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MATH 513: HONORS ALGEBRA II
UNIVERSITY OF MICHIGAN
WINTER 2012
LECTURE NOTES
PROF. KAREN SMITH DAVID J. BRUCE
- Introduction to Fields - 3/7/
For most people there mathematical career begins with calculus and the properties of the real numbers. The real numbers have many great properties. For example, we can add, multiple, and both of these operations behave nicely in that they are combinative, associative, and multiplication distributes over addition. However, the integers also have these things, but unlike the integers the real numbers have not only additive inverses, but also multiplicative inverses. For those of you have a course in analysis you might have learned that that the real numbers are what we call a field. Abstractly a field is a set F along with two binary operations (+, ×), which we generally refer to as multiplication and addition such that ∀a, b, c ∈ F:
(1) a + (b + c) = (a + b) + c Associativity of Addition
(2) a + 0 = 0 + a = a Additive Identity
(3) a + (−a) = (−a) + a = 0 Additive Inverses
(4) a + b = b + a Commutativity of Addition
(5) a · (b · c) = (a · b) · c Associativity of Multiplication
(6) a · 1 = 1 · a = 1, 1 6 = 0 Multiplicative Identiy (7) a · a−^1 = a−^1 · a = 1 Multiplicative Inverses
(8) a · b = b · a Commutativity of Multiplication
(9) a · (b + c) = a · b + a · c Distributive Law
As one notices a field carries with it a lot of structure, and the above axiomatic definition is somewhat unwieldily because of this. However, using the language we developed to talk about rings we can create a ”nicer” definition of a a field.
Date: March 18, 2012. 1
2 PROF. KAREN SMITH DAVID J. BRUCE
Definition 1. A field F is a commutative ring where every non-zero element has a multi- plicative inverse.
A somewhat useful characterization of a field is that a a field F has only contains two ideals namely (0) and F. This is easy to see since if A ⊂ F is a non-zero ideal then x ∈ A, and since ideas are closed under multiplication x−^1 x = 1 ∈ A, and so A is intact F.
You are probably already quite familiar a number fields. For example, as mentioned above R is a field as is C, and as is Q. The rationals are actually a particularly type of field, namely Q is what we previously called the field of fractions of Z. These are probably the most familiar fields and they have infinite in cardinality, however, there are many other more exotic fields, and even fields, which are finite in cardinality. For example, one can define
Definition 2. The characteristic of a field F - denoted char(F) - is the the smallest natural number p such that: 1 +︸.. .︷︷ + 1︸ p times
By convention if no such p exists we say that F has characteristic zero. The structure of a field means that the characteristic of a field is quite limited.
Proposition 1. The characteristic of a field F is either zero or a prime number.
There are may ways to prove this, and we will present two of them here. The first prove will use contradiction, while the second one will use some slick ring theory that makes the convention of characteristic zero make sense.
Proof. Assume that characteristic of F is not 0 and that p is not prime so there exists a, b ∈ Z+^ such that p = ab. Now by definition we know that:
0 = 1 +︸.. .︷︷ + 1︸ p times
= p = ab
However, since F is a field, and hence domain we know that ab = 0 means that either a = 0 or b = 0, and so: 1 +︸.. .︷︷ + 1 = 0 ︸ a times
= 0 or 1 +︸.. .︷︷ + 1 = 0 ︸ b times
But, a < p and b < p contradicting the fact that p is minimal, and so we see that p cannot be composite.
Proof. First let us recall that we have a unique ring homomorphism:
Z −−→φ F n 7 → 1 +︸.. .︷︷ + 1︸ n times
4 PROF. KAREN SMITH DAVID J. BRUCE
The kernel of π is in fact the ideal (xp−^1 +... + 1) so by applying the first isomor- phism theorem for rings we see that Q[x]/(fp) ∼= Q(ψp). As interesting exploration into this field one could consider whether or not Q[ψp] is dense in C.
So fair in our explorations of algebra the general pattern has been that we define some algebraic structure e.g. a ring, a group, a module, and then we consider subsets of these objects like ideals, normal subgroups, submodules, and try and find interesting relationships about them. Fields are different in that the story of fields has been the story of polynomials. The evolution from the integers to the rational numbers to the reals, and finally to the complex numbers can be viewed as a the quest to find solutions to polynomials. For example, it was the probably the ancients who discovered that the solution to a linear polynomial ax − b is b/a, but b/a is not always a integer. So why not through in all these solutions to the integers so that we can always solve linear polynomials. Similarly the hunt for solutions of polynomials like x^2 − 2 and x^2 + 1 motived the construction of the real and complex numbers respectively. So unlike many things in algebra instead of looking at smaller objects with fields we want to look at bigger extensions of fields because we often want for certain polynomials to have roots.
Given a field K we say K is a field extension of F is F is a subfield of K. Meaning that a K is field extension of F if and only if K is an F-algebra. So for example C is a field extension of the reals, which themselves are a field extension of the rational numbers. Given a field extension it is natural to wonder if there is a way to tell how big the field extension is relative to the ”base” field. To do this we shall define what we call the degree of a field extension:
Definition 3. If K is a field extension of a field F then the degree of a field extension - often denoted deg(K/F) or [K : F] - is the dimension of K as an F vector space.
Again it may beneficial to consider a examples of field extensions and to calculate their degree.
Example 3. :
- If we consider C as a vector space over R we know that it has dimension two since { 1 , i} form a basis for this space. Thus, C has degree 2 as a field extension of R.
- If we consider R as a Q-vector space we know we need an infinite number of vectors to span R since hitting the irrational numbers is very tough. Thus, as a field extension of Q, R has infinite degree.
- Consider Q[x]/(x^2 as a field extension of Q. Since x^2 is irreducible clearly this is field, and since it contains the constant polynomials it contains Q as a subfield. In addition, Q[x]/(x^2 ) is only contains polynomials of degree less than two so { 1 , x} is a basis for Q[x]/(x^2 ) as a Q-vector space.
MATH 513: HONORS ALGEBRA II UNIVERSITY OF MICHIGAN WINTER 2012 LECTURE NOTES 5
As or last example might have hinted at if we have a field F and extend it to F[x]/(p) there should be a nice relationship between the degree of p and the degree of the field extension. In fact this is precisely why we use the terminology degree of a field extension instead of using order or dimension.
Proposition 2. Let F be a field and p ∈ F[x] be an irreducible polynomial of degree d then F[x]/(p) is a field extension of F of degree d.
Proof. Since p is prime we know that (p) is a maximal ideal since if it where not then (p) ⊂ (a) and so a would divide p an thus p would not be irreducible. Thus, F[x]/(p) is in fact a field, and since it contains the constant polynomials F is a subfield.
We now need to find a basis for F[x]/(p) as a F-vector space. I claim that { 1 , x,... , xd−^1 } is in fact a basis for this vector space. To see that these are linearly independent not that if B(x) = λ 0 + λ 1 x + · · · + λd− 1 xd−^1 = 0 then B(x) takes one the value zero infinitely many times. However, since B(x) has degree d − 1 it can only take on zero d − 1 times unless all the λi’s are zero, and so these are linearly independent.
Now to show that these polynomials do in fact span F[x]/(p) let f ∈ F[x]. By the division algorithm we know that ∃q, r ∈ F[x] such that:
f = pq + r deg(r) < deg(p)
Thus, f = r, and since the degree of r is less than d clearly we can write using this basis.
Definition 4. Let K be a field extension of F then an element α ∈ K is algebraic over F if an of the following equivalent statements are true:
- ∃f ∈ F[x] − { 0 } such that f (α) = 0.
- [F(α) : F] is finite.
- F(α) ∼= F[x]/p.
If α ∈ K does not satisfy any of these conditions we say that α is transcendental over F.
Definition 5. The unique monic polynomial satisfying either condition 1 or 3 of Definition 4 is the minimal polynomial of α.
- Field Extensions - 3/9/
Some of the most famous problems in mathematics come from the geometers of ancient Greece. For example, geometers wondered whether or not one could using a compass and straightedge trisect an arbitrary angle, or square a circle. Many of these problems
- such as the two mentioned previously - went unsolved for centuries, however, with the development of modern algebra and the emergence of Galois Theory these questions began to be answered. We will take more about compass and straightedge constructions later,
MATH 513: HONORS ALGEBRA II UNIVERSITY OF MICHIGAN WINTER 2012 LECTURE NOTES 7
Proof. To prove this let A = {α 1 ,... , αn} be a basis for K over E, and B = {β 1 ,... , βm} be a basis for E over F. Taking x ∈ K we know we can write:
x = λ 1 α 1 + · · · + λnαn.
Since A is a basis for K. Since each λi ∈ E we can rewrite this as:
x = (a 11 β 1 + · · · + a (^1) m βm)α 1 + · · · + (an 1 β 1 + · + anm βm)αn.
Thus, the set AB spans K as an F vector space. Since the λi’s uniquely determine the a′ ij s and the λi’s are themselves uniquely determined by x we know that AB must also be linearly independent. Thus, AB is a basis of for K as an F-vector space, and has dimension mn so [K : F] = mn exactly as we had wished.
Now that we have that proposition let us consider the chain of field extensions:
Q ⊂ Q(
2) ⊂ Q( 4
Since ( 2 n
2)^2 n^ − 2 = 0 clearly each one of these extension is algebraic. Furthermore, it is pretty easy to see thet Q( 2 n
- has a degree of 2 over Q(
2(n − 1)). Thus, we have an infinite chain of algebraic field extension, but applying the previous proposition we know that this results in a field extension of infinite degree. So clearly algebraic field extension need not have finite degree.
However, what about the other direction of our question? Are field extensions of a finite degree necessarily algebraic extensions? As it turns out this is the case.
Proposition 4. Let K be a field extension of F of finite degree then K is algebraic over F.
Proof. Since K is a finite extension of F we know that as a F-vector space K let n be the dimension of K over F. Now let α ∈ K and consider the set:
A := { 1 , α, α^2 ,... , αn}.
Since A has n + 1 elements in it and K only has dimension n we know that this set must be linearly dependent over F. This means there exists λi ∈ F such that:
λnαn^ + · · · + α 0 = 0.
However, this means that α satisfies λnxn^ + · · · + λ 0 , and so α is in fact algebraic over F. So we can conclude that K is in fact a algebraic field extension of F.
- Introduction to Constructions - 3/12/
Thinking back to high school some of you might remember learning compass and straight- edge constructions. For those of you who don’t remember give two points p and q we can construct:
- A line passing through p and q.
8 PROF. KAREN SMITH DAVID J. BRUCE
- A circle centered at p passing through q or a circle centered at q passing through p.
These are the basic tools of construction, and from this we say that a point α is constructed if it the intersection of constructed lines or constructed circles.
As a historical aside one might wonder where this notion of construction comes from, and why it is relevant to the study of field extensions. In fact when these concepts did not originate in algebra, but instead come from us to the cult of Greek geometers. The ancient Greeks viewed numbers not in the formal modern sense, but instead as the lengths of line segments. So for them the goal of these constructions was to in fact ”create” the numbers. Using these basic tools the Greeks were able to create numerous useful processes such as:
- Creating a line perpendicular to another line passing through a point.
- Creating a line parallel to another line passing through a point.
- Translating an certain line segment anywhere in Euclidean space.
We will not prove these, and instead leave them to the dedicated reader to prove. So the question now becomes using these contractions what numbers could the Greeks construct? If we let p 0 and p 1 be points in the Euclidean plane we can create a line 0 through p 0 and p 1 and with out loss of generality can call the length of the line segment p 0 p 1 1. If we then create a circle centered at p 1 passing through p 0 we know that this circle will intersect 0 at a point p 2 which is a distance 2 from p 0. So by repeating this procedure it is fairly obvious that we can construct all of the positive integers.
Figure 1. Constructing the Integers
It is also possible to construct all the positive rational numbers. To see this agin let p 0 and p 1 be points in the plane and 0 be the line throughout then, and say that the line segment p 0 p 1 has length one. Then let us construct to circles C 1 and C 2 about p 0 and p 1 respectively. These circles intersect at two points q 0 and q 1 , and assuming you either proved the possibility of the above constructions, or simply believe us we can then construct a line 1 which is perpendicular to ` 0 and passes through q 0 and q 1. Let us call the intersection
10 PROF. KAREN SMITH DAVID J. BRUCE
Figure 3. Constructing the
- Constructible Numbers - 3/14/
In the previous section we discussed how the ancient Greeks viewed numbers, and how they where able to construct numbers using compasses and straightedges. Recall that if P be a collection of points in the Euclidean place then we say that another point p can be constructed in one step if from P if:
- p =
1 ∩ 2 where 2 and 1 are lines determined by P. - p ∈
∩ C where is a line and C is a circle determined by P. - p ∈ C 1 ∩ C 2 where C 1 and C 2 are circles determined by P.
The Greeks believed that starting from two points in the Euclidean plane and inductively using this type of construction they could ”create” all numbers, Now days we know that this cannot be true since this would mean that the real numbers are countable, which we know is not the case.
Having discussed how the Greeks thought about numbers let us jump forward to the time of Descartes and consider these topics in a new way by introducing the Cartesian plane. After having chosen a coordinate system we can define:
Definition 6. The collection of points P is defined over a field K. where K is a subfield of R ,if ∀p ∈ P p = (α, β) ∈ R and α, β ∈ K.
So for example, if we let P = Q^2 then P is obviously defined over Q since ∀(α, β) ∈ Q^2 by definition α, β ∈ Q. Or if we let P = {(0,
35 then P is defined over Q(
35, since both 0 and
35 are in Q(
- However, in this case P is not defined over Q since
√ 35 is not in
MATH 513: HONORS ALGEBRA II UNIVERSITY OF MICHIGAN WINTER 2012 LECTURE NOTES
In the previous section we discussed how the Greeks left many famous open questions such as wether or not one can trisect an arbitrary angle, or whether you can square a circle. Having introduced the idea of what is for a collection of points to be defined over a field we can now begin to work towards answer to these questions. In order to this we need a critical proposition.
Proposition 5. If P ⊂ R^2 is a collection of points defined over K and q = (α, β) ∈ R^2 constructed in one step from P then α and β each satisfy some quadratic polynomial in K[x].
So if we consider p 0 = (0, 0) and p 1 = (1, 0) we know that P = {p 0 , p 1 is defined over Q, and we can construct (. 5 , −
3 /2) is one step then we have the these points satisfy this polynomials:
x^2 −
and x^2 −
Before we go about proving this proposition we need to prove a useful little lemma, which at its heart is something that people learn in high school geometry.
Lemma 1. If p, q ∈ R^2 are distinct points defined over K then:
(1) The line through p and q has an equation with coefficients in K.
(2) Both circles determined by p and q have equations with coefficients in K.
Proof. Let p = (α, β) and q = (γ, δ) where {p, q} is defined over K meaning that α, β, γ, δ ∈ K. Now let us consider the two things we must prove starting by examining the line passing through p and q.
(1) First let us consider the case where α = γ. If this is the case then the line passing through p and q is vertical and so we know an equation for this line is x = α. Since α ∈ K we know in this case the lemma is true.
Now let us consider the the case where α 6 = γ. If this is the case we know that slope of the line passing through p and q is:
slope = m =
δ − β γ − α
Since K is a field we know that m ∈ K, and so when we write the equation for the line though p and q: y =
δ − β γ − α
(x − α) + β.
We see that the coefficients for the equations are in K and so have proven the first part of the lemma. (2) Now let us turn our attention to the second part of the lemma and the equations for the circles determine by p and q. These equations are: (x − α)^2 + (y − β)^2 = r^2 (x − γ)^2 + (y − δ)^2 = r^2.
MATH 513: HONORS ALGEBRA II UNIVERSITY OF MICHIGAN WINTER 2012 LECTURE NOTES
Expanding and subtracting these equations we see that: (− 2 a 1 + 2a 2 )x + (− 2 b 1 + 2b 2 )y + (a^21 + b^21 − (a 2 + b 2 )) = (r^21 − r^22 ).
So we see that we are left with the intersection of two lines, which we showed in the first case. Thus, we know that q is the solution to a quadratic polynomial.
So we see that as we wished a point is only constructible if its coordinates satisfy a degree two polynomial.
If we start with two points in the Euclidean plane we are thus able to construct a lot more. We say that a point is constructible if we can construct a line segment whose length is that number. Constructible numbers are all the coordinates of the constructible points. Since each construction has degree of either 1, 2, or 4 we know that constructible numbers have degree 2n^ for some n. With this in find we can finally prove one of the open problems we discussed in the last section. The Greeks wondered if it was possible to construct a square with the same area as a given circle. If we consider the unit circle we know that it has area π, but to construct a square with are π we need to be able to construct a square with side length
π. However, since π is transcendental, and thus π is not constructible, and so we cannot in fact square a circle.
Figure 4. Squaring A Circle: Source: Wikipedia
Another question we are now ready to tackle is whether given an arbitrary cube we can construct a cube with double the volume? If we start with a cube of which has side length 1 we want be able to construct a cube, which has volume 2. However, this would mean we need to construct sides of length
32, but since the minimum polynomial for
32 is x^3 − 2, and so it does not have degree 2n, and thus is not constructible. So just we square a circle we cannot double a cube.
14 PROF. KAREN SMITH DAVID J. BRUCE
- Galois Theory - 3/16/
Some of you might wonder why we have devoted so much time to going over constructions. Aside from the fact that they are interesting application of field extensions they give us a good motivation for our next topic - Galois Theory. Galois Theory arose out of the quest to find a general equation for 5th degree and higher polynomials. In order to begin our introduction into the subject we shall consider a few examples.
Let f be a polynomial in Q[x]. From previous work we know that f factors completely into linear terms over C. (x − λ 1 )(x − λ 2 ) · · · (x − λn).
However, C is much to big of field, and so instead let us consider Q ⊂ Q(λ 1 ,... , λn) ⊂ C. In this case we call Q(λ 1 ,... , λn) the splitting field of f.
For example let us consider, f (x) = x^4 − 4 x^2 − 5. Factoring f over C we see that:
f (x) = x −
5)(x +
5)(x − i)(x + i).
Since
5 and i are not elements of Q we know that the splitting field of f is Q(
5 , i). Since Q(
5 , i) has degree four over Q we know f has degree four.
We have seen that the degree is a very useful invariant of field, however, it does not necessarily tell us much about the field structure of the extension. Instead, it tells us more about the structure of our extension as a vector space, but it would be very useful to have a more subtle invariant of a field. One of Galois’s big contribution was the discovery of a more subtle invariant of a field called the Galois group. Before define the Galois Group we must first define what an F-automorphism.
Definition 7. Given a field extension F ⊂ K, an F-automorphism of the extension is
φ : K −−→K field homomorphism which satisfies φ(λ) = λ for all λ ∈ F.
Continuing our example from above if we let f (x) = x^4 − 4 x^2 − 5 we can think about the its Q-automorphisms. Clearly, an obvious Q-automorphism is the identity map:
φ : Q(
5 , i) −−→Q (
5 , i) x 7 −→ x.
The question now becomes what are the other Q-automorphisms. To answer this let us assume that φ is a Q-automorphism and that φ(
- = α. Using the fact that φ must be a homomorphism we know that:
φ(
- = φ(
5)φ(
- = αα = α^2.
Thus, since α^2 = 5 we know we only have two choices for α, namely +
5 and −
5. A
similarly argument will show that φ(i) must be either ±i, and so we have exactly four Q-automorphism of Q(
(5), i).
Q(
5 , i) −−→Q (
5 , i) Q(
5 , i) −−→Q (
5 , i).
