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On sums and products in C[x]

We show that under the assumption of a 24-term version of Fermat's Last Theorem, there exists an absolute constant c>0 such that if S is a set of n>n_0 positive integers satisfying |S.S|>n^2. In other words, we prove a weak form of the Erdos-Szemeredi sum-product conjecture, conditional on an extension of Fermat's Last Theorem. Unconditionally, we prove this theorem for when S is a set of n monic polynomials. We also prove an analogue of a theorem of Bourgain and Chang for the ring C[x].


Introduction
Suppose that S is a subset of a ring R (in our case, the real or complex numbers), and define S.S := {st : s, t ∈ S}, and S + S := {s + t : s, t ∈ S}.
Partial progress on this problem has been achieved by Erdős and Szemerédi [9], Nathanson [13], Ford [10], Elekes [8], and Solymosi in [16] and in the astounding article [17]. There is the work of Bourgain-Katz-Tao [5] and Bouragin-Glibichuk-Konyagin [4] extending the results to F p , and then also the recent work of Tao [18] which extends these results to arbitrary rings.
We will prove a theorem below (Theorem 1), which holds for the polynomial ring C[x] unconditionally; but, under the assumption of a certain 24-term version of Fermat's Last Theorem, it holds for Z + , the positive integers, as we will explain below. This unconditional result for C[x], for which there presently is no analogue for C, R, Z or F p (though, M.-C. Chang has a related sort of result for Z as we discuss in remarks below.), is as follows.
Remark 2. It is perhaps possible to replace the condition that the polynomials be "monic" with the condition that none is a scalar multiple of another; however, it will make the proof more complicated (if our method of proof is used). Now let us consider the following conjecture, which can be thought of as a 24-term extension of Fermat's Last Theorem.
Conjecture. For all m ≥ 1 sufficiently large, the only solutions to the Diophantine equation are the trivial ones; that is, solutions where each ε i x m i can be paired with its negative ε j x m j , so that x i = x j and ε i = −ε j .
Under the assumption of this conjecture, we have that Theorem 1 holds for when S is a set of positive integers (instead of monic polynomials); in other words, Theorem 2 Suppose that the above Conjecture is true. Then, there exists an absolute constant c > 0 such that the following holds for all sufficiently large sets S of n positive integers: |S.S| < n 1+c =⇒ |S + S| ≫ n 2 .
Because the proof of this theorem is virtually identical to the proof of Theorem 1, where every use of Theorem 4 below is simply replaced with the above Conjecture, we simply omit the proof.
We also prove the following theorem, which extends a result of Bourgain and Chang [3] from Z to the ring C[x].
Theorem 3 Given a real c ≥ 1 and integers ℓ, k ≥ 1, the following holds for all n sufficiently large: Suppose that S is a set of n polynomials of C[x], where none is a scalar multiple of another, and suppose that Then, where for fixed c and k, f (c, k, ℓ) → 0 as ℓ → ∞. Remark 1. It is possible to generalize our method of proof to the case of polynomials over F p [x], but in that context there are thorny issues concerning the vanishing of certain Wronskian determinants that make the problem difficult. There are also issues that come up in handling pth powers of polynomials, which the differentiation mapping sends to the 0 polynomial.

Remark 2.
If the analogue of this theorem for when S ⊆ C could be proved, then it would also provide a proof to the above, since we could locate an element α ∈ C such that the evaluation map α : f ∈ C[x] → f (α) ∈ C preserves the structure of the sums and products of these polynomials. Remark 3. Using the above results, one can easily prove analogues of them for the polynomial ring C[x 1 , x 2 , ..., x k ], simply by applying an evaluation map ψ : The proofs of both of the above theorems rely on the following basic fact about polynomials, using ideas of Mason [12], which we prove near the end of the paper.
Theorem 4 For every k ≥ 2, there exists an exponent M such that there are no polynomial solutions to where no polynomial is a scalar multiple of another.
Remark. Previously, it had been proved that such an equation has no solutions under the condition that all the polynomials are pairwise coprime.
Theorem 5 Suppose that S is a finite subset of an additive abelian group, and that |S + S| ≤ K|S|. Then, 2 Proof of Theorem 1 By invoking Theorem 4, let M ≥ 1 be the smallest value such that the polynomial equation has no solutions for m ≥ M, and k ≤ 24, assuming no f i is a constant multiple of another. And, when we generalize the present proof (of Theorem 1) to handle the proof of Theorem 2, we just assume that M is such that if m ≥ M, then (1) has no non-trivial solutions. We assume throughout that where c > 0 is some parameter that can be determined by working through the proof -the point is that, although c will be quite small, it will be possible to take it to be some explicit value. Let ε = c(M + 1), and note that by the Ruzsa-Plunnecke inequality (Theorem 5) we have that since |S.S| < n 1+c , |S M +1 | = |S...S| < n 1+ε .
(The sole use of ε > 0 in the rest of the proof is to simplify certain expressions.) Let us begin by supposing that Then, for all but at most o(n 2 ) pairs ( such that Let P denote the set of all such n 2 − o(n 2 ) pairs (x 1 , x 2 ). It is clear that there exists a bijection ϕ : P → P, where ϕ maps pairs having sum s to pairs having sum s, and yet where if Using such a pairing ϕ of pairs (x 1 , x 2 ), we then define a set of quadruples Note that |Q| = |P | ∼ n 2 ,

A lemma about quadruples
To proceed further we require the following lemma.
Lemma 1 There exists a 5-tuple a, b, c, d, t ∈ S such that for ≫ n 2−4ε quadruples Proof of the lemma. Let N denote the number of pairs (x 1 , t) ∈ S × S for which there are fewer than n 1−ε /40 pairs (a, t 1 ) ∈ S × S satisfying Clearly, Given x 1 ∈ S we say that t is "bad" if (x 1 , t) is one of the pairs counted by N; otherwise, we say that t is "good". From the bounds above, it is clear that more than 4n/5 values x 1 ∈ S have ≤ n/8 bad values of t; for, if there were fewer than 4n/5 such x 1 ∈ S, then ≥ n/5 have > n/8 bad values of t, which would show that N > n 2 /40, a contradiction.
It follows that more than 3n 2 /5 pairs (x 1 , x 2 ) ∈ S 2 have the property that both x 1 and x 2 have at most n/8 bad values of t. So, there are at least n − 2n/8 = 3n/4 values of t that are "good" for both x 1 and x 2 .
Clearly, then, by the pigeonhole principle, there are such that there are at least n/2 values of t that are good for all x 1 , x 2 , x 3 and x 4 at the same time (disguised in what we are doing here is the fact that ϕ is a bijection from P → P ). When t is good for all x 1 , x 2 , x 3 and x 4 , we say that it is "good" for the quadruple (x 1 , x 2 , x 3 , x 4 ).
By the pigeonhole principle again, there exits t ∈ S that is good for at least n 2 /10 quadruples of Q. Now suppose that (x 1 , x 2 , x 3 , x 4 ) ∈ Q is one of these ∼ n 2 /10 quadruples, for some particular value of t. We define Under the assumption that t is good, we have |A|, |B|, |C|, |D| ≥ n 1−ε /40. So, among the n 2 /10 quadruples for which t is good, by the pigeonhole principle again, there exists a, b, c, d ∈ S for which there are This completes the proof of the lemma.

Resumption of the proof of Theorem 1
Upon applying the previous lemma, let Q ′ denote the set of quadruples such that is one of the ≫ n 2−4ε quadruples satisfying the conclusion of the lemma. Note that Now we claim that we can find three quadruples and The reason that we can find such quadruples is as follows: Suppose we fix (t 1 , ..., t 4 ) to be any quadruple of Q ′ , and suppose that we pick (u 1 , ..., u 4 ) in order to attempt to avoid Let r = t 2 /t 1 ∈ S/S. If (5) holds for all (u 1 , u 2 ), it means that each pair (u 1 , u 2 ) must be of the form But, by Ruzsa-Plunnecke, so there are at most n 1+ε pairs among the first two entries of quadruples of Q ′ ; but, since there at ≫ n 2−4ε quadruples of Q ′ , and since the first two coordinates of the quadruple determine the second pair of coordinates (via the mapping ϕ) we clearly must not have that all pairs (u 1 , u 2 ) satisfy (5).
Before proceeding with the rest of the proof, it is worth pointing out that the sort of condition on our four-tuples that we cannot so easily force to hold is, for example, The reason for this is that we do not have a good handle on how many pairs (t 1 , t 3 ) or (u 1 , u 3 ) there are -t 3 (or u 3 ) may be related to t 1 (or u 1 ) in a completely trivial way, leading to few pairs.

A lemma about submatrices
Now we require a lemma concerning the matrix Proof of the lemma. We show that 3×3 matrices are non-singular via contradiction: Suppose, on the contrary, that some 3 × 3 submatrix is singular, and without loss assume (that it is the 'first' 3 × 3 submatrix): Expanding out the determinant into a polynomial in its entries, we see that it produces a sum of Mth powers equal to 0. Furthermore, since all the t i , u j , v k are monic, none is a scalar multiple of another, except for factors ±1. Since no non-trivial sum of 24 or fewer Mth powers of polynomials can equal 0, it follows that each must be matched with its negative, in order for this sum of Mth powers to equal 0.
Note, then, that there are 6 = 3! possible matchings that can produce a 0 sum. Consider now the matching (assuming we have taken Mth roots) This matching implies Some of the other possible matchings will lead to equations such as t 1 /t 2 = u 1 /u 2 , which we have said was impossible by design; but, there is one other viable chain of equations that we get, using one of these matchings, and that is In what follows, whether this chain holds, or (6) holds makes little difference, so we will assume without loss of generality that (6) holds, and will let r = t 3 /t 1 denote the common ratio, which we note is a rational function. It is clear that assuming that this matching holds, we can reduce the equations Since all the elements of S are monic, none can be 0, and so this column vector is not the 0 vector. It follows that the 3 × 3 matrix here is singular (from the fact that one of the 3 × 3 submatrices of T is singular, we just got that another was singular). Upon expanding the determinant of this matrix into a polynomial of its entries, in order to get it to be 0 we must have a matching, much like the one that produced (6). As before, we will get two viable chains of equations: Either or Let us suppose that (7) holds, and let s be the common ratio, which is of course a rational function. Then, it follows that are independent, this matrix is non-singular; so, the column vector here must be the 0 vector, which is impossible since b = 0.
So, (8) above must hold. Redefining s to be the common ratio t 4 /t 2 here, it leads to the equation Since the matrix is non-singular, it follows that a = cr M and b = ds M .

But this implies that
and So, (t 1 , t 2 , t 3 , t 4 ) couldn't have been a quadruple of Q ′ , because if so, then for some (x 1 , x 2 , x 3 , x 4 ) ∈ Q we would have had and then the equations (9) and (10) imply that a contradiction. This completes the proof of the lemma.

Continuation of the proof
In addition to the quadruples (t 1 , ..., t 4 ), (u 1 , ..., u 4 ), (v 1 , ..., v 4 ), let (w 1 , ..., w 4 ) be another quadruple (we will later show that we can choose it in such a way that we contradict the assumption |S + S| = o(n 2 )). We have that is singular, as the vector (a, b, −c, −d) (written as a column vector) is in its kernel. Expanding out its determinant, we find that it must be 0; and, we know from Theorem 4 that this is impossible, except if we can match up terms, as we did in the proof of Lemma 2. Just so the reader is clear, an example of just one equation from such a matching is perhaps, say, In total, there will be 12 = 4!/2 different equations that make up such a matching. Let us suppose that in a hypothetical matching we got an equation of the form and for our purposes we just need to write this as where (α, β) ∈ C[x]×C[x] can be any of at most 6 2 polynomials (according to which combination of i, j, k, i ′ , j ′ , k ′ is chosen). So, we would have w M 2 /w M 1 = α/β; that is, w 2 /w 1 takes on at most 36 possible values. What this would mean is that the pair (w 1 , w 2 ) is essentially determined by w 1 , and that it can take on at most 36n possible values -far too few to consume the majority of the ≫ n 2−4ε pairs (w 1 , w 2 ) making up the first part of a quadruple of Q ′ .
We conclude that all but O(n) of the quadruples (w 1 , w 2 , w 3 , w 4 ) ∈ Q ′ cannot lead to a solution to (11) under any matching; moreover, all but O(n) will also avoid αw M We also have that there are at most O(n) quadruples can lead to solutions to because it would imply that w 4 /w 3 is fixed, and we are back in the situation (12). Furthermore, there are at most O(n) quadruples leading to solutions to any of the following pairs: or We can also avoid a matching that produces three equations (indexed by j) of the form because it would mean that the 3 × 3 submatrix of T with the ith column deleted, is singular (these α j , β j have the property that the determinant of this submatrix is 3 j=1 (α j − β j )). Furthermore, we cannot even have a pair of equations of the type (17), for the same value i, because it would imply that in fact we get three equations upon taking a product of the two and doing some cancellation; for example, suppose that i = 4, and that we have two equations of (17) holding. Then, there is a matching between two pairs of terms, upon expanding the determinant of the following matrix in terms of its entries: The matching corresponds, say, to α 1 = β 1 and α 2 = β 2 ; and, say, these correspond to the equations Multiplying left and right sides together, and cancelling, produces and, multiplying both sides by v M 2 produces the missing matching which proves that the matrix (18) is singular. But this is impossible, since it contradicts Lemma 2. We conclude therefore that, as claimed, we cannot have even a pair of equations from (17) hold, for any i = 1, ..., 4.
We have eliminated a great many possible matchings that could occur in order that the matrix Γ be singular: Our 12 equations in a matching can include at most one of each the four types and so must include at least 8 of the form all the while avoiding pairs (13), (14), (15), and (16).
Let us consider what would happen if we had a pair or a pair both holding. Without loss of generality in what follows, we just assume that (20) holds. Then, we would have that the equation Now, if we had any other quadruple (z 1 , z 2 , z 3 , z 4 ) ∈ Q ′ that also satisfied we would likewise get and then we would get the equation or else this matrix is singular. If the former holds, it implies that which contradicts the hypotheses about the sets of quadruples Q and Q ′ . So, the matrix must be singular; in other words, But since by the Ruzsa-Plunnecke inequality |S/S| < n 1+ε , there can be at most n 1+ε such vectors (z 1 , z 2 ), given (w 1 , w 2 ). In particular, this means that there can be only very few quadruples (w 1 , w 2 , w 3 , w 4 ) ∈ Q ′ that satisfy (20) or (21), for any particular combination of α, β, α ′′ , β ′′ , α ′′′ , β ′′′ . Since there are at most 6 6 possibilities for all these, (they are products of entries from T ), we deduce that there are ≫ n 2−4ε − O(n 1+ε ) quadruples of Q ′ that do not satisfy a pair of equations of the sort (20) or (21). So, we may safely assume that (w 1 , ..., w 4 ) does not satisfy these equations.
We may assume, then, that all 8 of our equations of the type (19) involve the same pair of w i 's, and do not involve w 1 , w 2 or w 3 , w 4 . So, for example, our matching includes 8 equations of the form We have now exhausted all of the possibilities, reached a contradiction in each case, and so shown that Γ must be non-singular for some choice of (w 1 , w 2 , w 3 , w 4 ) ∈ Q ′ . This then means that we couldn't have had so many quadruples in Q ′ , and therefore Q. Therefore, |S +S| ≫ n 2 , and we are done.

Proof of Theorem 3
Let us suppose that ε > 0 is some constant that we will allow to depend on c, k, and a certain parameter t mentioned below, but is not allowed to depend on ℓ. This ε > 0 will later be chosen small enough to make our proofs work. Also, we suppose that where ℓ ≥ 1 is as large as we might happen to require, as a function of c and k (and implicitly, ε). 1 If (22) holds, it follows that n = |S| ≤ |S.S| ≤ · · · ≤ |S ℓ | < n c .
Letting M ≥ 1 be some integer depending on c and k that we choose later, we have that for ℓ large enough, it is obvious that for some t < ℓ/M, 2 Furthermore, if we let then since it follows that In other words, S t M S is not appreciably larger than S t M . Let us now define so that (23) becomes |RS| ≤ |R| 1+ε .
We now arrive at the following useful lemma.

Lemma 3 Suppose that (24) holds. Then, there exists
s ∈ S, and r ′ ∈ R, such that for at least n 1−O(tε) values s ′ ∈ S we have that there exists r ∈ R satisfying rs = r ′ s ′ .
Note. Since we can choose ε as small as desired in terms of t, we have that n 1−O(tε) is basically n 1−δ , where δ > 0 is as small as we might happen to require later on in the argument (δ is allowed to depend on c and k, but not on ℓ). Proof of the lemma. Inequality (24) easily implies that |{(s, s ′ , r, r ′ ) ∈ S 2 × R 2 : rs = r ′ s ′ }| ≥ |R| 1−ε n 2 .
So, extracting the pair s, r ′ producing the maximal number of pairs s ′ , r satisfying rs = r ′ s ′ , we find that this pair leads to at least such pairs (s ′ , r) ∈ S × R.
Let S ′ denote the set of s ′ produced by this lemma, for the fixed pair (r ′ , s). We will show that from which it will follow that thereby establishing Theorem 3, since the larger we may take ℓ ≥ 1, the smaller we may take δ > 0 (so, our f (c, k, ℓ) = O(δ) → 0 for fixed c, k as ℓ → ∞).
To prove (25), it suffices to show that the only solutions to are trivial ones: Suppose that, on the contrary, this equation has a non-trivial solution. Then, upon multiplying through by r ′ , we are led to the equation Cancelling the s's, and writing r i = y M i , y i ∈ S t , produces the equation In order to be able to apply Theorem 4 to this to reach a contradiction, we need to show that this equation is non-trivial, given that we have a non-trivial solution to (26). First observe that which means that certain of the y M i 's are scalar multiples of one another if and only if the corresponding x i 's are scalar multiples of one another, and in fact with the same scalars λ.
After cancelling common terms from both sides of (27), we move the remaining terms from among y M k+1 + · · · + y M 2k to the left-hand-side, writing We also must collect together duplicates into a single polynomial: Say, for example, y 1 = · · · = y j . Then, we collapse the sum y M 1 + · · · + y M j to (j 1/M y 1 ) M .
For M large enough in terms of k, Theorem 4 tells us that our collapsed equation can have no non-trivial solutions, and therefore neither can (27). Theorem 3 is now proved.

Proof of Theorem 4
The proof of this theorem will make use of the ideas that go into the proof of the so-called ABC-theorem, which is also known as Mason's Theorem [12] (see also [11] for a very nice introduction). Although there are versions of Mason's theorem already worked out for some quite general contexts, in our useage of the ideas that go into the proof of this theorem, we will need to allow some of the polynomials to have common factors. Our proof is similar in many respects to the one appearing in [1] and [2] -the only difference is that we consider polynomials where none is a scalar multiple of another, and impose no coprimeness condition.

The basic ABC theorem
Before we embark on this task, let us recall the most basic ABC theorem, and see its proof. Remark 1. This theorem easily implies that the "Fermat" equation has no solutions for n ≥ 3, except trivial ones: Suppose that at most one of f, g, h is constant, and that this equation does, in fact, have solutions. Letting f be the polynomial of maximal degree, we find that So, n ≤ 2 and we are done.
Proof. The proof of the theorem makes use a remarkably simple, yet powerful "determinant trick". First, consider the determinant Note that this matrix is a Wronskian.
Let us see that ∆ = 0: If ∆ = 0, we would have that Since A(x) and B(x) are coprime, we must have that Both of these are impossible, unless of course both A(x) and B(x) are constants. If both are constants, then so is C(x), and we contradict the hypotheses of the theorem. So, we are forced to have ∆ = 0. Now suppose that so that A, B, C have only the roots α 1 , ..., α k , with multiplicities a 1 , ..., a k , respectively. We will now see that To see this, note that for each i = 1, ..., k, (x − α i ) a i divides either A(x), B(x), or C(x), since all three are coprime. Note also that adding the first column of the matrix in (28) to the second does not change the determinant, so that (Note that here we have used the fact that differentiation is a linear map from the space of polynomials to itself.) Now using the fact that it follows that (x − α i ) a i −1 divides all the elements of some column of either the matrix (28), or (29). It follows that But we also have a simple upper bound on the degree of ∆, We will also require the following lemma.
Lemma 5 For k ≥ 2 and ε > 0 there exists M ≥ 1 such that the following holds: Suppose that are linearly independent over C. Then, the equation has no solutions with G satisfying the following two conditions: • deg(gcd(G, f j )) < εD/2k, for all j = 1, ..., k − 1.
Proof of the lemma. Suppose, in fact, that there are such solutions, where we will later choose M purely as a function of ε and k, in order to reach a contradiction. Then, consider the determinant .
The hypotheses of the lemma, along with Lemma 4, imply that ∆ = 0. It is easy to see, upon using your favorite expansion for the determinant, that On the other hand, the jth column of the matrix is divisble by And, upon adding the first k − 2 columns to the last column, we see that ∆ is also divisible by G M −k+2 . So, Using again the hypotheses of the lemma, we can easily bound the degree of this gcd from above by Combining our upper and lower bounds on deg(∆), we deduce .
This is impossible for M large enough in terms of ε and k, since deg(G) > εD, and our lemma is therefore proved.

An iterative argument
To apply the above lemmas, we begin by letting ε = 1, and suppose that f M 1 + · · · + f M k = 0 (32) has a non-trivial solution (none is a scalar multiple of another), where M can be taken as large as we might happen to need. And, assume that we have pulled out common factors among all the f i . We furthermore assume that any subset of size k − 1 of the polynomials f M Without loss of generality, assume that j = k − 1. Then, we may write , so that f M 1 + · · · + f M k−2 + G M 1 g k = 0. Note that this last term is non-zero by the assumption that no k − 1 of the polynomials f M i can be linearly dependent over C. Then, we set ε 1 = ε/2k, and observe that deg(G 1 ) > ε 1 D.
On the other hand, if (33) does not hold, then we proceed on to the next subsection. Now suppose that deg(gcd(f i , G 1 )) > ε 1 D/2(k − 1), for some i = 1, ..., k − 2, where here we redefine D to D = max(deg(f 1 ), ..., deg(f k−2 ), deg(G M 1 g k )/M). (Note that we still have deg(G 1 ) > ε 1 D). Without loss of generality, assume i = k − 2. Then, we may write where deg(G 2 ) > ε 1 D/2(k − 1), so that f M 1 + · · · + f M k−3 + G M 2 g k−2 = 0. Of course, we also have to worry about wether this final term is 0, but that is not a problem as it would imply that k − 1 of the polynomials f M i are dependent over C.
On the other hand, if (34) does not hold, then we proceed on to the next subsection.
We repeat this process, producing ε 2 , ε 3 , ..., and G 3 , G 4 , .... We cannot continue to the point where our equation is f M 1 + G M k−2 g 2 = 0, with G k−2 non-constant, because then we would have that f 1 has a common factor with G k−2 , and therefore all of f 1 , ..., f k would have to have a common factor. So, the process must terminate before reaching G k−2 .

Conclusion of the proof of Theorem 4
When we come out of the iterations in the previous section, we will be left with an equation of the form where D = max(deg(f 1 ), ..., deg(f J ), deg(G M k−J−1 g J−1 )/M)), and where γ(k) is a function depending only on k. Applying now Lemma 5 to this, we find that this is impossible once M is large enough. So, our theorem is proved.

Acknowledgment
We would like to thank Jozsef Solymosi for his suggestion to include Theorem 2.