Class Log
Wednesday 5/1/13. Example: The splitting field L of the quintic f(x)=x5-16x+2 over Q has Galois group G(L/Q) isomorphic to S5. Proposition: In the correspondence of the main theorem of Galois theory, normal subgroups H of G=G(L/K) correspond to intermediate fields M such that the field extension K c M is Galois. Moreover, G(M/K)=G(L/K)/G(L/M). Theorem: If f(x) in Q[x] is an irreducible polynomial of degree 5 with Galois group isomorphic to S5 or A5 then its roots cannot be expressed in terms of radicals.
Monday 4/29/13. Statement of insolubility of quintic by radicals (Galois' theorem). Lemma: Let K c C be a subfield of the complex numbers containing z=exp(2 pi i/p), p prime. Let a in K be an element such that a ≠ bp for any b in K. Let f(x)=xp-a in K[x]. Then f(x) is irreducible over K and the splitting field L of f(x) over K has degree [L:K]=p and Galois group G(L/K) isomorphic to Z/pZ. Lemma: Let f(x) in Q[x] be a irreducible polynomial of degree 5 such that exactly 3 of the roots of f(x) are real. Then the Galois group of the splitting field L of f(x) over Q is isomorphic to the symmetric group S5 on 5 letters.
Friday 4/26/13. Proof of converse statement from last time: if L is the splitting field of a polynomial f(x) in K[x] over K then the field extension K c L is Galois. Proof of the Main theorem of Galois theory. Example: Cyclotomic fields. Let p be a prime and z=exp(2 pi i/p). Then Q c Q(z) is a Galois extension of degree p-1 with Galois group isomorphic to Z/(p-1)Z. Application: The regular 17 gon is constructible using ruler and compass.
Wednesday 4/24/13. Proof of Lemma stated last time. Proof of the fixed field theorem: If L is a field and H is a finite group of automorphisms of L then the degree [L:LH] of L over the fixed field LH is equal to the order |H| of the group H. Corollary (of Lemma and FFT): If K c L is a Galois extension and f(x) is an irreducible polynomial in K[x] which has a root in L, then f splits completely (i.e. factors into linear factors) over L. Corollary: If K c L is a Galois extension then L is the splitting field of some polynomial f(x) in K[x] over K. Preliminary discussion of converse: If L is the splitting field over K of some polynomial f(x) in K[x], then the extension K c L is Galois, that is, the order |G(L/K)| of the Galois group equals the degree [L:K] of the extension.
Monday 4/22/13. Proof of the Theorem of the primitive element. Lemma: Let L be a field, H a group of automorphisms of L, and K=LH the fixed field of H. Suppose a in L is an element and let a1,..,ar be the orbit of a under the action of H on L. Then the irreducible polynomial of a over K is given by f(x)=(x-a1)...(x-ar).
Friday 4/19/13. Example: The Galois group of the splitting field L over Q of the polynomial f(x)=x4+x3+x2+x+1 is isomorphic to Z/4Z. We have L=Q(z) where z is a 5th root of unity, and the subgroup <2>={0,2} of Z/4Z corresponds to the intermediate field M=Q(z+z-1). Lemma: If K is a field of characteristic 0 and f is an irreducible polynomial in K[x], then f has no repeated roots in any extension L of K. Theorem of the primitive element: If K is a field of characteristic 0 and K c L is a field extension such that the degree [L:K] is finite, then there exists an element a in L such that L=K(a).
Wednesday 4/17/13. Example: The Galois group G(L/Q) of the splitting field L of the polynomial f(x)=x3-3x+1 over Q is identified with the alternating group A3 c S3, in particular, it is isomorphic to Z/3Z. Statement of the Main Theorem of Galois theory: Let K c L be a field extension and assume K c L is Galois, i.e., the degree [L:K] is finite and |G(L/K)|=[L:K]. Then there is a bijective correspondence F between subgroups H of the Galois group G=G(L/K) and intermediate fields M, K c M c L, given by F(H)=LH, the fixed field of H, with inverse F-1(M)=G(L/M), the Galois group of the extension M c L. This correspondence is inclusion reversing and satisfies [M:K]=[G:H]=|G|/|H|, where M=F(H)=LH. Examples: L is the splitting field of f(x) over K, where (1) f(x)=(x2-d)(x2-e) and [L:K]=4; (2) K=Q and (a) f(x)=x3-2; (b) f(x)=x3-3x+1 (as in the example above).
Monday 4/15/13. No class (Patriot's day).
Friday 4/12/13. Key Lemma: Let K c L be the splitting field of a polynomial f(x) in K[x] over K, so f(x)=(x-a1)..(x-an) for some a1,..,an in L and L=K(a1,..,an). Assume that f has no repeated roots, i.e., the elements a1,..,an in L are distinct. Then the Galois group G(L/K) of the extension permutes the roots a1,..,an of f and this defines an injective homomorphism of groups t: G(L/K) -> Sn from the Galois group to the symmetric group on n letters. Examples: f(x)=x2-d, (x2-d)(x2-e), and x3-2.
Wednesday 4/10/13. Let K c L be a field extension. We say that L is the splitting field over K of a monic polynomial f(x) in K[x] if (1) f(x)=(x-a1)..(x-an) for some a1,..,an in L and (2) L=K(a1,..,an). Theorem: A field extension K c L of finite degree [L:K] is a Galois extension iff L is the splitting field over K of some monic polynomial f(x) in K[x]. Corollaries:(1) If K c L c M is a tower of field extensions and K c M is Galois, then L c M is also Galois. (2) If K c L is a field extension of finite degree [L:K] then there exists a field extension L c M such that K c M is Galois. [WARNING: Given a tower of field extensions K c L c M, (a) K c M Galois does NOT imply K c L Galois, and (b) K c L and L c M Galois does NOT imply K c M Galois.]
Monday 4/8/13. If K c L is a field extension such that [L:K]=4 and L=K(sqrt(d),sqrt(e)) for some d,e in K, then the Galois group G(L/K) is isomorphic to the Klein 4-group (Z/2Z)2. Statement of the fixed field theorem: Let L be a field and G a finite group of automorphisms of L. Define the fixed field LG to be the set of a in L such that g(a)=a for all g in G. Then LG is a subfield of L and the degree [L:LG]=|G|, the order of G. Corollary: If K c L is a field extension of finite degree [L:K] then the order of the Galois group |G(L/K)| divides [L:K]. We say K c L is a Galois extension if |G(L/K)|=[L:K].
Friday 4/5/13. Let Q c L be the field extension generated by the roots of f(x)=x3-2, then [L:Q]=6. Let Q c M be the field extension generated by the roots of g(x)=x3-3x+1, then [M:Q]=3 (so M=Q(a) for any root a of g(x)). Until further notice, we assume that the fields we consider have characteristic 0. Let K c L be a field extension. An automorphism of L over K is an isomorphism f: L -> L such that f(a)=a for all a in K. The Galois group of L over K is the group of all such automorphisms f, denoted G(L/K). Example: If [L:K]=2 then L=K(sqrt(d)) for some d in K, f(a+b(sqrt(d)))=a-b(sqrt(d)) is an automorphism of L over K, the Galois group G(L/K)={1,f} (where 1 denotes the identity map L->L), and so G(L/K) is isomorphic to Z/2Z.
Wednesday 4/3/13. Trisection of angles with ruler and compass: Although there do exist angles which may be trisected (e.g. pi/2) it is impossible to do so in general (e.g. pi/3 cannot be trisected). Proof: An angle t can be constructed iff cos(t) is constructible. In particular pi/3 can be constructed, but pi/9 cannot -- because
cos(pi/3)=1/2, but cos(pi/9) has degree 3 over Q (with irreducible polynomial 8x3-6x-1) and so is not constructible. Galois theory (preliminary discussion).
Monday 4/1/13. If a real number a is constructible then so is sqrt(a). A real number a is constructible iff there is a tower of fields Q=K0 c K1 c .. c Kn=K such that a is in K and Ki+1=Ki(sqrt(di)) for some di in Ki for each i=0,1,..,n-1. Corollary: If a real number a is constructible then the degree of a over Q is a power of 2. Applications: (1) The cube root of 2 is not constructible. (2) For p prime, if the regular p-gon is constructible then p-1 is a power of 2.
Friday 3/29/13. Constructions using ruler and compass. Motivation: which regular polygons can be constructed? [Answer (Gauss): If p is prime, then the regular p-gon can be constructed iff p-1 is a power of 2. Examples: p=3,5,17,257,65537,..] Rules: (1) We are given two points initially; (2) Given two constructed points, we may construct the line joining the points, or the circle with center one point passing through the other point; (3) Intersection points of constructed lines and circles are understood to be constructed. A constructible number is by definition the distance between two constructible points. The constructible numbers form a subfield of the real numbers.
Wednesday 3/27/13. Examples of field extensions and computation of degree. The extension L/Q generated by the cube root of 2 and the 4th root of 3 has degree 12.
The extension L/Q generated by square root 3 and square root 5 has degree 4. The extension L/Q generated by all the roots of x3-2 has degree 6.
Monday 3/25/13. Let K be a field and L a field containing K. The degree of the field extension L/K, denoted [L:K], is the dimension of L as a vector space over K. If a is algebraic over K and the irreducible polynomial of a over K has degree n then [K(a):K]=n (because K(a) has basis 1,a,a2..,an-1 as a vector space over K). If [L:K]=1 then L=K. If [L:K]=2 then L=K(a) for some a in L such that a2 is in K. If we have a tower of fields K contained in L contained in M, then [M:K]=[M:L][L:K]. Corollaries: If [L:K]=n then every element a in L is algebraic over K of some degree d dividing n. If [L:K]=p is prime then every element a in L which is not in K has degree p over K and generates L over K, that is, K(a)=L.
Friday 3/15/13. If K is a field, L is a field containing K, and a is an element of L, then the subring K[a] generated by a over K is a field iff a is algebraic over K. In general we write K(a) for the subfield generated by a over K, (so K(a) is the fraction field of K[a], and K(a)=K[a] iff a is algebraic over K). Computing inverses in K[a] (for a algebraic) using the identification of K[a] with K[x]/(f) and the Euclidean algorithm in K[x]. Examples: a = square root 2, a = cube root 2.
If K is a field, L and M are fields containing K, and a in L and b in M are algebraic over K, then a and b have the same irreducible polynomial over K iff there is an isomorphism p: K(a) -> K(b) such that p(c)=c for all c in K and p(a)=b. Example: complex conjugation.
Wednesday 3/13/13. Let K be a field. The maximal ideals in K[x] are the principal ideals (f(x)) generated by irreducible polynomials (recall also that every ideal in K[x] is prinicipal). For f(x) in K[x] an irreducible polynomial, define the quotient ring L=K[y]/(f(y)). Then L is a field containing K and the element a=y+(f(y)) of L satisfies f(a)=0. The quotient ring L is a vector space over the field K with basis 1,y,y2,..,yn-1 where n is the degree of f; in particular, the dimension of L as a vector space over K equals n.
For K a field, L a field containing K, and a an element of L, we say a is algebraic over K if there is a nonzero polynomial g(x) in K[x] such that g(a)=0 (otherwise we say a is transcendental over K). In this case the irreducible polynomial of a over K is by definition the monic polynomial f(x) in K[x] of smallest degree such that f(a)=0. Then f is irreducible in K[x] and generates the kernel of the evaluation map p:K[x]-> L given by p(g(x))=g(a). So by the first isomorphism theorem we obtain an isomorphism p̅: K[x]/(f) -> K[a], where K[a]:=p(K[x]) denotes the image of the evaluation map p.
Monday 3/11/13. Application of Eisenstein's criterion: The polynomial xn-1+xn-2+..+x+1=(xn-1)/(x-1) is irreducible iff n is a prime. Given a polynomial f in K[x] for some field K, construction of a field L containing K such that f has a root in L (Preliminary discussion).
Friday 3/8/13. Proof of Gauss' Lemma. Eisenstein's criterion: If f=anxn+..+a0 is a polynomial with integer coefficients and p is a prime such that p does not divide the leading coefficient an, p divides the remaining coefficients an-1,..,a0, and p2 does not divide the last coefficient a0, then f is irreducible in Q[x]. Corollary: There exist irreducible polynomials in Q[x] of arbitrarily large degree.
Wednesday 3/6/13. Examples of factorizations of polynomials into irreducibles over various fields (Q,R,C,Z/pZ). Statement of Gauss' Lemma: If f is a polynomial with integer coefficients and we have a factorization f=gh where g and h are nonconstant polynomials with rational coefficients, then there is a factorization f=GH where G and H are nonconstant polynomials with integer coefficients. Application: If f=anxn+..+a1x+a0 is a polynomial with integer coefficients and p is a prime such that (1) p does not divide the leading coefficient an and (2) the reduction f̅ of f modulo p is irreducible in (Z/pZ)[x], then f is irreducible in Q[x].
Monday 3/4/13. If K is a field and f(x) in K[x] has degree 2 or 3, then f is irreducible iff there exists a in K such that f(a)=0. If f=anxn+..+a1x+a0 is a polynomial with integer coefficients, and c=a/b is a rational number in its lowest terms such that f(c)=c, then b divides the leading coefficient an of f and a divides the last coefficient a0 of f. If K is a field and f(x) in K[x] is a nonzero polynomial, then there is a unique factorization f=cp1p2..pr where c in K is a nonzero constant and p1,..,pr are monic irreducible polynomials.
Friday 3/1/13. Example: Computing the gcd of two polynomials f,g in K[x], K a field, using the Euclidean algorithm. Irreducible polynomials: For K a field, we say f in K[x] is irreducible if f is not constant and there does not exist a factorization f=gh with g,h not constant. Irreducible polynomials for K=C (complex numbers) and K=R (real numbers). Irreducible polynomials over the finite field K=Z/pZ. The "sieve of Eratosthenes" method for listing irreducible polynomials over a finite field (same method used for listing primes in Z). There are irreducible polynomials of arbitrarily large degree over K=Z/pZ (following Euclid's proof that there are infinitely many primes in Z).
Wednesday 2/27/13. Corollaries of division algorithm: If g in R[x] satisfies g(a)=0 then g is divisible by (x-a). If R is an integral domain then a polynomial g in R[x] has at most deg(g) roots. (Application: the group of units in the field Z/pZ is cyclic.) If K is a field then every ideal in K[x] is principal. The gcd of two polynomials f,g in K[x].
Monday 2/25/13. If R is an integral domain then so is R[x]. (More precisely deg(fg)=deg(f)+deg(g) for nonzero polynomials f,g in R[x] when R is an integral domain.)
The division algorithm: Given f,g in R[x] such that f is nonzero and its leading coefficient is a unit in R, there are q,r in R[x] such that g=qf+r with r=0 or deg(r)< deg(f). Example: A polynomial ring in two variables x,y can be studied via R[x,y]=(R[x])[y].
Friday 2/22/13. The prime subfield of a field: If K is a field, then the characteristic of K equals 0 or p (a prime) and we get an injective homomorphism Q -> K in the first case and Z/pZ -> K in the second. Theorem: A finite field has order pn for some prime p and positive integer n. Review of vector spaces from Math 235: Definition of a vector space V over a field F. Dimension of a vector space. If V is a vector space over a field F of dimension n then V is isomorphic to Fn. The ring R[x] of polynomials in a variable x with coefficients in a ring R (commutative with 1).
Wednesday 2/20/13. Proof of Chinese remainder theorem. Example: R[x]/(x2-5x+4) is isomorphic to R ⊕ R (where R is the ring of real numbers). The fraction field of an integral domain. Examples: The fraction field of the ring R[x] of polynomials in the variable x with real coefficients is the field R(x) of rational functions in the variable x with real coefficients. The fraction field of the Gaussian integers.
Tuesday 2/19/13. First isomorphism theorem without surjectivity assumption: If f : R -> S is a ring homomorphism then f(R) is a subring of S and there is an induced isomorphism of rings g : R/ker(f) -> f(R) given by g(a+ker(f))=f(a). Examples: If f : Z -> R is the ring homomorphism (with domain the set Z of integers) determined by f(1)=1 then ker(f)=nZ for some n ≥ 0, and we get an isomorphism g : Z/nZ -> f(Z). If f : R[x,y] -> R[t] is the ring homomorphism given by f(x)=t2, f(y)=t3 and f(a)=a for a in R (R = ring of real numbers), then ker(f) equals the principal ideal (x3-y2), the image f(R[x,y]) of f equals the subring S of R[t] consisting of polynomials p(t) such that p'(0)=0, and we get an isomorphism g: R[x,y]/(x3-y2) -> S. The ring Z[i]/(2+3i) is isomorphic to Z/13Z. The Chinese remainder theorem: Let R be a ring with 1 and I and J ideals of R such that I+J=R. Then the map f : R/I∩J -> R/I ⊕ R/J given by f(a + I ∩ J) = (a+I,a+J) is an isomorphism of rings. Example: R=Z, I=nZ, J=mZ, gcd(n,m)=1 gives the usual CRT isomorphism f: Z/nmZ -> Z/nZ ⊕ Z/mZ, f(a)= (a mod n, a mod m).
Monday 2/18/13. No class (President's day).
Friday 2/15/13. Example: The Frobenius homomorphism for (Z/pZ)[x]. The first isomorphism theorem for ring homomorphisms: If f : R -> S is a surjective ring homomorphism then there is an induced ring homomorphism g: R/ker(f) -> S given by g(a+ker(f))=f(a) which is an isomorphism (i.e. a bijective homomorphism). Examples: Evaluation at a real number a gives an isomorphism R[x]/(x-a)-> R, evaluation at i gives an isomorphism R[x]/(x^2+1)-> C, evaluation at 0 followed by reduction mod 2 gives an isomorphism Z[x]/(2,x)->Z/2Z.
Wednesday 2/13/13. The kernel of a ring homomorphism R -> S. The kernel is an ideal of R. Conversely, given an ideal I of R, the quotient map R -> R/I is a ring homomorphism with kernel I. If R is a ring with 1 then there is a unique ring homomorphism f: Z -> R with f(1)=1 (here Z denotes the ring of integers). We define the characteristic of R to be the integer n ≥ 0 such that ker(f) = nZ. If R is an integral domain then the characteristic of R is 0 or a prime p. If R is a commutative ring of characteristic p (a prime) then the map F : R -> R given by F(a)=ap is a ring homomorphism called the Frobenius homomorphism.
Monday 2/11/13. Correspondence theorem: The ideals of R/I are the quotients J/I where J is an ideal of R containing I. Corollary: R/I is a field iff I is a maximal ideal (i.e. I is not equal to R, and the only ideals containing I are I itself and R). Ring homomorphisms. Examples.
Friday 2/8/13. Examples of prime ideals: The prime ideals in Z are {0} and (p) for p a prime. The prime ideals in C[x] are {0} and (x-a) for a an element of C (here C=complex numbers). The prime ideals in Z[x] are more complicated: they include {0},(x), and (2,x) (which form a nested chain of length 2).
Wednesday 2/6/13. More examples of quotient rings: R[x]/(x-a) is isomorphic to R for a in R, R[x]/(x^2+1) is isomorphic to C (here R={real numbers}, C={complex numbers}). Ideals generated by several elements. An example of a non-principal ideal: I=(2,x) in Z[x]. If R is a commutative ring with 1, not the trivial ring, then R is a field iff the only ideals are {0} and R. Prime ideals. If R is a commutative ring with 1 and I is an ideal of R, R/I is an integral domain iff I is prime.
Monday 2/4/13. Ideals. Quotient rings. (Isomorphism of rings.) Examples. Principal ideals. For the ring Z of integers and the ring R[x] of real polynomials in a variable x, every ideal is principal.
Friday 2/1/13. Direct sum of rings. Subrings. Quotient rings (preliminary discussion).
Wednesday 1/30/13. Skew fields. Example: the quaternions. Units.
Monday 1/28/13. Nilpotent elements. Integral domains. Fields.
Friday 1/25/13. Formal definition of a ring. Zero divisors.
Wednesday 1/23/13. Examples of rings.