. Resultants Integral Bases Suppose f and g are polynomials in one or more variables including x) with integer coefficients. Say f = A 0 x m + A x m + + A m, g = B 0 x n + B x n + + B n, with A 0,..., A m and B 0,..., B n polynomials not involving x and with A 0 B 0 0. We recursively define the resultant of f and g with respect to x as 0 and Res x f, g) = A n 0 if m = 0, B0 m if n = 0, ) mn Res x g, f) if 0 < m < n, ) mn B0 n Res x g, h) if 0 < n m, with h = B 0 f A 0 x m n g having lower degree in x than f. If you look closely you can see that A 0 A m 0 0 0 0 A 0 A m 0 0 0 0 A 0 A m 0...... 0 0 0 A 0 A m Res x f, g) = det B 0 B n 0 0 0 0 B 0 B n 0 0 0 0 B 0 B n 0...... 0 0 0 B 0 B n with A 0,..., A m appearing in rows through n and B 0,..., B n appearing in rows n + through n + m. It follows that Res x f, g) is a polynomial in zero or more variables not including x) with integer coefficients. Integral Bases -May-0, 9:5 pm
When fx) and gx) are polynomials in Z[x] we have fx) = λ x α ) x α m ), gx) = µ x β ) x β n ), with α,..., α m, β,..., β n C, and the resultant has the charming property that ) Res x fx), gx) = λ n µ m m n j= k= α j β k ) = λ n m j= gα j). Even better, if fx) and gx) are monic polynomials in Z[x] then Res x fx), Resy gy), t xy )) = Resx fx), n k= t xβ k) ) = m j= n k= t α jβ k ) Z[t], Res x fx), Resy gy), t x y )) = Resx fx), n k= t x β k) ) = m j= n k= t α j β k ) Z[t]. Thus, if α and β are roots of monic polynomials in Z[x] then so are αβ and α + β. Exercise. Show that if α is a root of a monic polynomial in Z[x] and λ Z then λα is a root of a monic polynomial in Z[x]. Example. Let fx) = x, gy) = y + 5. Then ) r y t, x) = Res y gy), t x y 0 5 = det t x = x tx + t + 5, t x r x,y t) = Res x fx), ry t, x) ) 0 0 = det t t + 5 = t + 6t + 9. t t + 5 Integral Bases -May-0, 9:5 pm
. Algebraic Integers An algebraic integer is a root of a monic polynomial in Z[x]. For example, if k is an integer and α = + k + ) then α is a root of x x k, so α is an algebraic integer. Exercise: Exercise. Gauss s Lemma. Show that if fx) is a monic polynomial in Z[x] and gx) and hx) are monic polynomials in Q[x] and fx) = gx) hx) then gx) and hx) belong to Z[x]. Apply Gauss s Lemma to show that the monic) minimal polynomial of an algebraic integer has integer coefficients. The set of algebraic integers belonging to an algebraic number field K we denote by O K or simply O. From the discussion above it follows that O K is both a ring and a Z-module. Suppose β is a nonzero algebraic integer with hx) its minimal polynomial over Q and m the degree of hx), and let λ be a positive integer. Then the minimal polynomial of β/λ is λ m hλx), which will not be in Z[x] if λ m > h0). For example, if β is a root of x x then β/ is a root of x x, but the minimal polynomial of β/ is x x, so β/ is not an algebraic integer. Exercise: Vandermonde Determinants. Show that if then V = v v v n v v v n v v v n... v n v n v n n det V = j<k v k v j ). Let fx) be an irreducible monic polynomial in Z[x] of degree n, let α be a root of fx), and let O be the set of algebraic integers belonging to Qα). Integral Bases -May-0, 9:5 pm
Suppose β O. Then Z[α, β ] = j n, k n Z αj β k O and the elements α j β k, j n, k n, expressed as vectors over Q, can be reduced over Z note!) to give a Z-module basis { ω, ω,..., ω n } for Z[α, β ]. Since {, α, α,..., α n } is a vector-space basis for Qα) over Q there is a matrix T ω Q n n such that ω ω ω. = T ω ω n α α. α n Note that without loss of generality we may construct ω,..., ω n so that T ω is lowertriangular with positive diagonal entries, and so we do. And since it follows that T ω Z n n. α α. α n ω n. ω = T ω ω ω. Exercises. Exploit the fact that T ω is in lower-triangular form with positive diagonal entries to prove the following.. For k =,..., n there exists a positive integer d k such that T ω ) kk = /d k, and in particular d =. Use the fact that α k can be expressed uniquely as a Z-linear combination of ω,..., ω n.). d k is a multiple of d k for k =,..., n. Use the fact that if λ k d k + µ k d k = gcdd k, d k ) for integers λ k and µ k then λ k αω k +µ k ω k can be expressed uniquely as a Z-linear combination of ω,..., ω n.). d k ω k Z[α] for k =,..., n. Use the fact that αω k can be expressed uniquely as a Z-linear combination of ω,..., ω n.) Integral Bases -May-0, 9:5 pm
Decomposing fx) into linear factors as fx) = x α ) x α n ) we define tr hα) ) = hα ) + hα ) + + hα n ) for hx) Q[x]. If hα) O then ) Res x fx), t hx) = t hα ) ) t hα n ) ) = t n tr hα) ) t n + ± hα ) hα n ) Z[t] and therefore tr hα) ) Z. Now let A = α α α α n α α α αn, A T =... α n α n α n αn n α α α n α α α n α α α n... α n αn αn n so that A A T = tr) trα) trα ) trα n ) trα) trα ) trα ) trα n ) trα ) trα ) trα ) trα n+ )... trα n ) trα n ) trα n+ ) trα n ) and det AA T = j k α j α k ) = disc f = ) nn )/ Res x fx), f x)). Integral Bases 5 -May-0, 9:5 pm
Treating ω = ω α),..., ω n = ω n α) as polynomials in α we will write ω jk for ω j α k ). We define W = ω ω ω ω n ω ω ω ω n ω ω ω ω n, W T =... ω n ω n ω n ω nn ω ω ω ω n ω ω ω ω n ω ω ω ω n... ω n ω n ω n ω nn so that W W T = trω ω ) trω ω ) trω ω ) trω ω n ) trω ω ) trω ω ) trω ω ) trω ω n ) trω ω ) trω ω ) trω ω ) trω ω n )... trω n ω ) trω n ω ) trω n ω ) trω n ω n ) Z n n and hence det W W T Z. On the other hand, with d ω = d d n we have W = T ω A, det W W T = dett ω A A T T T ω) = det T ω ) det A) = d ω disc f, disc f = d ω det W W T, T ω = det T ω T ω ) = d ω T ω ) d ω Z n n. Proposition. If d is the largest square dividing disc f then Z[α] O d Z[α]. Integral Bases 6 -May-0, 9:5 pm
Example. Let fx) = x x +. Then disc f = 80 = 5, so that d = 6. Among the 5 distinct nonzero representatives of dz[α] modulo Z[α] there are five algebraic integers. ωα) Res x fx), ωx), t x ) α t 7t + + 6 α t t + + α t 0t + 5 + α t 6t + + 5 6 α t t + Row-reduction to lower-triangular form over Z gives 6 5 6 6 and it follows that {, + 6 α } is a Z-basis for O. Integral Bases 7 -May-0, 9:5 pm
Example. Let fx) = x 7x x. Then disc f = 7 = 6, so that d = = 8. Among the 5 distinct nonzero representatives of dz[α] modulo Z[α] there are seven algebraic integers. ωα) Res x fx), ωx), t x ) α + α t 9t 6t + α t 9t + t + α + α t 7t + 6t + α t 7t + t 5 + α t 5t + t + α t 5t + 0t 7 + α + α t t + 5t 7 Row-reduction to lower-triangular form over Z gives and it follows that {, + α, + α } is a Z-basis for O. Integral Bases 8 -May-0, 9:5 pm
. Computing Integral Bases Efficiently Often we need to work with the ring of integers of an algebraic number field and so we need an efficient way to find an integral basis for the field, i.e., a Z-module basis for its ring of integers. It is clear that the method sketched in the examples in the previous section is not efficient.) Let K be an algebraic number field of degree n with ring of integers O. An order of K is a subring of K that contains and is also an n-dimensional Z-module. Note that if K = Qα) with α O then Z[α] is an order of K, as is O itself. Let A be an order of K and let { ω,..., ω n } be a Z-basis for A. Exercises.. Prove that A O, as follows. For an arbitrary element α of A let M α = a a a n a a a n... Zn n a n a n a nn be given by αω j = a j ω + + a jn ω n for j =,..., n and let χ α x) = detxi M α ). Show that χ α x) is a monic polynomial in Z[x] with χ α α) = 0.. Apply the previous result with A = O to show that if β, γ K and β 0 and β γ k O for all k 0 then γ O. Integral Bases 9 -May-0, 9:5 pm
Let p be prime and define O p = { α O pα A }, R p = { β A β k pa for some k }, [R p /R p ] = { γ K γr p R p }. Exercises. For θ K let µ θ denote the minimal polynomial of θ over Q. Prove the following.. γr p R p = p, pγ, pγ,... A = γ O.. [R p /R p ] is an order of K.. A [R p /R p ] O p.. α k po for some k = µ α x) µ pβ x k ) with pβ = α k = µ α x) x m mod p) for some m. 5. µ α x) x m mod p) for some m = α m pa = α k po for some k. 6. R p = { α A α k po for some k }. 7. [R p /R p ] = { γ O γr p A }. Proposition. If A = O p then A = [R p /R p ]. Proof. There exists m such that R m p pa. Choose γ O p A, and while γr p A replace γ γα γr p A. It follows that after m such replacements we would have γ A. Therefore, after at most m replacements we will have γr p A, but with γ / A, so that [R p /R p ] A. Integral Bases 0 -May-0, 9:5 pm
Computing R p. Let q = p log p n. Then q n and, for α A, α k pa for some k α n pa α q pa. Take Then α = x ω + + x n ω n A. α p x p ωp + + xp nω p n x ω p + + x nω p n mod pa) and so Define the matrix T by Then α q x ω q + + x nω q n mod pa). ω q k = T jk ω j. α R p α q pa T jk x k 0 mod p) for j =,..., n. Solving this system of congruences gives R p. Computing T p. The computation of R p becomes burdensome when p is large. Instead, let T p = { α A trαβ) pz for all β A }. For i =,..., n let the matrix W i express the action of ω i on ω,..., ω n, i.e., ω i ω j = W i ) jk ω k for i, j =,..., n, and define the matrix T by T jk = trω j ω k ) = trw j W k ). Then, for α = x ω + + x n ω n A, we have α T p T jk x k 0 mod p) for j =,..., n. Solving this system of congruences gives T p. Integral Bases -May-0, 9:5 pm
Proposition. If p > n then R p = T p. Proof. It is clear that R p T p. Assume p > n and take α T p. Let µ α denote the minimal polynomial of α over Q. Since trα k ) 0 mod p) for all k > 0, it is a consequence of Newton s relations that µ α x) x m mod p), and therefore α R p. Computing [R p /R p ]. Let the matrices W,..., W n be defined as above. Take γ = y ω + + y n ω n QA = K and set C γ = y W + + y n W n. The matrix C γ expresses a Z-basis for γa in terms of ω,..., ω n. If the matrix J expresses a Z-basis for R p in terms of ω,..., ω n then JC γ J gives a Z-basis for γr p in terms of the basis for R p expressed by J, and therefore γr p R p JC γ J Z n n y i JW i J ) Z n n. Solving this set of n relations for the unknowns y,..., y n gives [R p /R p ]. Integral Bases -May-0, 9:5 pm