PUMaC 2015 · 加试 · 第 7 题

Problem 7.3 (Galois Automorphisms; 2, 8 ) . Let f ( x ) = x − 70 x + 25 . a) What are the roots of f ( x ) ? b) You are given that f ( x ) is a nice enough polynomial that Q [ x ] / ( f ( x )) is a Galois. Give four examples of Galois automorphisms—an isomorphism from one set to itself. (You may just tell us where each root is sent for each automorphism). √ √ √ √ √ √ √ √ Proof. a) We have that the roots of f are 20 + 15, 20 − 15, − 20 + 15, and 20 − 15. √ √ √ √ b) Let α = 20 + 15, and β = 20 − 15, here are all of them: σ ( α, − α, β, − β ) 7 → ( α, − α, β, − β ) , 1 σ ( α, − α, β, − β ) 7 → ( − α, α, − β, β ) , 2 σ ( α, − α, β, − β ) 7 → ( β, − β, α, − α ) , 3 σ ( α, − α, β, − β ) 7 → ( − β, β, − α, α ) . 4 The motivation for this comes from our example of Q [ i ]. A very natural thing to try is sending i 7 → − i . If we try a similar method here, we may guess and check how the signs flip. Before moving on though, we can go even further than what we have done here; we can adjoin multiple roots of many different polynomials at once to Q . For example, a classic example you may see if you study √ 3 2 3 mathematics more is adjoining to Q the roots of both x + x +1 and x − 2 at the same time to yield Q [ ω, 2] for ω a primitive 3rd root of unity. 8 Elliptic Curves and Galois Theory Galois theory is incredibly rich, but unfortunately there are details we must omit about the subject for the sake of time, and this is sufficient background; we can now relate Galois theory and elliptic curves. Suppose you have a general elliptic curve E and a general point P on it. We define a k -division point of P as some 2 point Q on E such that kQ = P . A fact of elliptic curves is that there are exactly k such k -division points in C (the coordinates of Q may be complex numbers), but likely many of them won’t be rational points. But examine for a moment such a non-rational point β such that kβ = P . Suppose we take the x and y k k coordinates of β and adjoined them to Q . What would we get? Going further, suppose we took all such k β such that kβ = P and adjoined to Q all of the x and y coordinates of these division points. We get k k i i some large field extension we label K . This directly gives us a way to use Galois theory in a way that gives k us information about our initial E and P . Take on faith that K / Q is indeed a Galois extension, and examine some Galois automorphism σ of this k extension; it acts on all these coordinates we just adjoined. Let ( a, b ) be one of the k -division points; then σ (( a, b )) = ( σ ( a ) , σ ( b )). Now we have a curious situation: we have found a Galois automorphism that acts on points on an elliptic curve! Weee. That these Galois automorphisms act on the set of k -division points is important. Can you visualize how they are acting? These functions send coordinate pairs, essentially vectors, to other vectors. This is quite similar to how matrices act on vectors! In fact, this leads to what is called a Galois representation: a homomorphism from the Galois group to a linear algebra construct. Here we become guilty of omitting some details, but it would take too much work to present in full rigor. But please take these two propositions to be true. 28 PUMaC 2015 Power Round Section 8 page 29 2 3 Proposition 33. Let E : y + y = x − x , P = (0 , 0) , and let K be the field described above, namely the k field extension of Q by adjoining all the coordinates of the k -division points of P . Then there is a surjective k k 2 k homomorphism from the Galois group Gal( K / Q ) to AGL ( Z / 2 Z ) = ( Z / 2 Z ) o GL ( Z / 2 Z ) . Denote k 2 2 k this by ϕ : Gal( K / Q ) → AGL ( Z / 2 Z ) . k 2 A quick word on notation here. We defined the semidirect product in Proposition 16, and here we see an k k 2 example of one. For an element of such a group AGL ( Z / 2 Z ), we will write it as ( ~ v, M ) where ~ v ∈ ( Z / 2 Z ) 2 k and M ∈ GL ( Z / 2 Z ). 2 2 3 Proposition 34. Let E : y + y = x − x and P = (0 , 0) be a point on E . Let be a prime larger than 37 . Then P has odd order in E ( F ) , the reduction of the curve to this finite field, if and only if for all k ∈ N , [ ] K / Q k k ∃ ( ~ v, M ) ∈ im( ) ⊂ AGL ( Z / 2 Z ) such that ~ v lies in the column space of M − I (the column space of 2 k 2 a matrix such as ( M − I ) is the set { ( M − I ) · ~ v, ~ v ∈ ( Z / 2 Z ) } ), where im is the image under the mapping [ ] K / Q k defined in Proposition 33. (This symbol is called the Artin symbol, which we unfortunately do not have the time to define thoroughly. It is, however, a subset of Gal( K / Q ) . Hint: most important for you, k the contestant, is Proposition 37.) Let’s parse this last proposition; recall from the interlude that prime divides some term of the sequence if and only if P has odd order in E ( F ). We saw earlier as well in problem 3.6 that this happens if and only i if for all integers i , there exists some element β ∈ E ( F ) such that 2 · β = P . Finally, this condition is i i equivalent to the latter part of the above proposition. If you are familiar with Galois theory, as a hint of why this might be true, the fact that such a β exists implies that it is fixed by the Fr¨ obenius automorphism. i k k 2 This leads to the fact that AGL ( Z / 2 Z ) acting on ( Z / 2 Z ) fixes some element ~ x . Thus ( ~ v, M )( ~ x ) := 2 M · ~ x + ~ v = ~ x = ⇒ ( M − I ) ~ x = − ~ v . From here it’s easy to see that ~ v ∈ im( M − I ) ⇔ − ~ v ∈ im( M − I ). k Definition 35. Let ( ~ v, M ) ∈ AGL ( Z / 2 Z ) be an element of the affine general linear group. We call ( ~ v, M ) 2 a ruminative element if ~ v is in the column space of M − I . For the observant reader, another way to describe this element ( ~ v, M ) is to say that M fixes a vector k 2 k k 2 ~ x ∈ ( Z / 2 Z ) where the action of AGL ( Z / 2 Z ) on ( Z / 2 Z ) is as described above. 2 Thus we have come from an original question about primes dividing terms of a sequence to a question about the column space of matrices. This latter is something that we can much more likely solve directly. (In general, this is a useful strategy. Linear algebra is a subject that is very well understood compared to other mathematical subjects. This is the motivation behind group representations, for example. In fact, that linear algebra is so well understood has given rise to the half-serious joke of dismissing a problem by saying, “it’s just linear algebra!”) For one final step before we try to use linear algebra to find a fraction, we present the Chebotarev Density Theorem. Theorem 36. Suppose K/ Q is a Galois extension with G := Gal( K/ Q ) where C ⊂ G is a conjugacy class [ ] K/ Q of G . Define π ( x ) := # { p ≤ x : p is a prime that is unramified in K and = C } . Then C p π ( x ) | C | C lim = . x →∞ π ( x ) | G | 29 PUMaC 2015 Power Round Section 8 page 30 (The definition of unramified is unimportant for us. In our specific case, this equivalently means primes that are greater than 37.) As written, this doesn’t seem to necessarily apply to anything we’ve written so far. However, one of the facts that we obscured in our presentation of the two propositions above is that the images of the Galois groups above are in fact images of conjugacy classes. The ultimate result of all of this is the following. ′ Proposition 37. Let π ( x ) denote the number of primes less than x , and let π ( x ) denote the number of [ ] K / Q k primes less than x that divide some term of the Tiger sequence. Let S represent , the conjugacy class k ′ k inside Gal( K / Q ) where the k was such that β such that 2 · β = P . Let S := im( S ) and AGL ( Z / 2 Z ) = k k k 2 k im(Gal( K / Q )) represent the images under the homomorphism ϕ : Gal( K / Q ) → AGL ( Z / 2 Z ) from above. k k 2 Then ′ π ( x ) | S | lim = lim x →∞ k →∞ π ( x ) | Gal( K / Q ) | k ′ | S | = lim . k k →∞ | AGL ( Z / 2 Z ) | 2 S here is equal to the Artin symbol of some prime number ` . In other words, the final fraction we have to compute is the last expression in the proposition above. The ′ ′ k best way to interpret S in the above is that S is the subset of AGL ( Z / 2 Z ) that consists of the ruminative 2 elements. Here is a final note on the previous two sections. We are not fully defining some definitions such as conjugacy classes and the Artin symbol. They are written here for full formality, but are not necessary for ′ you the contestants to fully understand. The only part of the previous parts to take away is that S is the k subset of AGL ( Z / 2 Z ) of elements that are ruminative, and we may calculate the number of such elements. 2 Hint: this interpretation is the only thing you have to take away from sections 7 and 8 in order to solve section 9. 30 PUMaC 2015 Power Round Section 9 page 31 9 Final Fraction Thus we are only left with calculating the density! Here are the final steps. Definition 38. Let v : Z → Z be a function such that v ( n ) = m , where m is the exponent of p in the p p prime factorization of n . For example, v (24) = 3 , v (8) = 0 , and v ( − 25) = 2 . 2 3 5 k Proposition 39. Let M ∈ GL ( Z / 2 Z ) be a matrix such that v (det( M − I )) = r . Then the number of 2 2 2 k − r elements in the column space of M − I is 2 . (Two notes: we do not regard the cases where det( M − I ) = 0 , k and by definition, det( M − I ) is reduced to be the integer n such that 0 ≤ n < 2 and n ≡ det( M − I ) k (mod 2 ) where det( M − I ) is evaluated in Z .)