MA3H1 TOPICS IN NUMBER THEORY PART III

MA3H1 TOPICS IN NUMBER THEORY PART III SAMIR SIKSEK 1. Congruences Modulo m In quadratic recirocity we studied congruences of the form x 2 a (mod ). We now turn our attention to situations where is relaced by a ower of. We shall need the following lemma whose roof is an easy exercise, but try out a few examles first to convince yourself that it is true. Lemma 1.1. Let f(x) Z[X] and let n > 0 be an integer. Then f (n) (X)/n! has integer coefficients. Next is Hensel s Lemma which is the main result of this section. Theorem 1.2. (Hensel s Lemma) Let f(x) Z[X]. rime and m 1. Suose a Z satisfies f(a) 0 (mod m ), f (a) 0 (mod ). Then there exists some b Z such that (1) b a (mod m ), f(b) 0 (mod m+1 ). We say that we lift a to a solution modulo m+1. Proof of Hensel s Lemma. By Taylor s Theorem Let be a f(a + x) = f(a) + f (a)x + f (2) (a) x 2 + + f (n) (a) x n 2! n! where n is the degree of f (note that all higher derivatives vanish). We want b to satisfy two conditions, one of them that b a (mod m). Let us write b = a + m y where the integer y will be determined later. Then f(b) = f(a) + m f (a)y + 2m (integer). Since f(a) 0 (mod m ) we have f(a) = m c where c is an integer. Thus f(b) = m (c + f (a)y) + 2m (integer). Date: December 11, 2009. 1

2 SAMIR SIKSEK Note that m+1 2m. To make f(b) 0 (mod m+1 ) it is enough to choose y so that (c + f (a)y). In otherwords, we want y so that f (a)y c (mod ). But f (a) 0 (mod ) and so is invertible modulo. Let h satisfy hf (a) 1 (mod ). Then we choose y = hc and take b = a hc m and then both congruences in (1) are satisfied. The roof of Hensel s Lemma is constructive; this means that it can be used to solve congruences modulo rime owers. You need to ractice Hensel s Lemma a few times to get the hang of it. The following examle will hel show you how. Examle 1.1. Solve the congruence x 2 2 (mod 7 3 ). Answer: It is easy to solve x 2 2 (mod 7) by trying all the values modulo 7. We get that x 3, 4 (mod 7). Note that obviously if u is a solution then u is also a solution. Next we solve x 2 2 (mod 7 2 ). Note that any solution must also satisfy x 2 2 (mod 7) and so x 3, 4 (mod 7). Suose first that x 3 (mod 7). Then x = 3 + 7y where y is an integer. Substituting in x 2 2 (mod 7 2 ) we obtain or equivalently 9 + 42y + 49y 2 2 (mod 7 2 ) 7(1 + 6y) 0 (mod 7 2 ) or equivalently 1 + 6y 0 (mod 7), so y 1 (mod 7), so we obtain that x = 3 + 7y 3 + 7 = 10 (mod 7 2 ). Similarly, if x 4 (mod 7) then x 39 (mod 49) (which is the same as 10 modulo 49). Now let us solve x 2 2 (mod 7 3 ). Then x 10, 39 (mod 7 2 ). Suose first x 10 (mod 7 2 ). Then x = 10 + 7 2 z for some integer z. Hence 100 + 2 10 7 2 z + 7 4 z 2 2 (mod 7 3 ). Note 100 2 = 98 = 2 7 2. Thus 7 2 2(1 + 10z) 0 (mod 7 3 ). This is equivalent to 1 + 10z 0 (mod 7) which gives z 2 (mod 7). Hence x = 10 + 7 2 z = 108 (mod 7 3 ). Similarly starting form x 39 (mod 7 2 ) would give x 235 (mod 7 3 ). In the above examle, we note that to obtain a solution modulo 7 2 we had to add 7y = 1 7 and to obtain a solution modulo 7 3 we had to add a 7 2 z = 2 7 2. We can continue this calculation and write u our solutions in the following suggestive manner:

TOPICS IN NUMBER THEORY 3 m solutions to x 2 2 (mod 7 m ) 1 ±3 2 ±(3 + 7) 3 ±(3 + 7 + 2 7 2 ) 4 ±(3 + 7 + 2 7 2 + 6 7 3 ) 5 ±(3 + 7 + 2 7 2 + 6 7 3 + 7 4 ) We are writing solutions as a series in owers of 7 with coefficients between 0 and 6. This suggests very much an analogy with decimal exansions. We immediately begin to wonder if the series converges in any sense. Of course it does not converge in the sense of 1st year analysis as the owers of 7 are tending to infinity. However we will change our notion of large and small to make it converge. 2. -Adic Absolute Value Before we define the -adic absolute value, it is worth recalling ord and its roerties. Remember that if is a rime and α is a non-zero rational, then ord (α) is the unique integer such that α = ord(α) a, a, b Z, a, b. b We defined ord (0) = +. Recall also that one formulation of the Unique Factorization Theorem says that any non-zero rational α can be written as (2) α = ± P ord(α), where P is the set of all rimes. Of course only finitely many of the exonents ord (α) are non-zero, so the roduct makes sense. Definition. Let be a rime and α a non-zero rational number. We define the -adic absolute value of α to be α = ord(α). We define 0 = 0 which is consistent with our convention that ord (0) = +. Examle 2.1. Let α = 50/27. Then 2 1 = 2 3 3 = 3 α = 5 2 = 5 1 2, 3, 5. Now evaluate P α. What do you notice.

4 SAMIR SIKSEK Examle 2.2. Notice that r = r, so owers of with ositive exonent are actually small. It now looks likely that the series where we seems to be exanding 2 as a owerseries in 7 does converge. We will come to that soon, but first we need some roerties of the -adic absolute value. Theorem 2.1. Let be a rime and α, β Q. Then (i) α 0. Moreover, α = 0 iff α = 0. (ii) αβ = α β. (iii) α + β max{ α, β }, with equality if α β. Inequality (iii) is called the ultrametric inequality. Notice that it imlies the triangle inequality α + β α + β but is actually much stronger. Proof of Theorem 2.1. We ll leave (i) and (ii) as exercises. (iii). Recall the following roerty of ord : (3) ord (α + β) min{ord (α), ord (β)} with equality if ord (α) ord (β). Write r = ord (α), s = ord (β), t = ord (α + β), and suose that r s. Then t min(r, s) = r. Hence α + β = t r = max{ r, s } = max{ α, β }. Let s do Now suose that α β. Then r s which means that r s. Hence ord (α) ord (β) and we have equality in (3). Hence t = r and so α + β = t = r = max{ r, s } = max{ α, β }. Examle 2.3. The triangle inequality is true for the -adic absolute value, so everything you roved reviously for the ordinary absolute value using the triangle inequality also holds for the -adic absolute value. But the ultrametric inequality is much stronger. Notice the following striking consequence of the ultrametric inequality. Let C be a constant and a rime. Consider the set {α Q : α C}. This is a disc centred at the origin. However, the ultrametric inequality tells us that if we add two elements in this disc, we stay inside it. Comare this with what haens if you add two elements of the disc in the comlex lane {α C : α C}.

TOPICS IN NUMBER THEORY 5 Here it is easy to add two elements in the disc so that you leave the disc. The triangle inequality for the usual absolute value will tell you that if α C and β C then α + β 2C, so you can see that the ultrametric inequality is much stronger than the triangle inequality. Theorem 2.2. (The Product Formula) Let α be a non-zero rational number. Then α P α = 1, where P is the set of rimes. Proof. Prove this using (2). Notice that all but finitely many terms in the roduct are 1, so the roduct makes sense. 3. Convergence Definition. We say that the series {a n } n=1 converges -adically to a Q if lim a n a = 0. n We can also exress this in terms of esilons: the series {a n } n=1 converges to a Q if for every ɛ > 0, there is some N such that for all n N, we have a n a < ɛ. A series j=1 a j converges -adically if the sequence of artial sums s n = n j=1 a j converges -adically. Examle 3.1. Let a Q. It is easy to see that the constant sequence {a} n=1 converges -adically to a. Examle 3.2. The sequence { n } n=1 converges to 0 -adically since n 0 = n 0 as n. Examle 3.3. Consider 5-adically the series The n-th artial sum is 1 + 5 + 5 2 + 5 3 +. s n = 1 + 5 + + 5 n 1 = 5n 1 5 1 = 5n 4 1 4. As 5 n 0, it seems that the sequence of artial sums is converging to 1/4. Let s check this: s n ( 1/4) 5 = 5 n /4 5 = 5 n 0 as n. Hence {s n } n=1 converges 5-adically to 1/4 and we can write 1 + 5 + 5 2 + = 1 4.

6 SAMIR SIKSEK Examle 3.4. Now consider 7-adically the same series 1 + 5 + 5 2 + 5 3 +. Now the artial sums are exactly the same as before, and find that s n ( 1/4) 7 = 5 n /4 7 = 1 1 as n. This shows that the series does not 7-adically converge to 1/4. Does it converge to something else, or not converge at all? We ll answer this question shortly. Definition. A sequence {a n } n=1 of rational numbers is -adically null if it -adically converges to 0. A sequence {a n } n=1 of rational numbers is -adically Cauchy if lim a m a n = 0. m,n Examle 3.5. As we saw reviously, the sequences {0} n=1 and { n } n=1 converge -adically to 0 and so are both null. The following lemma will give us lots of examles of Cauchy sequences. Lemma 3.1. If the sequence of rational numbers {a n } n=1 converges -adically then it is -adically Cauchy. Proof. Suose {a n } converges -adically to a Q. Then lim n a n a = 0. Now a m a n = (a m a) (a n a) = max{ a m a, a n a } using the ultrametric inequality. Hence a m a n 0 as m, n, which is what we wanted to rove. Notice that the roof is almost the same as the roof you saw in first-year analysis with the usual absolute value. The only difference is that the triangle inequality is relaced by the ultrametric inequality. What about the converse of Lemma 3.1. Does every -adically Cauchy sequence of rationals converge to a rational number? If you recall our earlier examle where we were solving x 2 2 (mod 7 n ), we seemed to be constructing a 7-adically Cauchy sequence that converges to 2 which is not rational. So it seems that the converse of Lemma 3.1 does not hold unless we relace the rationals by something bigger. We know that with the usual absolute value a sequence is Cauchy if and

TOPICS IN NUMBER THEORY 7 only if it converges; but here we are talking about real numbers, not just rational numbers. For examle, you that the sequence ( a n = 1 + 1 ) n n is a Cauchy sequence of rational numbers that converges to e which is not rational but real. But what is a real number? The best way to define real numbers is to say that a real number is simly a Cauchy sequence of rational numbers! Think about it. This motivates our next definition. Definition. A -adic number α is a -adically Cauchy sequence {a n } n=1 of rational numbers. We write Q for the set of -adic numbers. We identify Q as a subset of Q via the ma (4) Q Q, a {a} n=1. Let s go back to the reals for a moment to make sure that our definition makes sense. We said that a real number is simly a Cauchy sequence of rationals. So e is just the sequence (1 + 1/n) n. But there are other sequences converging to e. For examle, take the artial sums of the series 1 + 1 1! + 1 2! + 1 3! +. So to say that a real number is a Cauchy sequence seems an ambiguous way to define real numbers. However, the ambiguity disaears as soon as we adot the convention that two Cauchy sequences define the same real number if their difference is a null sequence. We do the same in the -adic setting. Definition. We say that two -adic numbers {a n } and {b n } are equal if the difference {a n b n } is -adically null. Examle 3.6. Via the identification (4) we think of 0 Q to be the same as the zero sequence {0} in Q. Now the { n } and {0} are both -adically null sequences and we have that 0 = {0} = { n } = any null sequence of rationl numbers. Lemma 3.2. Suose that the sequence of rational numbers {a n } converges -adically to a Q. Then in Q lim n a n = a = {a n } n=1. Proof. What is the lemma saying? There is no doubt that lim n a n = a. Now a Q and via the identification (4) we can write a = {a} n=1. So what we re asked to rove that the sequences {a n } n=1 and {a} n=1

8 SAMIR SIKSEK are the same in Q. In other words, they differ by a -adically null sequence. But this true: lim n a n a = 0 as {a n } n=1 converges to a -adically. This comletes the roof. Lemma 3.2 gives a hint of how to define limits of -adically Cauchy sequences that don t seem to have a rational limit. Definition. Suose that {a n } n=1 is a -adically Cauchy sequence of rational numbers. We define the limit lim n a n = {a n } n=1. There is no misrint in this definition! A -adically Cauchy sequence converges to a -adic number that haens to be the sequence itself. This solves the convergence roblem and, by Lemma 3.2, is consistent with the case where the sequence does converge to a rational number. It might be said that this is a cowardly way of solving the issue of -adically Cauchy sequences for which there is no rational limits. But mathematics is full of such cowardice. For examle, to square-root 2 we introduce the symbol 2 and work with it. Everytime we square this symbol we relace it with 2. This does not tell us what the square-root of 2 is, but is a convenient sychological way of avoiding answering the question. Likewise, the only difficulties with acceting the above definition are urely sychological, and at any rate, it is rather late in the term to dro MA3H1 and take u something else. 4. Oerations on Q Of course Q would not be very interesting if it was a set with no additional structure. In fact we can define addition and multilication on Q in a natural way: {a n } + {b n } = {a n + b n } and {a n } {b n } = {a n b n }. One must check that these oerations are well-defined. For a start we want to make sure that the sequences {a n + b n } and {a n b n } are -adically Cauchy so that we are staying in Q. We also want to check that if {a n } and {a n} differ by a -adically null sequence and if {b n } and {b n} differ by a -adically null sequence then {a n + b n } and {a n + b n} differ by a -adically null sequence and {a n b n } and {a nb n} differ by a -adically null sequence. These we ll leave as relatively easy exercises. We also want to check that the usual roerties of addition and multilication hold (commutativity, associativity, distributivity of multilication over addition); again these are easy exercises.

TOPICS IN NUMBER THEORY 9 What about division? Here there is a slight difficulty. We might want to define {a n }/{b n } = {a n /b n }. Of course we will exclude the case when {b n } is -adically null. But even if {b n } is not null, it might contain some zeros. We might then say, ignore them, after all ignoring finitely many terms in a sequence is not going to affect its limit. But what if {b n } has infinitely many zeros? Well that can t haen and to show that we need a lemma. Lemma 4.1. Let {a n } be a sequence of rational numbers that is - adically Cauchy. Then the sequence { a n } converges to some element in the set {0} { r : r Z}. Proof. Note that the convergence we re talking about in the second sentence of the lemma is covergence with resect to the usual absolute value. Now certainly a n is in the set {0} { r : r Z}, and it s easy to see that any Cauchy subsequence of {0} { r : r Z} must actually converge to some element of this set. Thus all we have to show is that { a n } is Cauchy with resect to the usual absolute value. Now it is an easy exercise to check that a b a b. Hence 0 am a n am a n. As {a n } is -adically Cauchy, lim m,n a m a n = 0. Hence by the Sandwich Theorem, lim am a n = 0. m,n This shows that the sequence { a n } is Cauchy with resect to the usual absolute value and comletes the roof. Lemma 4.2. Let {b n } be a -adically Cauchy sequence of rational numbers that is non-null. Then the sequence contains at most finitely many zero elements. Proof. By the revious lemma, b n has a limit, which is either zero, or a ower of. However, as {b n } is non-null, this limit must be nonzero. Now if the sequence contains infinitely many zeros then { b n } contains infinitely many zeros and hence a subsequence converging to zero. This contradicts that fact that { b n } converges to a non-zero limit. The above lemma allows us to define division. If {a n } and {b n } are elements of Q and {b n } 0 (i.e. non-null) then there is some N such that for n N, b n 0 and we define {c n } = {a n }/{b n } by assigning

10 SAMIR SIKSEK c n randomly for n < N and letting c n = a n /b n for n N. Note that {c n } {b n } agrees with {a n } excet for finitely many terms and so their difference is null; in other words {c n } {b n } = {a n } in Q. Theorem 4.3. Q is a field containing Q as a subfield. Proof. The roof is an easy but slightly lengthy verification which we leave as an exercise. We can extend the -adic absolute value to Q as follows. Definition. Let α Q be reresented by the -adically Cauchy sequence of rationals {a n }. We define the -adic absolute value of α by α = lim a n. n Note that the limit exists by Lemma 4.1 and is equal to some element of the set {0} { r : r Z}, but we still need to show that α is well-defined in the sense that if {a n } = {b n } in Q then lim a n = lim b n. n n The assumtion that {a n } = {b n } in Q means that the difference {a n b n } is -adically null. Now by the triangle inequality a n = (a n b n ) + b n a n b n + b n. Hence lim a n lim a n b n + lim b n. n n n As {a n b n } is null, lim n a n b n = 0, so lim a n lim b n. n n Swaing the rôles of the as and bs in the above argument gives lim b n lim a n. n n Hence lim b n = lim a n, n n which shows that α is well-defined for α Q. 4.1. Proerties of -adic absolute value. Now that we have defined the -adic absolute value on Q, it is natural to ask if it has the same roerties it had on Q, and it does. Theorem 4.4. Let be a rime and α, β Q. Then (i) α 0. Moreover, α = 0 iff α = 0. (ii) αβ = α β. (iii) α + β max{ α, β }, with equality if α β.

TOPICS IN NUMBER THEORY 11 Proof. The roof follows by choosing -adically Cauchy sequences of rationals reresenting α and β and then using the definition of in terms of these sequences and Theorem 2.1. We leave this as an exercise. 5. Convergence of Series The ultrametric inequality has a dramatic effect of making the convergence of series very easy to check. Theorem 5.1. Let be a rime. The series j=1 a j converges - adically if and only if lim j a j = 0. We know that with the usual absolute value the theorem is true only in the left to right direction. The famous counterexamle being the harmonic series 1 + 1 2 + 1 3 + 1 4 +, which diverges even though lim j 1/j = 0. Working -adically, we don t need any of the comlicated convergence tests of first-year analysis the theorem makes it all very easy! Proof of Theorem 5.1. Suose that lim n a n = 0. All we have to do is to show the the sequence of artial sums s n = n j=1 a j is Cauchy. A Cauchy sequence converges to some element of Q (which haens to equal the sequence itself). Now suose m > n. Then s m s n = a n+1 + a n+2 + + a m = max n+1 j m a j, by the ultrametric inequality. For any ɛ > 0, there is some N such that if j N then a j < ɛ. Hence if m, n N then s m s n < ɛ, roving that the sequence {s n } is -adically Cauchy. 6. -adic Integers Definition. The set of -adic integers Z is defined by Z = {α Q : α 1}. So Z looks like a circle of radius 1 centred at the origin. In this sense, the following theorem is striking. Theorem 6.1. Z is a ring and contains Z as a subring. Proof. Note first that if a Z then ord (a) 0 and so a = ord(a) 1. Hence Z Z. To comlete the roof we must show that Z is a ring. You don t have to know anything about rings excet the definition. Note that Z Q and Q is already a field. So we have to show that Z is

12 SAMIR SIKSEK closed under addition and multilication (it already contains 0 and 1). But if α, β Z then αβ = α β 1 1 = 1, and α + β max{ α, β } max{1, 1} = 1. Hence αβ, α + β Z. Lemma 6.2. If {a n } is a -adically Cauchy sequence with a n Z then lim n a n is in Z. Conversely, any α Z is the limit of such a sequence. Proof. Suose {a n } is a -adically Cauchy sequence with a n Z and let α = lim n a n. Then a n 1 and so α = lim n a n 1 which shows that α Z. The converse is harder. Suose α Z. Now Z Q and so α = lim n a n where {a n } is a -adically Cauchy sequence of rational numbers, but there is a riori no reason for the a n be integral. We will construct a -adically Cauchy sequence {b n } where the b n are in Z and {a n b n } is a -adically null sequence. Then α = lim n b n as required. Now lim a n = α 1. n Consider ord (a n ). If there are infinitely many n such that ord (a n ) 1 then there are infinitely many n such that a n and this contradicts the above. Hence there is some N such that ord (a n ) 0 for all n N. So we can write a n = u n v n where u n, v n Z, with v n. Since v n, we know that v n is invertible modulo n. Let v n w n 1 (mod n ), where w n Z and write b n = u n w n Z. Then a n = u n /v n u n w n = b n (mod n ) and so a n b n n. This comletes the roof. 7. Hensel s Lemma Revisited Corollary 7.1. Let be a rime. Let f(x) be a olynomial with integer coefficients. Suose that there is an integer a such that f(a) 0 (mod ), f (a) 0 (mod ). (i) Then there is a sequence {a m } m=1 such that a 1 = a, and (5) f(a m ) 0 (mod m ), a m+1 a m (mod m ).

TOPICS IN NUMBER THEORY 13 (ii) The sequence {a m } m=1 converges to α Z and α satisfies f(α) = 0. Proof. We start with a 1 = a and aly Hensel s Lemma (Theorem 1.2) with m = 1. We obtain an a 2 such that f(a 2 ) 0 (mod ), a 2 a 1 (mod ). Suose now we have constructed a 1, a 2,..., a k to satisfy (5). that Note a k a k 1 (mod k ), a k 1 a k 2 (mod k 1 ),..., a 2 a 1 (mod ). Then certainly a k a 1 = a (mod ) and so f (a k ) f (a) 0 (mod ). We can now aly Hensel to obtain a k+1. This comletes the roof of (i). Let us rove (ii). We want to rove that {a n } converges -adically. Write b 1 = a 1, b 2 = a 2 a 1, b 3 = a 3 a 2,.... Then a n = b 1 + b 2 + + b n. Hence the sequence {a n } converges -adically iff the series b m converges -adically. But b m = a m a m 1 (m 1), since a m a m 1 (mod m 1 ). Thus {a n } converges in Q. As a n Z, we know from the revious lemma that {a n } converges to some α Z. Now 1 ( ) f(α) = f lim a n = lim f(a n ) = 0, n n since f(a n ) n. Corollary 7.2. Let b Z be non-zero and an odd rime. ( ) Then b is a square in Z if and only if b = 2r c c where r Z and = 1. ( ) Proof. Suose b = 2r c c where r Z and = 1. All we have to ( ) show is that c is a square in Z. Let f(x) = X 2 c. Since = 1, there is some a 0 (mod ) such that c a 2 (mod ). Hence f(a) 0 (mod ), f (a) = 2a 0 (mod ). By the above corollary to Hensel s Lemma, there is some α Z such that f(α) = 0, so α 2 = c as required. 1 For a olynomial f, the equality c is an easy exercise. f (lim n a n ) = lim n f(a n)

14 SAMIR SIKSEK Let us rove the converse. Suose b is a square in Z, say b = β 2 where β Z. Write b = s c where b. Then s = b = β 2. Now β is a ower of, so s is even, say s = 2r. So b = 2r c is a square in Z. Hence c is a square in Q. Say c = γ 2 with γ Q. But 1 = c = γ 2, so γ Z. Let {a n } be a Cauchy sequence of integers converging to γ. Then {a 2 n} coverges to c. Hence there is some N such that for n N, a 2 n c 1, which can be rewritten as ( This shows that c c a 2 n (mod ). ) = 1. Deciding which integers are squares in Z 2 is a little more tricky, and needs a stronger version of Hensel s Lemma. Theorem 7.3. (Hensel s Lemma Strong Version) Let f(x) Z[X]. Let be a rime and m 0. Suose a Z and write k = ord (f (a)). Suose that m 1 and Then there exists b Z such that f(a) 0 (mod m+2k ). b a (mod m+k ), f(b) 0 (mod m+1+2k ). Proof. Write b = a + m+k y where y Z is yet to be determined. By Taylor s Theorem, f(b) = f(a) + m+k y f (a) + 2m+2k (integer). We can write f(a) = m+2k c, where c, d Z and d. Then f (a) = k d f(b) = m+2k (c + dy) + 2m+2k (integer). To comlete the roof of the theorem, all we have to do is to choose y Z so that (c + dy). In other words we want dy c (mod ), and we can do this as d is invertible modulo ; this is where we use the fact that d. We can now imrove on Corollary 7.1 as follows.

TOPICS IN NUMBER THEORY 15 Corollary 7.4. Let be a rime. Let f(x) be a olynomial with integer coefficients, and k 0 an integer. Suose that there is an integer a such that f(a) 0 (mod 2k+1 ), ord (f (a)) = k. (i) Then there is a sequence {a m } m=1 such that a 1 = a, and f(a m ) 0 (mod m+2k ), a m+1 a m (mod m+k ). (ii) The sequence {a m } m=1 converges to α Z and α satisfies f(α) = 0. The roof is an easy modification of the roof of Corollary 7.1 using the strong version of Hensel s Lemma and we leave it as an exercise. Corollary 7.5. Let b Z be non-zero. Then b is a square in Z 2 if and only if b = 2 2r c where c 1 (mod 8). Proof. The roof is almost the same as the roof of Corollary 7.2 but uses Corollary 7.4 instead. Let us show that if c 1 (mod 8) then c is a square in Z 2. Let f(x) = X 2 c. Then f(1) 0 (mod 2 3 ), ord 2 (f (1)) = 1, so taking k = 1 in Corollary 7.4 shows that there is some α Z 2 such that f(α) = 0. Then c = α 2 as required. The rest of the roof is an exercise. 8. The Hasse Princile Let f Z[X 1,..., X n ]. We want to know if the equation f(x 1,..., X n ) = 0 has a solution in integers. As Z Z for all rimes and as Z R we know the following: f = 0 has a solution in Z n = { f = 0 has a solution in Z n for all rimes f = 0 has a solution in R n. Is the converse statement true? The converse statement is called the Hasse Princile. It is true for many classes of olynomials, and for these classes of olynomials we say that the Hasse Princile holds. But it is false for many other classes of olynomials and for those we say that the Hasse rincile fails. Here is a counterexamle to the Hasse rincile for olynomials in 1 variable. Examle 8.1. Let f(x) = (X 2 2)(X 2 17)(X 2 34). Show that f(x) = 0 is a counterexamle to the Hasse rincile. Answer: Basically we are asked to show that f(x) = 0 has solutions in Z for all rimes and in R, but has no solutions in Z. It clearly and

16 SAMIR SIKSEK has solutions in R, which are ± 2, ± 17, ± 34, and clearly it has no solutions in Z as none of these roots are integral. Now 17 1 (mod 8) and so by Corollary 7.5 17 = α 2 for some ( α Z 2. Then f(α) = 0, so f(x) = 0 has a solution in Z 2. Also 2 17) = 1, so 2 is a square in Z17 by Corollary 7.2, and so f(x) = 0 has a solution in Z 17. Suose that 2, 17. We want to show that f(x) = 0 has a solution in Z. Equivalently, we want to show that at least one of 2, 17, 34 is a square in Z. Suose that 2, 17 are not squares in Z. By Corollary 7.2, ( ) ( ) 2 17 = 1, = 1. But multilying we obtain ( ) 34 = ( 2 ) ( ) 17 = 1. Hence 34 is a square in Z and so f(x) has a solution in Z. This comletes the roof. 8.1. A Bivariate Counterexamle to the Hasse Princile. Let f(x, Y ) = 2Y 2 X 4 + 17. We claim that f(x, Y ) = 0 is a counterexamle to the Hasse Princile. Clearly f(x, Y ) = 0 has solutions in R; for examle (X, Y ) = ( 4 17, 0). The standard roof that f(x, Y ) = 0 has solutions in Z for all rimes uses some advanced results and we shall omit it. How do we show that f(x, Y ) = 0 has no solutions in Z? Suose x, y Z such that f(x, y) = 0. Clearly y 0, and f(x, y) = f(x, y), so we can assume that y > 0. Moreover, from 2y 2 = x 4 17, if 17 x or 17 y then 17 divides both and 17 2 17 which is imossible. Thus 17 x, 17 y. Suose is an odd rime divisor of y. Then x 4 17 (mod ). So certainly 17 is a square modulo, and we can write ( ) 17 = 1. By the Law of Quadratic Recirocity (as 17 1 (mod 4)), ( 17) = 1. Write y = 2 r r i i where the i are distinct odd rime divisors of y. Then ( i ) = 1 17

TOPICS IN NUMBER THEORY 17 by the above, and ( ) 2 = 1, 17 as 17 1 (mod 8). Thus ( ( ) r y 2 ( i ) ri = = 1. 17) 17 17 Hence y z 2 (mod 17). From 2y 2 = x 4 17 we obtain that 2z 4 x 4 (mod 17). We said earlier that 17 x, y, so z 0 (mod 17). Thus z is invertible modulo 17. Let w xz 1 (mod 17). Then w 4 2 (mod 17). However, by trying the values w = 0, 1,..., 16 we find that the congruence w 4 2 (mod 17) does not have any solutions, giving us a contradiction. Samir Siksek, Mathematics Institute, University of Warwick, Coventry, CV4 7AL, United Kingdom E-mail address: samir.siksek@googlemail.com