Algebra. Übungsblatt 12 (Lösungen)

Transcription

1 Fakultät für Mathematik Sommersemester 2017 JProf Dr Christian Lehn Dr Alberto Castaño Domínguez Algebra Übungsblatt 12 (Lösungen) Aufgabe 1 Berechnen Sie Minimalpolynome f i, i = 1,, 4, der folgenden komplexen Zahlen 1 α 1 = α 2 = α 3 = α 4 = Es bezeichne k i den Zerfällungskörper von f i Man berechne die Grade [k i : Q] für i = 1,, 4 Ist Q(α i ) = k i? Für welche i ist Q Q(α i ) eine normale Körpererweiterung? Lösen sie dieselbe Aufgabe, wenn wir den Grundkörper Q durch k = F 5 ersetzen Hierbei sind die α i als im algebraischen Abschluss k von k liegend zu verstehen Was sind nun die Grade bzw Beweis In this exercise we can learn that the meaning of expressions like 2 goes beyond real numbers If we are working over Q, then the symbol above means the usual and known positive square root of two, but in a general field k, it means one of the two elements of k such that its square is two What could be 2 in F 7? 1 From the lecture (example IV113a) we already know that the minimal polynomial of α 1 is f 1 (x) = x 4 4x 2 +2 (we can see its irreducibility in Q[x] applying Eisenstein s criterion with p = 2, for instance) f 1 is a biquadratic polynomial, so it is easy to find its roots Namely, they are x 1 = 2 + 2, x 2 = 2 + 2, x 3 = 2 2, x 4 = 2 2 We have the obvious inclusions Q Q(α 1 ) k, where [Q(α 1 ) : Q] = 4 Clearly x 2 belongs to Q(α 1 ), but it is not so clear that x 3 and x 4 are there too However, we

2 could notice that α 1 x 3 = 4 2 = 2 = α 2 1 2, so x 3 Q(α 1 ) and then, k 1 = Q(α 1 ), having in conclusion that [k 1 : Q] = 4 and the extension Q Q(α 1 ) is normal Over F 5, the story does not change that much We can check that the polynomial x 2 2 has no roots in F 5, so 2 / F 5 and then, the calculation of the minimal polynomial would be approximately the same as before, getting f 1 = x 4 + x as a candidate In order to see that it is irreducible, since it has no roots in F 5 we can assume the existence of two degree two polynomials x 2 +ax+b and x 2 +cx+d (monic because so is f 1 ) such that their product is f 1 We obtain the following equations: a + c = 0 b + ac + d = 1 ad + bc = 0 bd = 2 From the first equation we have c = a If a = 0, then from the second d = 1 b and so, b(1 b) = 2 from the fourth, but that is impossible Then a cannot vanish, so we deduce from the third equation that b = d, but then in the fourth we would have that b 2 = 2, which we already know it is impossible in F 5 In conclusion, f1 is irreducible Its roots are the analogous of the ones before, ± 2 ± 2, and the formula α 1 x 3 = 4 2 = 2 = α is still valid, so the extension F 5 F 5 (α 1 ) is normal and [k 1 : F 5 ] = 4 2 From the relation (α 2 2 3) 2 = 3 we get that α 2 is a root of the polynomial f 2 = x 4 6x 2 +6, which is irreducible in Q[x] by virtue of Eisenstein s criterion with p = 2 or p = 3 The roots of f 2 are x 1 = 3 + 3, x 2 = 3 + 3, x 3 = 3 3, x 4 = 3 3 As we did in the previous point, x 3 and x 4 will belong to Q(α 2 ) if and only if α 2 x 3 = 6 belongs to Q(α 2 ) as well Since 3 = α2 2 3, we can wonder whether 2 lives in Q(α 2 ) or not Now, from exercise 736 we know that [Q( 2, 3) : Q] = 4, so 2 Q(α 2 ) if and only if the fields Q( 2, 3) and Q(α 2 ) are the same, because [Q(α 2 ) : Q] = 4 as well Assume α 2 = a + b 2 + c 3 + d 6 Q( 2, 3) for some a, b, c, d Q Then squaring both sides of the equality we get the following system of equations: If a = 0, then the system simplifies to a 2 + 2b 2 + 3c 2 + 6d 2 = 3 ab + 3cd = 0 2ac + 4bd = 1 ad + bc = 0 2b 2 + 3c 2 + 6d 2 = 3 cd = 0 4bd = 1 bc = 0,

3 from which we can deduce that c = 0, d = 1/4b and 2b 2 + 3/(8b 2 ) = 3 This last equation can be written as a polynomial one, associated with the polynomial 16t 4 24t 2 + 3, which is irreducible (Eisenstein s criterion, p = 3), so it has no roots and then a 0 Then we can write d = bc a, and b = 3cd a = 3bc2 a 2 Then, either b = 0 or (a/c) 2 = 3, but that last thing is impossible, because 3 / Q Hence b = d = 0 and we obtain the new system } a 2 + 3c 2 = 3 2ac = 1 As before, it yields the equation a 2 + 3/(4a 2 ) = 3, coming from the polynomial 4t 4 12t 2 + 3, which is again irreducible (Eisenstein s criterion, p = 3) and thus has no solution Summing up, α 2 / Q( 2, 3), so Q(α 2 ) k 2, and then the extension Q Q(α 2 ) is not normal Now consider the minimal polynomial of x 3 over Q(α 2 ) Since 3 Q(α 2 ), we can take g = x 2 (3 3), which is irreducible (x 3 / Q( 3) as 3 / Q) Then the degree of the extension Q(α 2 ) Q(α 2, x 3 ) is two and in conclusion, [k 2 : Q] = [k 2 : Q(α 2 )][Q(α 2 ) : Q] = 8 Over F 5, things are quite different Performing the same calculations as over Q, we find that α 2 is a root of f2 = x 4 x 2 + 1, but that polynomial is reducible, and the product of the irreducible polynomials x 2 + 2x 1 and x 2 + 3x 1 (one can see that solving the system arising from imposing an equality of the form f 2 (x) = (x 2 + ax + b)(x 2 + cx + d)) Then α 2 will be a root of one of those, so in this case k 2 = F 5 (α 2 ), because quadratic extensions are split (once we find a root of a degree two polynomial we have already splitted it into linear factors) and thus [k 2 : F 5 ] = 2 If you, dear and curious reader, want to know which of the two polynomials has α 2 as a root, use that in F 5, α 2 2 = (2 3) = ( 3 + 3) 2 3 Clearly in this case the minimal polynomial of α 3 is f 3 = x 3 2, and then, [Q(α 3 ) : Q] = 3 The other two roots of f 3 are x 2 = α 3 ζ and x 3 = α 3 ζ 2, ζ being a primitive cubic root of one It is clear that k 3 = Q(α 3, ζ) Now, since the minimal polynomial of ζ is x 2 + x + 1 = (x 3 1)/(x 1) 1, the extension Q Q(ζ) has degree two, so if ζ belonged to Q(α 3 ), the degree [Q(α 3 ) : Q] should be even, and that is not the case Therefore, the extension Q Q(α 3 ) is not normal The minimal polynomial of ζ over Q(α 3 ) is again x 2 + x + 1, so in the end, [k 3 : Q] = [Q(α 3, ζ) : Q(α 3 )][Q(α 3 ) : Q] = 6 Over F 5 the situation is much simpler Namely, 3 3 = 2 (actually x x 3 is bijective in F 5 ), so 3 2 F 5 2 and then, k 3 = F 5 ( 3 2) = F 5, so the minimal polynomial is x + 2, the extension is normal and the degree is one 1 Do you remember that it is irreducible? No? Go to exercise We are implicitly fixing 3 2 to be the cubic root of 2 in F 5 that also belongs to F 5

4 4 α 4 is a root of f 4 = (x 2) 2 2 = x 2 4x + 2, which is irreducible, not having roots in Q The other root of f 4 is 2 2, which is 2 (α 4 2), so k 4 = Q(α 4 ), that is, the extension Q Q(α 4 ) is normal and [k 4 : Q] = 2 Over F 5, α 4 is again a root of f 4 = x 2 + x + 2, which has no roots on F 5 either, so it is irreducible and thus the minimal polynomial of α 4 Following the same argument as before, the extension F 5 F 5 (α 4 ) is normal and [k 4 : F 5 ] = 2 Aufgabe 2 Zeigen Sie, dass Q = {x C x algebraisch über Q} ein algebraischer Abschluss des Körpers der rationalen Zahlen ist Beweis It is clear from the definition that the extension Q Q is algebraic Now we should check that Q is algebraically closed as well Let K be the algebraic closure of Q in C, and take any element α K Since it must be algebraic over Q, there will exist some a 0,, a n 1 Q such that α n + a n 1 α n a 1 α + a 0 = 0, for some natural n Now note that the extension Q(a 0,, a n 1 ) Q(a 0,, a n 1, α) is finite of degree n at most, but since every a i is algebraic over Q, the extension Q Q(a 0,, a n 1 ) is also finite and then [Q(a 0,, a n 1, α) : Q] <, so α is algebraic over Q, and hence belongs to Q As a conclusion, Q = K, as we wanted to prove Aufgabe 3 Es sei k der Zerfällungskörper über Q von f = x 3 6x 6 und α die eindeutig bestimmete reelle Lösung von f = 0 Ferner setzen wir ϱ = 1+i 3 Zeigen Sie, 2 dass k = Q(α, ϱ) gilt Beweis In this exercise, as well as in many other examples, it is useful to play a bit with the roots of the polynomial, and multiply them, or raise them to some number, to find which elements belong to the fields under consideration Sometimes it does not lead to any good, but sometimes yes, as here Another way of writing ϱ is e 2πi/3 ; now it is easier to see that ϱ is a primitive cubic unit root Thanks to example IV113c from the lecture, we know that the other roots of f are β = ϱ ϱ and γ = ϱ ϱ 3 2 Obviously, k = Q(α, β, γ), so we have to prove that Q(α, β, γ) = Q(α, ϱ) On one hand, α 2 = , so 3 2 = α 2 α 4 Q(α) Since every root of f can be written in terms of 3 2 and ϱ, k Q(α, ρ) = Q( 3 2, ϱ) On the other, analogously as before, β 2 β 4 = ϱ and γ 2 γ 4 = ϱ 3 2, so ϱ = (β 2 β 4)/(γ 2 γ 4) k Then, Q(α, ϱ) k and we are done

5 Aufgabe 4 Zeigen Sie, dass je zwei Körper mit 9 Elementen zueinander isomorph sind Beweis Let K be a field with nine elements Since it is finite, it must have positive characteristic, otherwise it would contain Q Then there is a prime p for which F p K, and since K is finite again, that extension must be finite and then, K can be seen as a vector space over F p Thus there must be some r 1 such that K = p r In conclusion, p = 3 and r = 2 Since K is a quadratic extension of F 3, we can express it as a quotient ring of the form F 3 [x]/(p(x)), where p(x) is an irreducible polynomial in F 3 [x] of degree 2 There are nine polynomials of degree 2 in F 3 [x], and only three of them irreducible: p 1 (x) = x 2 + 1, p 2 (x) = x 2 + x + 2, and p 3 (x) = x 2 + 2x + 2 If we build an isomorphism from F 3 [x]/(p i (x)) to F 3 [x]/(p j (x)) for any i and j we will be done In order to do so, we just need such isomorphisms for (i, j) = (1, 2), (2, 3) Consider the automorphism of F 3 [x] given by the change of variable x x 1 (with inverse automorphism given by x x + 1) Together with its inverse homomorphism exchanges p 1 and p 2, and p 2 and p 3, inducing thus an isomorphism between F 3 [x]/(p 1 (x)) and F 3 [x]/(p 2 (x)) and F 3 [x]/(p 2 (x)) and F 3 [x]/(p 3 (x)), as we wanted A much more general and conceptual proof is the following, valid to prove the existence and uniqueness of every finite field The beginning is the same; any field of nine elements must be an extension of F 3 Now, given a field of nine elements, the eight units of the multiplicative group are a root of the polynomial x 8 1, and hence of p(x) = x 9 x, together with zero On the hand, we can consider then the splitting field of p over F 3 Now, the sum, product, opposite and inverse of every root is also a root 3, so the set of roots of p is actually a subfield of nine elements of F 3 that must be the splitting field of p Finally, the splitting field of a polynomial is unique, and we are done 3 The most difficult to prove is the sum: if a and b are roots of p in F 3, then (a + b) 9 = a + b because we work in a characteristic three field, so a + b is also a root of p