitive characters of modulus R by the standard notation * χ(mod R) . Definition 2.3 (Dirichlet L-functions). Let χ be a Dirichlet character. The associated L-function, L(s, χ), is defined for Re(s) > 1 by This has an analytic continuation to either C or C\{1} depending on the character.
In this paper, we will prove the following two main results. The first one is an exact formula for the second moment of Dirichlet L-functions in function fields.
The next result, is an asymptotic formula for the fourth moment of Dirichlet L-functions associated to primitive Dirichlet characters.

Function Field Background
We provide some definitions and results relating to function fields that are needed in this paper. Many of these results are well known and so we do not provide a proof. Some proofs can be found in Rosen's book [Ros02], particularly chapter 4.
Definition 3.2 (ω Function). For all R ∈ A\{0} we define ω(R) to be the number of distinct prime factors of R.
Definition 3.3 (Ω Function). For all R ∈ A\{0} we define Ω(R) to be the total number of prime factors of R (i.e. counting multiplicity).
It is not hard to show that Definition 3.5. For all R ∈ A with deg R ≥ 1 we define p − (R) to be the largest positive integer such that if P | R then deg P ≥ p − (R). Similarly, we define p + (R) to be the smallest positive integer such that if P | R then deg P ≤ p + (R).
Proposition 3.6 (Orthogonality Relations). Let R ∈ M. Then, Proposition 3.7. Let R ∈ M and let A, B ∈ A. Then, * Proof. This follows easily from Proposition 3.7 when we take A, B = 1.
For a character χ we will, on occasion, write the associated L-function as where we define for all non-negative integers n and all characters χ.
Suppose χ is the character of modulus 1 and Re(s) > 1. Then, L(s, χ) is simply the zeta-function for the ring A. That is, We note further that The far-RHS provides a meromorphic extension for ζ A to C with a simple pole at 1. The following Euler product formula will also be useful for Re(s) > 1. Now suppose that χ 0 is the trivial character of some modulus R and Re(s) > 1. It can be shown that So, again, the far-RHS provides a meromorphic extension for L(s, χ 0 ) to C with a simple pole at 1.
Finally, suppose that χ is a non-trivial character of modulus R and Re(s) > 1. It can be shown that This is just a finite polynomial in q −s , and so it provides a holomorphic extension for L(s, χ) to C. Proposition 3.9 (Functional Equation for L-functions of Primitive Characters). Let χ be a primitive character of some modulus R = 1. If χ is even, then L(s, χ) satisfies the function equation and if χ is odd, then L(s, χ) satisfies the function equation where |W (χ)| = 1.
A generalisation of the proposition above appears in Rosen's book [Ros02, Theorem 9.24 A].
Proposition 3.10. Let χ a primitive odd character of modulus R. Then, where we define Proof. The functional equation for odd primitive characters gives us that Taking the squared modulus of both sides gives us that We note that both sides are equal to |L(s, χ)| 2 , and so by the linear independence of powers of q −s we can take the terms n = 0, 1, . . . , deg R − 1 on the LHS and the terms n = 0, 1, . . . , deg R − 2 on the RHS to give Hence, Proposition 3.11. Let χ a primitive even character of modulus R = 1. Then, Proof. The functional equation for even primitive characters gives us that For any primitive character χ 1 of modulus R = 1, we define L −1 (χ 1 ) := 0 and recall that L deg R (χ 1 ) = 0. If we define Similarly as in the proof of Proposition 3.10, we take the squared modulus of both sides, and we take the terms n = 0, 1, . . . , deg R from the LHS and the terms n = 0, 1, . . . , deg R − 1 from the RHS to give We now take s = 1 2 and simplify to get , and similarly, It is convenient to define In this section we state and prove some results for the functions µ, φ and ω that are required for the proofs of the main theorems. We will need the following well-known theorem.
Theorem 4.1 (Prime Polynomial Theorem). We have that where the implied constant is independent of q. We reserve the symbol c for the implied constant.
We will also need the following two definitions.
Definition 4.2 (Radical of a Polynomial, Square-free, and Square-full). For all R ∈ A we define the radical of R to be the product of all distinct monic prime factors that divide R. It is denoted by rad(R). If R = rad(R), then we say that R is square-free. If for all P | R we have that P 2 | R, then we say that R is square-full.
Definition 4.3 (Primorial Polynomials). Let (S i ) i∈Z >0 be a fixed ordering of all the monic irreducibles in A such that deg S i ≤ deg S i+1 for all i ≥ 1 (the order of the irreducibles of a given degree is not of importance in this paper). For all positive integers n we define We will refer to R n as the n-th primorial. For each positive integer n we have unique non-negative integers m n and r n such that where the Q i are distinct monic irreducibles of degree m n + 1. This definition of primorial is not standard. Now, before proceeding to prove results on the growth of the ω and φ functions, we note that for all R ∈ A\{0} . The first equation holds for all s ∈ C. The second holds for all s ∈ C\{0} and is obtained by differentiating the first with respect to s.
Also, for all square-full R ∈ A\{0} we have that The first equation holds for all s ∈ C. The second holds for all s ∈ C\{1} and is obtained by differentiating the first with respect to s.

Lemma 4.4. For all positive integers n we have that
Proof. By (4) and the prime polynomial theorem, we see that By taking logarithms of both equations above, we deduce that Using this result we can obtain results about the growth of the ω function.
Lemma 4.5. For all primorials R n with n > 1 we have that Proof. The cases when m n = 0 are not difficult. We proceed with the cases where m n ≥ 1. Using the prime polynomial theorem, we have that We see that Whereas, for the main term we apply the more precise calculations Hence, (9) gives We also have that and so where we have used Lemma 4.4. The proof follows.
Lemma 4.6. We have that Proof. This lemma follows from Lemma 4.5 if we also prove that To this end, consider the case where R is a square-free monic with ω(R) ≥ 3. Suppose R has n prime divisors. Then, For the second relation we used Lemma 4.5. For the third relation we used the fact that for all q the function x log q x is increasing at x > e and that log q |R n | ≥ ω(R n ) ≥ 3 > e.
Now consider the more general case where R is monic and ω(R) ≥ 3. We have that where the second relation follows from (11), and, again, the third relation follows from the fact that for all q the function x log q x is increasing at x > e.
Finally consider the case where R is monic with deg R ≥ 4 and ω(R) ≤ 2. We have that So, we have proved (10), and this completes the proof.
Remark 4.7. We note that there is an analogous result to Lemma 4.6, in the number field setting: We now work towards obtaining a result on the growth of the φ function.
Lemma 4.8. For all integers n ≥ 2 we have that Proof. On one hand, we have that Taking the exponential of both sides gives On the other hand, because n ≥ 2, we have that and so Taking exponential of both side gives and for infinitely many R ∈ A we have that where a and b are positive constants which are independent of q and R.
Proof. For (12) we need only prove the case where R is square-free (and deg R ≥ 1). Assuming this, we have that In fact, we need only prove the inequality for the case where R is a primorial, as the square-free case follows from this. Indeed, if R is square-free with n ≥ 1 prime factors, then So, it suffices to prove (12) for the primorials. It is clear that it is also sufficient to prove (13) for the primorials as there are infinitely many of them, and this is exactly what we will do. We first proceed with (12). We have that By Lemma 4.8, we see that where for the second to last relation we used the fact that e Hence, we deduce that where a = 7c + 2. Finally, we apply Lemma 4.4 and rearrange to see that For (13), we proceed in a similar fashion. We have that By Lemma 4.8 again, we see that Also, by (14), we see that where b = 7c. Finally, we apply Lemma 4.4 and rearrange to see that and for infinitely many R ∈ A we have that where c and d are positive constants which are independent of q and R.
Proof. From Corollary 3.8, we have that Now, by similar means as in Proposition 4.9, we can obtain that for infinitely many R ∈ A, where c and d are independent of q and R.
Remark 4.11. By similar, but simpler, means as in the proof of Propositions 4.9 and 4.10, we can show that Finally, we prove the following four lemmas.
Lemma 4.12. For all non-negative integers k we have that Proof. Again, it suffices to prove both results for the primorials. From previous results in this section, we can see that For the second result, we note that it is sufficient to show that which is equivalent to which we have proved.
Lemma 4.13. Let R ∈ M. We have that Proof. As we have done in Lemma 4.6 and Proposition 4.9, we can reduce the proof to the case where R is a primorial. That is, we need only prove that To this end, we recall that #P m = q m m + O q m 2 m . From this, we can deduce that there is a constant c ∈ (0, 1), which is independent of q, such that #P ≤m ≥ cq m 2 for all positive integers m. In particular, if we take m = ⌈ 2 log q log n c ⌉, then #P ≤m ≥ n . So, where the second relation follows from the prime polynomial theorem.
Lemma 4.14. We have that where the last relation uses the fact that φ(N ) ≫ |N | log q log q |R| (see Proposition 4.9).
Lemma 4.15. We have that Proof. For square-free N we have that

The Second Moment
We now proceed to prove Theorem 2.4.
Proof of Theorem 2.4. We have that * The second equality follows from Proposition 3.7. For the last equality we note that if R is square-full, EF = R, and µ(E) = 0, then F and R have the same prime factors. Therefore, if we also have that (A, R) = 1 and B ≡ A(mod F ), then (B, R) = 1. Continuing, * The last equality follows from the fact that F and R have the same prime factors, and so, if µ(G) = 0, Hence, By applying this to (15), and using (5) to (8), we see that *

The Brun-Titchmarsh Theorem for the Divisor Function in F q [T ]
In this section we prove a specific case of the function field analogue of the generalised Brun-Titchmarsh theorem. The generalised Brun-Titchmarsh theorem in the number field setting was proved by Shiu [Shi80]. It gives upper bounds for sums over short intervals and arithmetic progressions of certain multiplicative functions. We will look at the case where the multiplicative function is the divisor function in the function field setting.
The main results in this section are the following two theorems.
Theorem 6.1. Suppose α, β are fixed and satisfy 0 < α < 1 2 and 0 < β < 1 2 . Let X ∈ M and y be a positive integer satisfying β deg X < y ≤ deg X. Also, let A ∈ A and G ∈ M satisfy (A, G) = 1 and deg G < (1 − α)y. Then, we have that Intuitively, this seems to be a good upper bound. Indeed, all N in the sum are of degree equal to deg X, and so this suggests that the average value that the divisor function will take is deg X. Also, there are q y 1 |G| ≈ q y 1 φ(G) possible values for N in the sum.
Theorem 6.2. Suppose α, β are fixed and satisfy 0 < α < 1 2 and 0 < β < 1 2 . Let X ∈ M and y be a positive integer satisfying β deg X < y ≤ deg X. Also, let A ∈ A and G ∈ M satisfy (A, G) = 1 and deg G < (1 − α)y. Finally, let a ∈ (F q ) * . Then, we have that Our proofs of these two theorems are based on Shiu's proof of the more general theorem in the number field setting [Shi80]. We begin by proving preliminary results that are needed for the main part of the proofs.
The Selberg sieve gives us the following result. A proof is given in [Web83]. We then have that #S Q,z :=# S\ ∪ P |Qz S P = #{A ∈ S : (P | A and P ∈ Q) ⇒ deg P > z} Corollary 6.4. Let X ∈ M and y be a positive integer satisfying y ≤ deg X. Also, let K ∈ M and A ∈ A satisfy (A, K) = 1. Finally, let z be a positive integer such that deg K + z ≤ y. Then, Proof. Let us define Then, we have that which is what we want to bound.
This follows from the fact that K and D are coprime and that deg K where |c D | ≤ 1. Therefore, we have ω(D) = 1 and |r(D)| ≤ 1 for all D | Q z . We also have that ψ(D) = φ(D) for square-free D.
We can now see that and we have that To this we apply Lemma 4.15 and the fact that Also, we have that The result now follows by applying Theorem 6.3.
The proof of the following corollary is almost identical to the proof above.
Corollary 6.5. Let X ∈ M and y be a positive integer satisfying y ≤ deg X. Also, let K ∈ M and A ∈ A satisfy (A, K) = 1. Finally, let z be a positive integer such that deg K +z ≤ y, and let a ∈ (F q ) * . Then, Lemma 6.6. Let w be a positive integer. We have that where the implied constant can be taken to be independent of q and will be denoted by d.
Proof. By using the prime polynomial theorem, we have that The result now follows.
Lemma 6.7. Let 0 < α, β < 1 2 , let z > q be an integer, and let w(z) := log q z. Then, as z → ∞, where d is as in Lemma 6.6. In particular, this implies that (under the condition that z > q).
Proof. Let δ > 0. We will optimise on the value of δ later. We have that where the last two relations follow from the Taylor series for the exponential function. Continuing, where the last inequality follows from Lemma 6.6. By using the definition of w(z), we have that and if we take Lemma 6.8. Let z and r be a positive integers satisfying r log q r ≤ z. Then, Proof. Let 3 4 ≤ δ < 1. We will optimise on the value of δ later. We have that where the last relation uses the Taylor series for the exponential function. Note that where the last relation uses the fact that δ ≥ 3 4 . Also, we can write 1 and that where the second-to-last relation follows from a similar calculation as (18).

We substitute (17), (18), and (19) into (16) to obtain
We can now take δ = 1 − r log q r 4z (by the conditions on r given in theorem, we have that 3 4 ≤ δ < 1, as required). Then, Proof of Theorem 6.1. We will need to break the sum into four parts. First, we define z := α 10 y. Now, for any N in the summation range, we can write N = P 1 e 1 . . . P j e j P j+1 e j+1 . . . P n en (20) where deg P 1 ≤ deg P 2 ≤ . . . ≤ deg P n and j ≥ 0 is chosen such that deg P 1 e 1 . . . P j e j ≤ z < deg P 1 e 1 . . . P j e j P j+1 e j+1 .
We will consider the following cases: Case 1: We have that We note that and so d(D) ≤ 2 20 α . Hence, We can now apply Corollary 6.4 to obtain where the second-to-last relation uses the fact that deg G ≤ (1 − α)y and z = α 10 y.
Case 2: Suppose N satisfies case 2. Then, the associated P j+1 (from (20)) satisfies P j+1 e j+1 | N , deg P j+1 ≤ 1 2 z, and deg P j+1 e j+1 > 1 2 z. For a general prime P with deg P ≤ 1 2 z we denote e P ≥ 2 to be the smallest integer such that deg P e P > 1 2 z. We will need to note for later that Let us also note that for N with deg N ≤ deg X we have that where the last relation follows from the fact that z = α 10 y and deg G ≤ (1 − α)y.
Case 3: Suppose N satisfies case 3. For the case where z ≤ q we have that w(z) = 1, meaning that the only possible value N could take is 1. At most this contributes O(1).
So, suppose that z > q, and so w(z) = log q z. Case 3 tells us that 1 2 z < deg B N ≤ z and Hence, as z −→ ∞, where the second-to-last relation follows from Lemma 6.7, and the last relation uses the fact that deg G ≤ (1 − α)y and z = α 10 y.
Case 4: The case z < 1 is trivial, and so we proceed under the assumption that z ≥ 1. We have that We now divide p − (D N ) into the blocks 1 r+1 z < p − (D N ) ≤ 1 r z for r = 2, 3, . . . , r 1 where .
and so where a = 2 20 αβ . So, continuing from (24), where X B is a monic polynomial of degree deg X − deg B such that deg X − BX B < y, and A B is a polynomial satisfying A B B ≡ A(mod G).
Corollary 6.4 tells us that where the last relation follows from the fact that deg B ≤ z, z = α 10 y, and deg G ≤ (1 − α)y. Hence, continuing from (25): Finally, we wish to apply Lemma 6.8. This requires that r log q r ≤ z. Now, when 1 ≤ z ≤ q we have that w(z) = 1 and r 1 = z. Hence, r log q r ≤ z log q q = z. When z > q we have that w(z) = log q z and r 1 = z w(z) . Hence, r log q r ≤ z log q z (log q z − log q log q z) ≤ z, since z > q. Hence, The proof now follows from (21), (22), (23), and (26).
Proof of Theorem 6.2. The proof of this theorem is almost identical to the proof of Theorem 6.1. Where we applied Corollary 6.4, we should instead apply Corollary 6.5. Also, the calculations respectively.

Further Preliminary Results
Lemma 7.1. Let c be a positive real number, and let k ≥ 2 be an integer. Then, Proof. We will first look at the case when y ≥ 1. Let n be a positive integer, and define the following curves: We can see that For the third integral we note that when s ∈ l 3 (n) we have |y s | ≤ 1 (since Re s ≤ 0 and y ≥ 1). Hence, Now we will look at the case when 0 ≤ y < 1. Again, let n be a positive integer, and define the following curves: l 3 (n) := c + ne it : t ∈ −π 2 , π 2 (orientated clockwise), L(n) :=l 1 (n) ∪ l 2 (n).
We can see that The limit of the first integral is equal to zero by the residue theorem, because there are no poles inside L(n). The limit of the second integral is also zero, and this can be shown by a method similar to that applied for the curve l 3 (n) in the case y ≥ 1. So, for 0 ≤ y < 1 we deduce that c+i∞ c−i∞ y s s k ds = 0.
We now have the following proposition.
Proposition 7.2. Let R ∈ M and let x be a positive integer. Then, Proof. For all positive integers x we have that By (5), (6), and Lemma 4.13, we see that .
Whereas, when x < deg R, we have that The proof follows. Then, for all R ∈ A and j = 1, 2, 3, 4 we have that Remark 7.4. We must mention that, in the lemma and the proof, the implied constants may depend on j, for example; but because there are only finitely many cases of j that we are interested in, we can take the implied constants to be independent.
Proof. First, we note that where We note further that For all R ∈ A and k = 0, 1, 2, 3 it is not difficult to deduce that g (k) The function log x k+1 x−1 is decreasing at large enough x, and the limit as x −→ ∞ is 0. Therefore, there exist an independent constant c ≥ 1 such that for k = 0, 1, 2, 3 and all A, Hence, taking n = ω(R), we see that where we have used the prime polynomial theorem and Lemma 4.4.
Then, we have that STEP 2: For the first integral in (32) we note that F (1 + s) y R s s 3 has a fifth-order pole at s = 0 and double poles at s = 2mπi log q for m = ±1, ±2, . . . , ±n. By applying Cauchy's residue theorem we see that If we apply the product rule for differentiation, then one of the terms will be 1 4! lim Now we look at the remaining terms that arise from the product rule. By using the fact that ζ A (1+s) = 1 1−q −s and the Taylor series for q −s , we have for k = 0, 1, 2, 3, 4 that Similarly, By (35), (36), and Lemma 7.3, we see that the remaining terms are of order STEP 2.2: Now we look at the remaining residue terms in (33). By similar (but simpler) means as above can show that STEP 2.3: By (33), (37) and (38), we see that STEP 3: We now look at the integrals over l 2 (n) and l 4 (n). For all positive integers n and all s ∈ l 2 (n), l 4 (n) we have that F (s + 1)y R s = O q,R (1). One can now easily deduce for i = 2, 4 that Lemma 7.6. We have that x 2 + 3q + 1 2q x + 1.

From this we easily deduce that
Proof. For s > 1 we define We can see that By comparing the coefficients of powers of q −s , we see that Lemma 7.7. Let R ∈ M. We have that Proof. We have that where the last relations follows from Lemma 7.6. We also note that This proves the first relation in the lemma. The second relation follows from Lemma 4.11.
Lemma 7.8. Let F, K ∈ M, x ≥ 0, and a ∈ F q * . Suppose also that 1 2 x < deg KF ≤ 3 4 x. Then, Proof. We have that, where N ′ , G ′ , K ′ are defined by HN ′ = N, HG ′ = G, HK ′ = K. Continuing, we have that The third relation holds by Theorem 6.2 with β = 1 6 and α = 1 4 (one may wish to note that (K ′ F, G ′ ) = 1 and that the other conditions of the theorem are satisfied because 1 2 x < deg KF ≤ 3 4 x). The last relation follows from Lemma 4.14.
Lemma 7.9. Let F, K ∈ M and x ≥ 0 satisfy deg KF < x. Then, Proof. The proof is similar to the proof of Lemma 7.8. We have that where we define X := T x−deg H . We can now apply Theorem 6.1 to obtain that Lemma 7.10. Let F ∈ M and z 1 , z 2 be non-negative integers. Then, for all ǫ > 0 we have that Proof. We can split the sum into the cases deg AC > deg BD, deg AC < deg BD, and deg AC = deg BD with AC = BD. The first two cases are identical by symmetry.
When deg AC > deg BD, we have that AC = KF + BD where K ∈ M and deg KF > deg BD. Furthermore, from which we deduce that z 1 +z 2 2 < deg KF ≤ z 1 + z 2 ; and from which we deduce that deg BD = z 1 + z 2 − deg KF .
When deg AC = deg BD, we must have that deg AC = deg BD = z 1 +z 2 2 (in particular, this case applies only when z 1 + z 2 is even). Also, we can write AC = KF + BD, where deg KF < deg BD = z 1 +z 2 2 and K need not be monic.
STEP 2: We now consider the case when z 1 + z 2 > 19 10 deg F . . This allows us to apply Lemma 7.8 for the first relation below.

STEP 2.3:
We now look at the sum Proof. The case where a = 1 is just Lemma 7.10. The proof of the case where a = 1 is very similar to the proof of Lemma 7.10. In fact it is easier, because the the case where deg AC = deg BD cannot exist: We would require that AC and BD are both monic, but also require that at least one of AC and BD have leading coefficient equal to a = 1.
where the second-to-last relation uses the following:

The Fourth Moment
We now proceed to prove Theorem 2.5. In the proof we implicitly state that some terms are of lower order than the main term and that is easy to check. We do not give the justification explicitly, although all the results one needs for a rigorous justification are given in Section 4.
Proof of Theorem 2.5. Let χ be a Dirichlet character of modulus R. By Propositions 3.10 and 3.11, we have that We will show that * χ mod R |a(χ)| 2 has an asymptotic main term of higher order than * χ mod R |b(χ)| 2 and * χ mod R |c(χ)| 2 . From this and the Cauchy-Schwarz inequality, we deduce that * gives the leading term in the asymptotic formula.
where the last equality follows from Proposition 3.7. Continuing, * STEP 1.1: We will look at the first term on the far-RHS. Since AC = BD, we can write A = F U, B = F V, C = GV, D = GU , where F, G, U, V are monic and U, V are coprime. Let us write N = U V , and note that there are 2 ω(N ) ways of writing N = U V with U, V being coprime. Then, where z R ′ := deg R − log q 9 ω(R) .
Let us look at the first term on the far-RHS of (44). We apply Proposition 7.2. When x = z R −deg N 2 and deg N ≤ z R ′ , we have that 2 ω(R) x q x = O(1). Hence, where the last equality follows from Proposition 7.5 and Lemma 7.7.
Now we look at the second term on the far-RHS of (44). Because z R ′ < deg N ≤ z R , we have that deg F ≤ log q 3 √ 2 ω(R) . Using this and Proposition 7.2, we have that Also, by similar means as in Lemma 7.6, we can see that Hence, where, again, we have used Propositions 7.2 and Lemma 7.7.
as deg R → ∞. The second relation follows from Lemma 7.10 with F := R. This can be applied because z 1 + z 2 ≥ 2z R = 2 deg R − 2 log q 2 ω(R) > 19 10 deg R for large enough deg R.   .
For the first term on the far-RHS we have that For the second term we have that where we have used Lemma 7.10. Hence, Similarly, by using Lemma 7.11 for the even case, we can show, for a = 0, 1, 2, 3, that Hence, by using the Cauchy-Schwarz inequality, we can deduce that * χ mod R |c(χ)| 2 ≪ φ * (R) P prime P |R 1 − |P | −1 3 1 + |P | −1 (deg R) 3 ω(R).
STEP 4: From steps 1 to 3, and the use of the Cauchy-Schwarz inequality (as described at the start of the proof), the result follows.