Galois Theory Flashcards
Show that the splitting field of x^4+ax^2+b (if it is separable) has Galois group a subgroup of D8. Show that if alpha is a root, so is sqrt(b)/alpha. Give examples where the splitting field is C_4, C_2xC_2, or D_8.
- The four roots are a, -a, b, -b with all four distinct. So if we place these four roots on a square, with the negatives on diagonals, we see that permutations of the roots must fix the diagonals (i.e., wherever a goes, -a goes to the opposite corner). In other words, permutations naturally inject into the symmetries of a square, i.e., D8.
- If alpha is a root, sqrt(b)/alpha is also a root. Note that sq
State and prove the Tower Law. Show that alpha algebraic over K iff K(alpha)/K finite. Show that the algebraic numbers form a field. Show that if alpha is algebraic, K(alpha)/K is algebraic. Show that being algebraic is transitive over fields.
- Let M/L/K. Then [M:K]=[M:L][L:K], interpreting the product of infinite dimensions as infinite.
Proof:
- If M/L is infinite dimensional, there are x_n in M linearly independent over L hence over K. So M/K is infinite dimensional.
- If L/K is infinite dim, then since M contains L, M/K is infinite dim.
- If both are finite dim, pick a basis v_i of M/L and u_j of L/K. For x in M, x = sum a_i v_i, a_i in L. Then a_i = sum b_ij u_j, b_ij in K. Thus x = sum b_ij u_j v_i. So the set {u_jv_i} is spanning. Moreover, it is lin indep, since if x=0, then a_i=0, which implies b_ij=0. So this is a basis of the required size.
Let L/K, and alpha, beta in L have degrees m,n over K, where m,n coprime. Then [K(alpha, beta):K]=mn
Proof: we have mn dividing [K(a,b):K]. Now if the min poly of alpha over K is f of degree m, then the min poly of alpha over K(b) is degree at most m. So [K(alpha,beta):K] <= mn. Therefore equality.
Define a splitting field. Prove that splitting fields exist and are unique up to K-isomorphism.
Existence: Let f in K[X]. If deg(f)=1, K is a splitting field of f. Otherwise, pick some irred factor g of f. Then let K’ = K[T]/(g) a field extension of K. Note that K’=K(alpha), where alpha is t+(g) and that in K’, f factors as f(X)=(X-alpha)f’(X) for some f’(X). By induction, we have a splitting field K’(a1, a_2,…, a_k)=K(a,a1,a_2,…,a_k) of f’ over K’. Then this is a splitting field of f over K.
Uniqueness:
1. Pick F,F’ s.t. K <= F <= L, K <= F’ <= L’, there exist psi: F–>F’ and iso, and [F:K] is maximal.
2. Say F!=L. So some root alpha of f exists in L but not F. Let g the min poly of alpha over K. So f=gh, for g,h polys in K.
3. Under the isomorphism psi, we get a root of psi(g) in L but not F’.
Proof: Note that psi extends to an iso from F[T]–>F’[T]. So psi induces an iso F(alpha) = F[T]/(g) –> F’[T]/(psi(g)). Note that f = psi(f)=psi(g)psi(h), so we can pick a root of psi(g) alpha’ in L’. Then F’[T]/(psi(g)) = F’(alpha’), noting that psi(g) is irred as g is. So F(alpha) iso to F’(alpha’).
Show that any finite field has prime character p and size p^n. Show that such fields exist and are unique up to isomorphism. Show that F_{p^k} lives in F_{p^n} iff k divides n. Show that an irred deg d poly divides x^(p^d)-x. Show that the set of irreducible polynomials dividing x^(p^n)-x is precisely the set of irred polys with degree dividing n.
Show that an automorphism group between finite fields is generate by the frobenius map.
- Trivially the char(K) is prime. Then K is a finite extension of F_p hence we have |K|=p^n =q.
- Note that x^q-x splits over any field L of size q. Then note that x^q-x is separable (as qx^{q-1}-1 has not roots), so it has |q| roots in L. Thus L is the set of roots of x^q-x. In particular, L is the splitting field of x^q-x. Thus we have uniqueness. For existence, we simply define L to be the splitting field of x^q-x and note it therefore has size q.
- If F_{p^k} lives in F_{p^n}, then k|n by tower law. Conversely, x^(p^n)-x is divisible by x^(p^k)-x. So the roots of the latter (namely F_{p^k}) live in the former (namely F_{p^n}).
To prove the divisibility, not x-1 | x^t-1. So it suffices to show p^k-1 | p^n-1. But this follows from the same result.
4.
Define what it means for alpha to be separable / K, and L to be separable over K. Show that if char(K)=0, all extensions are separable. Give an example of L/K not separable.
Let K=Fp[X], L = Fp[x^(1/p)]. Then x^(1/p) has min poly T^p-x = (T-x)^p which is not separable.
Let alpha sep/K. Then K(alpha) sep/K.
Key Lemma:
Let alpha alg/K with min poly f. Then if M/K is any field over which f splits, |Hom_K(K(alpha),M)| is precisely the number of distinct roots of f. Hence, f is separable iff |Hom_K(K(alpha),M)|= degf.
- Let beta in K(alpha) of deg m over K. Let alpha be deg n over K(beta). Let h the min poly of alpha over beta which is separable.
- Let M a field over which both the min poly of alpha and beta (over K) split.
- Hom_K(K(alpha),M) —> Hom_K(K(beta),M) given by restriction.
- The fibers of the above map have size precisely [K(alpha):K(beta)] whenever alpha algebraic over K(beta).
i. e.,given phi on the RHS, how many extensions are there to the LHS? alpha must go to a root of phi(h), of which there are precisely (as h separable) n distinct ones. Thus LHS <= RHS * n. - We have LHS = nm, and RHS <= m. Thus RHS=m, and we are done!
Show that if f,g polys over K, and L/K, then gcd_L(f,g)=gcd_K(f,g) (!)
Show that the lcm of separable polys is separable.
- If L/K and f,g polys over K, then gcd_K(f,g)=gcd_L(f,g).
Proof: h_1 = gcd_K(f,g), h_2 = gcd_L(f,g). Then h_1 | f,g in K hence h_1 | f,g in L. Hence h_1 | h_2 in L. But h_1=pf+qg for polys p,q in K. So h_2|h_1 in L. So h_1=h_2 (up to scaling). - Pick a field in which all polys separate.
State and prove the Theorem of the Primitive Element. Show that the condition of separability is necessary.
Let L/K finite, with L=K(alpha_1,…,alpha_n, beta) with alpha_i separable / K. Then L=K(gamma) for some gamma.
- If K finite, then L finite, then choose gamma a generator of the (cyclic) multiplicative group L*.
- It suffices for n=1.
- Pick gamma=beta+c*alpha, c in K. Enough to show alpha in K(gamma).
- Let f the min poly of alpha / K. g the min poly of beta / K, M splitting field of fg. So f(x)=(x-alpha1)…(x-alpha_n) in M[X], g(x)=(x-beta_1)…(x-beta_m).
- Let h(x)=g(gamma-cx) in K(gamma)[X]. Note that h(alpha) = g(beta)=0. So x-alpha | gcd(h,f). It’s enough to show gcd(h,f)=x-alpha, since then x-alpha is the lin. combo of polys in K(gamma).
- This happens iff alpha_i not a root of h for i >= 2, i.e., gamma-c*alpha_i not beta_j for any j. There are only finitely many choices of c for which this doesn’t hold. Since K is infinite, we are done.
- Let K=F_p(x,y), L=F_p(x^(1/p),y^(1/p)).
- [L:K]=p^2. But any element gamma of L satisfies gamma^p in K (since this holds on the generators). And so the min poly of gamma has degree <= p. So [K(gamma):K] <= p.
Let M/L, L/K finite extensions.
M/L, L/K separable iff M/K separable
M/K separable immediately implies M/L and L/K are separable.
If M/L and L/K are separable, L=K(alpha), M=L(beta). So M=K(alpha,beta)=K(gamma). It suffices to show gamma is separable.
Let T a field in which “everything” splits.
alpha sep / K ==> [K(alpha):K] distinct K-homo’s K(alpha)–>T
beta sep / K(alpha) ===> [K(alpha,beta):K(alpha)] disctinct homo’s K(alpha,beta)–>T extending a given homo K(alpha)–>T.
Therefore there are [K(gamma):K] distinct K-homo’s to T.
Define a primitive n’th root. When does a field K have a primitive n’th root? Show that cyclotomic extensions inject into a certain group via the reciprocity map. Show that this might not be a surjection. Describe this injection in this case of finite groups. Show that this is a surjection for K=Q, n=p prime. Give a condition for this map to be a surjection.
Now let K=Q, e a prim n’th root of 1.
(a) Show that there is a unique subfield over which Q(e) is a quadratic extension and find it.
(b) If n=p is an odd prime, show that there is a unique subfield quadratic over q and find it.
chi is a surjection iff G acts transitively on the primitive roots of unity.
(a) Let eta = eps+eps^{-1}, noting then that x^-eta x+1 has eps as a root. Then note that complex conjugation fixes eta but (and hence all of Q(eta)) but not eps, hence eps is not in Q(eta). So [Q(eps):Q(eta)]=2.
(b) Method 1: Note that G=(Z/p)^x = , sigma order p-1. Let index 2 in (Z/p)^x. By inspection eta=(eps+eps^r+eps^(r^2)+..) is fixed by r, and (x-eta)(x-eta^{bar})=x^2+x-|eta|^2.
|eta|^2=(p-1)/2+
Then is the unique index 2 subgroup. Hence there is a unique quadratic extension of Q. We note that -1 is not in , so (Z/p)^x is the union of and -.
By inspection, we see that eta=(eps+eps^r+eps^(r^2)+..) is fixed by and that the orbit of eta under (Z/p)^x is thus eta and eta^{bar}. So eta is a root of (x-eta)(x-eta^{bar})
Hence its min poly over Q is at most a quadratic. But we can also see that eps^{-1}+eps^{-r}+… is in the orbit of eta but not equal to eta.
Define the n’th cyclotomic polynomial over a field K, where char(K)=0 or char(K) doesn’t divide n. Explain how we can find Phi_n given Phi_d for each d dividing n. What is Phi_1? Explain why we can regard Phi_d as a poly in Z[X]. Compute Phi_n for some small n. Show that the reciprocity map is surjective iff phi_n is irred. Compute phi_p(x) for prime p.
Prove that Phi_n is irred (!) Thus the reciprocity map is an isomorphism over Q.
We define Phi_n as the product (x-eps), where eps are the primitive n’th roots of 1. Then note that the Galois group Gal(K(eps_n)/K) fixes the primtive roots of 1, hence Phi_n is a poly in K. Note that Phi_d divides x^n-1 whenever d|n. Note every root of x^n-1 is a primitive d’th root for some d|n (namely the order of that root). Hence, x^n-1 is the product of Phi_d, for d|n. Phi_1(x)=x-1. Thus by induction, we have Phi_n(x) an integer poly.
phi_1(x)=x-1 phi_2(x)=(x^2-1)/(x-1)=x+1 phi_3(x)=x^2+x+1 phi_4(x)=x^2+1 phi_5(x)=x^4+x^3+x^2+x+1.
- Suppose phi_n=fg,where f has positive degree. Then, f has a root eps.
- It suffices to show eps^k is a root of f for all k coprime to n. So it suffices to show eps^p a root of f for all p not dividing n.
- Suppose otherwise. Thus eps^p is a root of g but not f. Work in F_p. Then eps a root of g(x^p). But this is jut g(x)^p. So g(x) and f(x) have a common root in F_p, hence x^n-1 is not separable. But p divides n, so x^n-1 is separable, contradiction.
- Define a Kummer extension L/K.
- Inject Gal(L/K) into Z/nZ and list the possible groups.
- Give an equivalent condition for the group the map to be surjective.
- Let K a field containing a prim n’th root (so char(K)=0 or char(K) doesn’t divide n).
A Kummer extension of K is K(n’th root of a) for some a in K. - Fixing a prim root eps, and a root of x^n-a, say alpha we get that the roots are precisely alpha * eps^i for i in Z/nZ. So we can clearly inject Gal(L/K) into Z/nZ. Hence G=Z/dZ for some d dividing n.
- The map is surjective iff |Gal(L/K)|=n iff [L:K]=n iff x^n-a is irred / K.
Show if K contains a prim n’th root of 1, and L/K a Galois extension with Gal(L/K)=Z/nZ, then L is a Kummer extension (!)
- Let G= and regard sigma as a linear map of the vector space L/K. Since the polynomial x^n-1 kills sigma and x^n-1 factors into linear factors / K (as K has a prim n’th root of 1), we see that sigma is diagonalisable.
- So pick an eigenbasis with eigenvalues lambda_1, …, lambda_n. Note lambda_i are n’th roots of 1, and some lambda_i is primitive (prove that the eigenvalues form a subgroup of L^x hence cyclic. If the order of the subgroup is less than n, then the order of sigma is < n). So pick the corresponding eigenvec gamma. So sigma(gamma^n)=lambda^n gamma^n = gamma^n. Hence a=gamma^n is in K. Note that the orbit of gamma under is gamma*eps, for each n’th prim root eps. Hence L=K(gamma).
Show if G is a finite subgroup of Aut(L), then L/L^G is
- Algebraic, with every element of order dividing |G|
- Separable
- Finite
- State and prove Artin’s Theorem. Define a Galois extension and show that L/L^G is a Galois extension.
Do what is natural for 1,2.
Let K=L^G.
3. Pick alpha in L s.t. [K(alpha):K] is maximal, noting that [K(alpha):K] divides G by 1. Pick beta in L. Suppose that K(alpha) isn’t L.
[K(beta,alpha):K(alpha] <= [K(beta):K] <= |G|. So [K(beta,alpha):K] <= |G|[K(alpha):K] < |G|^2 < infinity.
By the primitive element theorem, we let K(alpha,beta)=K(gamma). So K(gamma)=K(alpha) by maximality of K(alpha). Thus beta is in K(alpha).
4 (i) [L:L^G]=|G|, (ii) Aut(L/L^G)=G
(i) Write L=K(gamma). If g in G fixes gamma, g fixes L, so g is id. Thus the orbit of gamma has size |G| as required.
(ii) Certainly G <= Aut(L/L^G). Thus L^(Aut(L/L^G)) <= L^G. But L^G <= L^(Aut(L/L^G)). Thus L^G=L^(Aut(L/L^G)).
Since Aut(L/L^G) is finite (as [L:L^G] finite), by (i) |Aut(L/L^G)|=[L:L^(Aut(L/L^G))]=[L:L^G]=|G|. So done.
