Test 2: Mon Apr 17 in class. Come 5 minutes early for extra time.
Topics:
Sylow Theorems 1,2,3
Presentation of groups (generators and relations), Cayley diagrams
Rings and ideals, basic properties
Adjoining an element to a field:
F(α) = F[α] = F[x]/(f(x))
Field extensions and degree.
Homework:
I will not collect homework (except problems marked with ♠), but it will be the basis of the weekly quizzes (usually on Fridays). You may hand in Extra Credit problems at any time during the course, no matter when they were posted. Exercise and page numbers refer to the Artin text.
The kernel of φ is Ker(φ) := { r ∈ R s.t. φ(r) = 0 } .
Problems
Prove that R/K is a well-defined commutative ring: that is, if r1 = r'1 and
r2 = r'2 mod K , then
r1 + r2
= r'1 + r'2
and r1 . r2
= r'1 . r'2 .
This is not immediately obvious: for example, you must show that if r'1 = r1 + k1 and
r'2 = r2 + k2 then
r1r2
- (r1+k1)(r2+k2) ∈ K .
Show that there is a natural onto ring homomorphism π : R → R/K defined by π(r) = r + K . We have Ker(π) = K .
Let φ : R → R' be any ring morphism. Show that Ker(φ) is an ideal.
Consider a quotient R/K = F[x] / (f(x)) , where F is a field, f(x) ∈ F[x] is an irreducible polynomial of degree n which generates the ideal:
(f(x)) := { q(x) f(x) s.t. q(x) ∈ F[x] } .
is the ideal generated by f(x).
Then for α := x , we have f(α) =
f(x) = 0 ∈ R/K
The standard form of a polynomial a(x) is its remainder under division by f(x): that is, we write a(x) = q(x)f(x) + r(x) , where deg(r) < n , and note that a(α) = r(α) .
To add, subtract, or multiply standard forms, perform the operation in F[x], then reduce to standard form.
To divide in the quotient ring, use the Euclidean algorithm to write
m(x)a(x) + n(x)f(x) = gcd(a(x),f(x)) = 1 ∈ F[x],
then write 1/a(α) ≡ m(α) .
Practice computing the four field operations in the ring Q[α] := R/K = Q[x]/(x3 + x + 1) .
If you are ambitious, obtain closed formulas for addition, subtraction, multiplication, and reciprocals of standard forms.
Answer:
Show that the ideals generated by 0 and by 1 are:
(0) = {0} and (1) = R . Every ideal J ⊂ R contains
(0) ⊂ J .
Prove the following facts about quotient rings & fields.
The commutative ring R is a field ⇔ the only ideals of R are K = (0) and K = R .
The natural homomorphism π : R → R/K gives a one-to-one correspondence between the ideals J~ ⊂ R/K and the ideals J with K ⊂ J ⊂ R .
The quotient R/K is a field ⇔ K is a maximal ideal , meaning there is no ideal larger than K but smaller than all of R.
Let R = F[x] , the polynomial ring with coefficients in a field F.
Show that we have containment of ideals (f(x)) ⊂ (g(x)) ⇔ g(x) divides f(x) in F[x] .
Show that (f(x)) is a maximal ideal ⇔ f(x) is an irreducible polynomial, meaning f(x) has no factors in F[x] except 1 and f(x) itself (and constant multiples).
For α an algebraic number, we define its minimal polynomial f(x) ∈ Q[x] as the monic polynomial of smallest degree such that f(α) = 0 . Show that any minimal polynomial is irreducible. Then use part (b) to show that the ring
Q[α] := Q[x]/(f(x)) is actually a field.
Using the polynomial division algorithm, practice computing the four field operations for:
The field with 4 elements, F4 := F2[x]/(x2 + x + 1), where
F2 = Z2 , the field with 2 elements. Make addition and multiplication tables, and compute 1/a for each a ∈ F4 .
Find an irreducible polynomial of degree 3 over F2 , and construct a field with 8 elements F8 . Repeat the above exercises for this field.
HW 4/14
Quiz: Field extensions, degree [L:F]. Reference: Ch 13.1--13.4.
A field extension is a field contained in a larger field, F ⊂ L . Forgetting the multiplication between elements of L, we can consider L as a vector space over F,
and we can define the degree:
[L : F] := dimF(L) .
That is, [L : F] = n , the number of elements in {α1,...,αn} , an F-basis of L.
This means every element α ∈ L has a unique expression as a linear combination of αi with F coefficients:
α =
c1α1 + ... + cnαn
with ci ∈ F .
Usually we will take the first basis vector to be α1 = 1 ∈ L . Note that F is considered in two ways: F is the field of coefficients, and F = F.1 is an "axis" inside L.
We use the following notations for the smallest ring and the smallest field containing F and an element α :
However, we have shown that if α is the root of a polynomial equation f(α) = 0 for f(x) ∈ F[x] , then the ring F[α] ≈ F[x]/(f(x)) is already a field. Thus F[α] = F(α) ,
and we will usually write F(α) for this field.
Multiplicative property of degree:
If F ⊂ E ⊂ L is a double extension of fields, then:
[L : F] = [L : E] [E : F]
Show that if f(x) ∈ F[x] is the minimal (irreducible) polynomial of some α, then
F[α] has basis { 1 , α , α2 , ... , αn-1 } , so that [F(α) : F] = n .
Show that if F ⊂ E ⊂ L is a double field extension,
E has an F-basis {α1,...,αn} ,
and
L has an E-basis {β1,...,βm} ,
then L has an F-basis
{ αi βj
with i = 1,...,n and j = 1,...,m }
Use this to prove the multiplicative property of degree.
Let F ⊂ L be an extension of degree n. Then any element α ∈ L generates a double field extension F ⊂ F(α) ⊂ L with:
n = [L : F(α)] [F(α) : F]
Thus α must satisfy an equation of degree
[F(α) : F] , which divides n.
Verify this for the degree 2 extension Q ⊂ Q(√3) . That is, for each element α = a + b√3 ∈ Q(√3) , find its minimal polynomial f(x) ∈ Q[x], which must be of degree 1 or 2.
Hint: An element α = a ∈ Q has minimal polynomial f(x) = x - a . For any other element, the elements 1, α, α2 must be linearly dependent over Q , meaning there is an expression
c0 + c1α + c2α2 = 0 .