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Home , Compass (drawing tool), Constructible number, Doubling the cube, Euclidean algorithm, Parallelogram law, Squaring the circle

and Constructions and Abstract

Introduction

Around 300 BC, wrote a series of 13 books on and theory. These books are collectively called the Elements and are some of the most famous books ever written about any subject.

In the Elements, Euclid described several “ruler and compass” constructions. By ruler, we mean a with no marks at all (so it does not look like the with centimeters or inches that you get at the store). The ruler allows you to draw the (unique) line between two (distinct) given points. The compass allows you to draw a with a given point as its center and with radius equal to the distance between two given points.

But there are three famous constructions that the Greeks could not perform using ruler and compass:

• Doubling the : constructing a cube having twice the volume of a given cube. • Trisecting the : constructing an angle 1/3 the measure of a given angle. • : constructing a with equal to that of a given circle.

The Greeks were able to construct several regular polygons, but another famous problem was also beyond their reach:

• Determine which regular polygons are constructible with ruler and compass.

These famous problems were open (unsolved) for 2000 years! Thanks to the modern tools of , we now know the solutions:

• It is impossible to double the cube, trisect the angle, or square the circle using only ruler (straightedge) and compass. • We also know precisely which regular polygons can be constructed and which ones cannot.

(Not everyone seems to be aware of this, however. To this day, around the world occasionally receive communications from people claiming to have found a method to trisect the angle, for example).

The idea of using algebra to bear on geometry problems, and vice-versa, is very beautiful and powerful. You are already familiar with , where you introduce a system of coordinates (Cartesian or otherwise) that allows you to describe many geometric objects and their measures (lengths, , volumes, angle measures) through algebraic equations and manipulations.

We will embark on a road that also uses algebra to solve these famous old problems from classical Greek geometry. But the algebra that we need is more sophisticated. It deals with generalized algebraic structures and their properties, and we call it abstract algebra.

Some of the objects from abstract algebra that we will meet along our journey are groups, rings, fields, rings of , extensions, algebraic and transcendental , and vector spaces.

(Vector spaces are also protagonists in the subject called linear algebra, which studies linear transformations between vector spaces).

Overview

• We will start by reviewing some of Euclid’s constructions that you learned in high school geometry.

• Next, we will identify the points in the with the complex numbers, and we will specify precisely what it means to construct a point (number) with ruler and compass and what it means for a number to be constructible.

• Then we will take a long dive into abstract algebra and introduce the objects mentioned above, together with the relevant facts that we will need.

• Finally, we will use these algebraic tools to describe properties of the constructible numbers and to show why the three famous constructions (doubling the square, trisecting the angle, squaring the circle) are impossible.

• If time allows, we will say a few words (without any technical details) about the solution of the other problem, namely determining precisely which regular polygons can be constructed.

Constructions from High School Geometry

In your high school geometry class, you probably learned several of Euclid’s ruler and compass constructions. Here are some that you should know how to do: (1) Bisect a . (2) Draw the perpendicular to a given line through a given point on the line. (3) Draw the perpendicular to a given line through a given point not on the line. (4) Bisect an angle. (5) Copy a given angle so its vertex is a given point and one of its rays lies on a given line (and we can also choose the half-plane of this line on which to copy the angle). (6) Draw the line to a given line through a given point not on the line (the says this line is unique). (7) Draw a triangle that is similar to a given triangle, with a prescribed side. (8) Subdivide a line segment into equal parts, for any 1.

𝑛𝑛 𝑛𝑛 ≥ For example, constructions (1) – (3) all rely on the fact that the perpendicular bisector of a line segment is also the set of all points that are equidistant from its two endpoints. This fact is easily proved by properties of isometries (distance- preserving transformations of the plane, also known as rigid motions) or by triangle congruence criteria. This fact allows you to construct the perpendicular bisector of a line segment very easily:

Therefore, you not only bisected the line segment, which is construction (1), but you also constructed a line perpendicular to a given line. Question 1: Can you modify or add to this construction to perform (2) and (3)? Construction (4), bisecting an angle, can be accomplished by making use of the SSS triangle congruence criterion:

Question 2: Can you prove that the angle bisector of an angle (of measure less that 1800) is the ray (inside the angle and with initial point at the vertex) whose points are equidistant from the sides of the angle?

Construction (5), copying an angle, also relies on SSS:

Constructions (6) and (7) are both based on construction (5). The construction of parallel lines (6) relies on the theorem that if a transversal to two lines produces congruent corresponding , then the two lines are parallel:

The construction of similar triangles (7) relies on the AA criterion for triangle similarity. It is based on copying two angles whose vertices are the endpoints of a given line segment.

Finally, construction (8) is based on construction (6) of parallel lines and on easy proportions resulting from similar triangles.

Question 3: Explain how to use ruler and compass to subdivide a line segment into equal parts.

𝑛𝑛

The Set of Constructible Numbers

As you know, we can identify the Euclidean plane with the set of all complex numbers: we set up a , for example a system of Cartesian coordinates, and then every point in the plane can be uniquelyℂ identified with its coordinates ( , ), or equivalently with the = ( , ) = + : 𝑃𝑃 𝑎𝑎 𝑏𝑏 𝑧𝑧 𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑖𝑖𝑖𝑖

With this identification, points in the plane are complex numbers to us, and constructing a point is the same as constructing a complex number.

But what exactly does it mean to construct a point or number? We should give a precise definition. We are allowed the following constructions of lines and :

R: Given two distinct points , , we can draw the line that passes through them.

For this, of course, we use the𝑧𝑧1 ruler.𝑧𝑧2 C: Given a point and two distinct points , , we can draw the circle with center and radius | |. 𝑧𝑧1 𝑧𝑧2 𝑧𝑧3 For this,𝑧𝑧1 of course, we𝑧𝑧 2use− 𝑧𝑧the3 compass.

From these lines and circles, their points of (when not empty) give us new points as follows:

= the point of intersection of two distinct lines constructed as above.

𝑃𝑃𝑙𝑙𝑙𝑙 = the point(s) of intersection of a line and a circle constructed as above. 𝑃𝑃𝑙𝑙𝑙𝑙 = the point(s) of intersection of two distinct circles constructed as above. 𝑃𝑃 𝑐𝑐𝑐𝑐 Now we can use these new points to construct more lines and circles, which in turn yield new points, etc.

Definition: The point is constructible if there is a finite sequence of ruler and compass constructions using R, C, , and that begins with 0 and 1 and ends with . 𝑧𝑧 ∈ ℂ 𝑃𝑃𝑙𝑙𝑙𝑙 𝑃𝑃𝑙𝑙𝑙𝑙 𝑃𝑃𝑐𝑐𝑐𝑐 𝑧𝑧 We define = { is constructible}.

ℭ 𝑧𝑧 ∈ ℂ ∣ 𝑧𝑧 This is the set of all constructible numbers.

Examples: • Every is constructible ( is the set of all the ).

• Every “Gaussian𝑚𝑚 ∈ ℤ ” +ℤ [ ] is constructible. • The points ± are constructible:𝑚𝑚 𝑖𝑖𝑖𝑖 ∈ ℤ they𝑖𝑖 are the intersection points of the two 1 3 circles with radius√ 1 having centers at 0 and 1, respectively. 2 𝑖𝑖 2 • is constructible if and only if (its complex conjugate) is constructible. This is because of symmetry about the -axis. Any construction that starts with 0 and 1 and𝑧𝑧 ends with can be reflected 𝑧𝑧in an obvious way about the -axis to end with . This will also be clear once we establish𝑥𝑥 a few basic facts about constructible numbers (they 𝑧𝑧form a “field”, and = + is constructible 𝑥𝑥if and only if both 𝑧𝑧 and are constructible). 𝑧𝑧 𝑎𝑎 𝑖𝑖𝑖𝑖 𝑎𝑎 𝑏𝑏 Question 4: Explain how to construct = + where , .

𝑧𝑧 𝑚𝑚 𝑖𝑖𝑖𝑖 𝑚𝑚 𝑛𝑛 ∈ ℤ Example: Since we can copy any given angle on a given initial ray, the regular -gon is constructible if and only if = / is constructible: 2π𝑖𝑖 𝑛𝑛 •𝑛𝑛 If the regular -gon is constructible,𝜉𝜉𝑛𝑛 then𝑒𝑒 we can construct an angle of measure 2 / using the center and two adjacent vertices of the -gon (if the center is not yet a given point,𝑛𝑛 we can easily construct it, for example, as the intersection of the bisectors𝜋𝜋 𝑛𝑛 of the interior angles of the -gon). Then we copy𝑛𝑛 this angle so that it has vertex at the origin and one of its rays is the positive -axis. We can also arrange it so that the other ray lies on the upper 𝑛𝑛half plane. Then the intersection of that other ray with the is . 𝑥𝑥

• If is constructible, then𝜉𝜉𝑛𝑛 we can construct the angle of measure 2 / whose vertex is 0 and whose rays go through 1 and respectively. Copying this angle 𝑛𝑛 repeatedly𝜉𝜉 using always 0 as the vertex, and then intersecting the resulting𝜋𝜋 𝑛𝑛 rays 𝑛𝑛 with the unit circle, we obtain all the th roots𝜉𝜉 of 1, namely 1, , , … , . Joining these points in the obvious way, we obtain a regular -gon.2 𝑛𝑛−1 𝑛𝑛 𝜉𝜉𝑛𝑛 𝜉𝜉𝑛𝑛 𝜉𝜉𝑛𝑛 𝑛𝑛 Remark # 1: Playing by the Rules

We only allow exact constructions, and, by definition, they must be achieved in a finite number of steps. Also, we do not accept any other instruments except for the unmarked ruler and the compass. If we ease any of these restrictions, then more constructions become possible. That is, more numbers can be obtained.

Take for example the problem of trisecting the angle.

• If we allow a ruler that has two points marked on it, together with a compass, then we can trisect any angle: Say the two marked points on the ruler are a distance apart. Given an angle with measure , draw the circle with center and radius . Let be the 𝑟𝑟 point where the ray meets the circle and let be the point where the ray ∠𝐴𝐴𝐴𝐴𝐴𝐴 𝜃𝜃 𝑂𝑂 𝑟𝑟 𝑋𝑋 opposite to meets the circle. �𝑂𝑂𝑂𝑂����⃗ 𝑌𝑌 �𝑂𝑂𝑂𝑂����⃗ Now place the ruler with its edge on and one mark on the ray at a point . Slide it until the other marked point lies on the circle at a point . Then has �����⃗ measure /3: 𝑋𝑋 𝑂𝑂𝑂𝑂 𝐷𝐷 𝐸𝐸 ∠𝐸𝐸𝐸𝐸𝐸𝐸 𝜃𝜃

Question 5: Show that has measure /3, where is the measure of .

∠𝐸𝐸𝐸𝐸𝐸𝐸 𝜃𝜃 𝜃𝜃 ∠𝐴𝐴𝐴𝐴𝐴𝐴 • With (unmarked) ruler and compass only, we can also approximate the trisection of any angle to any desired (but not exact) accuracy:

The geometric series + + + converges to 1 = . Given an angle of 1 1 1 4 1 1 measure , we can bisect4 16it repeatedly64 ⋯, as many times1 −as4 we3 want. So we can find an angle of measure for any 1. Since we can copy (and therefore add) 𝜃𝜃 𝜃𝜃 angles, we can construct𝑛𝑛 an angle of measure as close to /3 as we want, by using 2 𝑛𝑛 ≥ sufficiently many terms of this geometric sequence. 𝜃𝜃

• If we allow constructions with infinitely many steps, then the above geometric sequence can be used to trisect any angle.

If you have a compass and a ruler with just two marks on it, not only you can trisect the angle (as we just saw), but you can also double the cube:

In the picture, is a regular hexagon with side length 1, and = 1.

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑋𝑋𝑋𝑋 Question 6 (Challenge): (a) Show that = 2 / = 2. 3 1 3 (b) Explain how𝑌𝑌𝑌𝑌 you can use√ compass and a marked ruler with two marks at a distance 1 from each other to construct a segment of length of length 2. 3 (c) Explain why the cube can be doubled with compass and a twice-marked√ ruler.

Remark # 2: Constructions from a Given Set of Points

If, instead of starting our constructions from the initial set of points {0, 1}, we start from an arbitrary finite set that contains 0 and 1, then we can talk about statements like “bisecting a given angle” or “bisecting a given segment” without having to worry about whether𝑃𝑃 this given angle or segment is constructible.

This is in fact the more natural setting for the high school constructions that we reviewed above. For example, we can bisect any given segment regardless of its length, because if the segment is given, then we can use its length in our construction. The length of the segment may not be a : it may be impossible to construct it if we start from 0 and 1. But we do not care because this number is just given to us, so we can open the compass to this length to draw a circle.

The entire theory of ruler and compass constructions works in the same way if we start from 0 and 1 or if we start from a bigger set of given points, even if the set of constructible points gets bigger. 𝑃𝑃

Having a bigger finite set of initial points does not change the impossibility of trisecting the angle or or squaring the circle, because these problems ask for a method that will trisect any angle, or double any cube, or square any circle. We can obviously trisect some angles (for example, 1800 or 900), and if we start with more given points, we may be able to trisect more angles. But there will always be angles that we cannot trisect.

Overall Strategy: a Look Ahead

From now on, things will get a little more abstract and sometimes technical. So it is a good idea to keep in mind our objective and an overall strategy to get there.

• We want to prove that certain constructions are impossible to achieve with (unmarked) ruler and compass. • It is easy to see that this is equivalent to proving that certain numbers are not constructible. • Using tools of abstract algebra, we will eventually find a property that a number must possess in order for it to be constructible. • Finally, we will show that certain numbers do not possess this property, and therefore are not constructible, and therefore certain constructions are impossible to achieve by ruler and compass.

We need to develop the algebraic tools needed to find that useful property that all constructible numbers must possess, and this will take a significant amount of work. But as a reward, the concepts from abstract algebra that we will discuss are of paramount importance in all branches of higher .

Groups

Recall that a on a set is a function : × . We use the notation ( , ) = . 𝑆𝑆 ∗ 𝑆𝑆 𝑆𝑆 → 𝑆𝑆 Definition:∗ 𝑎𝑎A 𝑏𝑏𝑎𝑎 is∗ a𝑏𝑏 set and a binary operation on such that • Associativity: ( ) =𝐺𝐺 ( ) , , ∗ 𝐺𝐺 • Identity: 𝑎𝑎 ∗such𝑏𝑏 ∗that𝑐𝑐 𝑎𝑎 ∗ 𝑏𝑏=∗ 𝑐𝑐 ∀=𝑎𝑎 𝑏𝑏 𝑐𝑐 ∈ 𝐺𝐺 • Inverses: ∃ 𝑒𝑒 ∈ 𝐺𝐺 , 𝑒𝑒 ∗ such𝑔𝑔 that𝑔𝑔 ∗ 𝑒𝑒 𝑔𝑔 ∀=𝑔𝑔 ∈ 𝐺𝐺 = −1 −1 −1 ∀ 𝑔𝑔 ∈ 𝐺𝐺 ∃ 𝑔𝑔 ∈ 𝐺𝐺 𝑔𝑔 ∗ 𝑔𝑔 𝑔𝑔 ∗ 𝑔𝑔 𝑒𝑒 Examples: • , , and (the sets of all integers, rational numbers, real numbers, and complex numbers respectively) are all groups under the operation of addition. ℤ ℚ ℝ ℂ • However, none of these sets is a group under multiplication. Why? • If we remove 0 from each of these sets, we get the sets , , and . Three of them are groups under multiplication, and one is not. Which∗ ∗ one∗ is not?∗ ℤ ℚ ℝ ℂ • Let > 1 be an integer. The set = {0, 1, 2, … , 1} under addition modulo is a group. 𝑛𝑛 ℤ𝑛𝑛 𝑛𝑛 − 𝑛𝑛• The set ( ) of × matrices with real entries is not a group under multiplication, but its subset consisting of all invertible matrices is a group. 𝑀𝑀𝑛𝑛 ℝ 𝑛𝑛 𝑛𝑛

Example: A very important example of a group is the following. Remember that a function is a bijection if it is both injective (one-to-one) and surjective (onto). Let be any nonempty set. The set of all bijections , which are called permutations of , forms a group under the operation of composition of functions. The 𝐴𝐴identity element is the identity function on , and𝐴𝐴 →the𝐴𝐴 inverse of a bijection is its inverse as a function.𝐴𝐴 This group is usually denoted by . 𝐴𝐴 𝑆𝑆𝐴𝐴 In particular, if is the set {1, 2, 3, … , }, then the resulting group is called the group of permutations on letters, or more commonly, the symmetric group on letters, and we denote𝐴𝐴 this group by 𝑛𝑛. 𝑛𝑛 𝑛𝑛 𝑆𝑆𝑛𝑛 The permutations of a mathematical object that preserve some feature(s) is a group, and groups of this kind arise in many branches of mathematics.

For example, the subset of all permutations of the plane that preserve the distance between points is a group, because composition of functions preserves this feature, as do inverse functions. This is the group of all isometries or rigid motions of the plane.

Definition: if the operation of a group is commutative, i.e., = , , then we say that is abelian. 𝐺𝐺 𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏 ∀𝑎𝑎 𝑏𝑏 ∈ 𝐺𝐺 𝐺𝐺 Examples: • The group of permutations of a set (having more than 2 elements) is not abelian, because composition of functions is not commutative. • The group of invertible × matrices under matrix multiplication is not abelian either, because matrix multiplication is not commutative. 𝑛𝑛 𝑛𝑛 • The other groups that we mentioned above are all abelian, because addition and multiplication of complex numbers is commutative, and so is addition modulo .

𝑛𝑛 Question 7: Prove that in any group ,

(a) The identity element is unique.𝐺𝐺 (b) Given , is 𝑒𝑒unique.∈ 𝐺𝐺 −1 𝑔𝑔 ∈ 𝐺𝐺 𝑔𝑔 Hint: The standard way to prove the uniqueness of an object having some property is to show that if and are two such objects, then = .

For part (a), show𝐴𝐴 that if𝐵𝐵 and are both identity elements𝐴𝐴 𝐵𝐵 of , then = . ′ ′ For part (b), show that if 𝑒𝑒 and 𝑒𝑒 are both inverse elements of 𝐺𝐺, then 𝑒𝑒 = 𝑒𝑒 . ′ ′ ℎ ℎ 𝑔𝑔 ℎ ℎ Rings and Fields

For us, a will always mean a “ with unity”. This means that multiplication is commutative and there is a multiplicative identity. If you take a class in abstract algebra, you will see a more general definition.

Definition: A ring is a set with two binary operations, + (addition) and (multiplication), such that 𝑅𝑅 ⋅ • is an abelian group under addition. We denote the additive identity by 0.

• 𝑅𝑅Multiplication is associative: ( ) = ( ) , , • Multiplication is commutative:𝑎𝑎 𝑎𝑎 𝑐𝑐= 𝑎𝑎 𝑏𝑏 𝑏𝑏 , ∀𝑎𝑎 𝑏𝑏 𝑐𝑐 ∈ 𝑅𝑅 • Multiplicative identity: 1 𝑎𝑎 such𝑎𝑎 that𝑏𝑏𝑏𝑏 1 ∀𝑎𝑎= 𝑏𝑏 ∈ 𝑅𝑅 • The distributive law holds:∃ ∈(𝑅𝑅+ ) = +𝑎𝑎 𝑎𝑎 ∀𝑎𝑎, ∈, 𝑅𝑅 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎 ∀𝑎𝑎 𝑏𝑏 𝑐𝑐 ∈ 𝑅𝑅 Note that we usually denote simply by .

𝑎𝑎 ⋅ 𝑏𝑏 𝑎𝑎𝑎𝑎 Examples: • , , and are all rings (with the usual addition and multiplication).

• ℤ ℚ is ℝa ring ℂwith addition and multiplication modulo . • ℤIf𝑛𝑛 is a ring, the set of all polynomials with coefficients𝑛𝑛 in is denoted by [ ]. This is also a ring. We will be very interested in this kind of ring, especially when is 𝑅𝑅a field (which we will define below). 𝑅𝑅 𝑅𝑅 𝑡𝑡

𝑅𝑅 Question 8: Prove that if is a ring, then 0 = 0 .

𝑅𝑅 𝑎𝑎 ∀𝑎𝑎 ∈ 𝑅𝑅 Two nonzero elements , of a ring such that = 0 are called zero .

A ring with no zero divisors𝑎𝑎 𝑏𝑏 is called an 𝑎𝑎𝑎𝑎 . So in an , = 0 = 0 or = 0 (the converse is true in any ring by Question 8).

𝑎𝑎 𝑎𝑎 ⟹ 𝑎𝑎 𝑏𝑏 A unit in a ring is an element that has a multiplicative inverse. That is, it is an element for which there is an element such that = 1.

We denote𝑢𝑢 by the additive inverse of𝑣𝑣 an element𝑢𝑢 𝑢𝑢 and by or the 1 multiplicative inverse of an element . −1 −𝑎𝑎 𝑎𝑎 𝑢𝑢 𝑢𝑢 𝑢𝑢 Question 9: (a) Prove that , , and are integral domains.

(b) Is an integralℤ ℚ ℝ domain?ℂ How about ? Explain your answers. ℤ5 ℤ6 Definition: A field is a ring with 1 0 whose nonzero elements are all units. That is, every nonzero element has a multiplicative inverse. ≠ If is a field, then the set = {0} of all nonzero elements is an abelian group under multiplication. ∗ 𝐾𝐾 𝐾𝐾 𝐾𝐾 −

Question 10: Prove that every field is an integral domain.

Examples: • , and are fields, but is not.

• ℚIf ℝ is prime,ℂ then is a ℤfield. Can you prove this? But if is composite, then is not a field. In fact, it is not even an integral domain. Can you prove this? 𝑝𝑝 ℤ𝑝𝑝 𝑛𝑛 ℤ𝑛𝑛 The Complex Numbers

Since we identified the Euclidean plane with the field of complex numbers, this is the stage in which we will work for ruler and compass constructions. Here are some things you should remember about : ℂ

ℂ • How to perform arithmetic in , both in rectangular and in polar coordinates. You should know the geometric properties of the arithmetic operations. Remember that = cos + sin . ℂ 𝑖𝑖𝜃𝜃 𝑒𝑒 𝜃𝜃 𝑖𝑖 𝜃𝜃 • The or magnitude | | of a complex number . If = + = ( , ), then | | = + is the distance from to the origin O. 𝑧𝑧 𝑧𝑧 𝑧𝑧 𝑎𝑎 𝑖𝑖𝑖𝑖 2 2 𝑎𝑎 𝑏𝑏 𝑧𝑧 √𝑎𝑎 𝑏𝑏 ∈ ℝ 𝑧𝑧 • Conjugation: if = + = ( , ), then its conjugate is = = ( , ). Remember that = | | , and therefore = for 0. 𝑧𝑧 𝑎𝑎 𝑖𝑖𝑖𝑖 𝑎𝑎 𝑏𝑏 | | 𝑧𝑧̅ 𝑎𝑎 − 𝑖𝑖𝑖𝑖 𝑎𝑎 −𝑏𝑏 2 −1 𝑧𝑧̅ 2 Also, + =𝑧𝑧𝑧𝑧̅ + 𝑧𝑧 and = 𝑧𝑧. 𝑧𝑧 𝑧𝑧 ≠

𝑧𝑧��1�����𝑧𝑧�2� 𝑧𝑧�1 𝑧𝑧�2 �𝑧𝑧�1��𝑧𝑧�2� 𝑧𝑧�1𝑧𝑧�2 • th roots: every 0 has exactly distinct th roots, located at the vertices of a regular -gon centered at O. If 0 = , let / be the unique positive 𝑛𝑛 ≠ 𝑧𝑧 ∈ ℂ 𝑛𝑛 𝑛𝑛 (that implies real, of course) th root of > 0 and𝑖𝑖𝑖𝑖 let 1 =𝑛𝑛 . Then the th roots 𝑛𝑛 ≠ 𝑧𝑧 𝑟𝑟𝑒𝑒 𝑟𝑟 2𝜋𝜋 / 𝑖𝑖 𝑛𝑛 of are = , , , … , . 𝑛𝑛 𝜃𝜃 𝑛𝑛 𝑟𝑟 𝜉𝜉 𝑒𝑒 𝑛𝑛 1 𝑛𝑛 𝑖𝑖𝑛𝑛 2 𝑛𝑛−1 𝑧𝑧 𝑟𝑟 𝑒𝑒 𝑤𝑤 𝑤𝑤𝜉𝜉𝑛𝑛 𝑤𝑤𝜉𝜉𝑛𝑛 𝑤𝑤𝜉𝜉𝑛𝑛 • In particular, the th roots of 1 are 1, , , … , . 2 𝑛𝑛−1 𝑛𝑛 𝜉𝜉𝑛𝑛 𝜉𝜉𝑛𝑛 𝜉𝜉𝑛𝑛 Finally, the complex numbers have a formidable property: is algebraically closed. This fact is also called the Fundamental Theorem of Algebra (FTA), and it means that every non-constant with coefficientsℂ in has a root, or zero, in . ℂ ℂ An immediate consequence of FTA is that every non-constant polynomial with complex coefficients factors into linear factors over . Another immediate consequence is that a polynomial of degree with complex coefficients has exactly zeros, counting multiplicity. ℂ 𝑛𝑛 𝑛𝑛 Note that and are not algebraically closed. For example, the polynomial 2 has rational coefficients but no rational roots. The polynomial + 1 has real2 coefficientsℚ ℝbut no real roots. In , any (non-constant) polynomial 2has a zero, 𝑡𝑡no −matter how large its degree. 𝑡𝑡 ℂ

The Ring of Polynomials

If is a ring, the ring of polynomials over in the indeterminate is

𝑅𝑅 [ ] = { ( ) = + + +𝑅𝑅 + | ,𝑡𝑡 } 2 𝑛𝑛 ≥0 The are𝑅𝑅 c𝑡𝑡alled the𝑝𝑝 𝑡𝑡 coefficients𝑎𝑎0 𝑎𝑎1 𝑡𝑡of 𝑎𝑎(2)𝑡𝑡. We⋯ say that𝑎𝑎𝑛𝑛𝑡𝑡 ( 𝑛𝑛) ∈ ℤ [ ] 𝑎𝑎is𝑖𝑖 a∈ polynomial𝑅𝑅 over , or with coefficients in . 𝑎𝑎𝑖𝑖 𝑝𝑝 𝑡𝑡 𝑝𝑝 𝑡𝑡 ∈ 𝑅𝑅 𝑡𝑡 𝑅𝑅 𝑅𝑅 Sometimes we will simply write instead of ( ), when the indeterminate is understood. 𝑝𝑝 𝑝𝑝 𝑡𝑡

Example: The polynomial = ( ) = 3 7 + + 2 3 is a polynomial over , since its coefficients are all integers.5 That3 is, 2 [ ]. 𝑝𝑝 𝑝𝑝 𝑡𝑡 𝑡𝑡 − 𝑡𝑡 𝑡𝑡 𝑡𝑡 − Sinceℤ , we also have [ ], 𝑝𝑝[∈]ℤ, 𝑡𝑡 [ ]. ℤ ⊆ ℚ ⊆ ℝ ⊆ ℂ 𝑝𝑝 ∈ ℚ 𝑡𝑡 𝑝𝑝 ∈ ℝ 𝑡𝑡 𝑝𝑝 ∈ ℂ 𝑡𝑡 The usual polynomial addition and multiplication make [ ] a ring. The additive identity is 0, the zero polynomial, and the multiplicative identity is the constant polynomial 1. 𝑅𝑅 𝑡𝑡 We will be mostly interested in the ring [ ] where is a field. Note that [ ] is an integral domain but not a field. Why? 𝐾𝐾 𝑡𝑡 𝐾𝐾 𝐾𝐾 𝑡𝑡

Definition: Let 0 [ ]. If = + + + where 0, then the degree of is . We write = . 𝑛𝑛 ≠ 𝑝𝑝 ∈ 𝑅𝑅 𝑡𝑡 𝑝𝑝 𝑎𝑎0 𝑎𝑎1𝑡𝑡 ⋯ 𝑎𝑎𝑛𝑛𝑡𝑡 𝑎𝑎𝑛𝑛 ≠ 𝑝𝑝 𝑛𝑛 𝜕𝜕𝑝𝑝 𝑛𝑛 We can either say that the zero polynomial has no degree, or we can define 0 = , with the symbol obeying the rules < , + = and ( ) + ( ) = . Then we can easily see that 𝜕𝜕 −∞ −∞ −∞ 𝑛𝑛 −∞ 𝑛𝑛 −∞ ∀Claim:𝑛𝑛 ∈ ℤ If , −∞ [ ], −then∞ ( −+∞ ) max{ , }, and if has no zero divisors, then ( ) = + . 𝑝𝑝 𝑞𝑞 ∈ 𝑅𝑅 𝑡𝑡 𝜕𝜕 𝑝𝑝 𝑞𝑞 ≤ 𝜕𝜕𝑝𝑝 𝜕𝜕𝑞𝑞 𝑅𝑅 𝜕𝜕 𝑝𝑝𝑝𝑝 𝜕𝜕𝑝𝑝 𝜕𝜕𝑞𝑞 A constant polynomial is a polynomial of degree 0, or the zero polynomial. A linear polynomial is a polynomial of degree 1. A quadratic, cubic, quartic, and quintic polynomial is a polynomial of degree 2, 3, 4, and 5 respectively.

Divisibility in [ ]

𝑲𝑲 𝒕𝒕 Let be any field. If you want, you can just think that = .

Division𝐾𝐾 : Let , [ ] with 0. There𝐾𝐾 existℂ unique polynomials , [ ] such that = + and < . 𝑓𝑓 𝑔𝑔 ∈ 𝐾𝐾 𝑡𝑡 𝑓𝑓 ≠ 𝑞𝑞( 𝑟𝑟is∈ the𝐾𝐾 𝑡𝑡 and 𝑔𝑔 is the𝑓𝑓𝑓𝑓 𝑟𝑟 𝜕𝜕𝑟𝑟 when𝜕𝜕𝑓𝑓 you divide by ). 𝑞𝑞 𝑟𝑟 𝑔𝑔 𝑓𝑓 The Algorithm in [ ] works just like in , except that instead of requiring the remainder to be smaller than the , we require it to have degree smaller than that of the divisor.𝐾𝐾 𝑡𝑡 ℤ Division of one polynomial by another also works almost exactly as division with remainder in .

ℤ Question 11: Find the quotient and remainder when dividing = + 5 by = + 7. 7 3 3 𝑔𝑔 𝑡𝑡 − 𝑡𝑡 𝑓𝑓 𝑡𝑡 Definition: Let , [ ]. We say that divides , or is a factor of , or is a divisor of , or is a multiple of , if [ ] such that = . 𝑓𝑓 𝑔𝑔 ∈ 𝐾𝐾 𝑡𝑡 𝑓𝑓 𝑔𝑔 𝑓𝑓 𝑔𝑔 𝑓𝑓 We use the𝑔𝑔 notation𝑔𝑔 | when divides𝑓𝑓 ∃ℎ ∈ and𝐾𝐾 𝑡𝑡 when 𝑔𝑔 does𝑓𝑓ℎ not divide . 𝑓𝑓 𝑔𝑔 𝑓𝑓 𝑔𝑔 𝑓𝑓 ∤ 𝑔𝑔 𝑓𝑓 𝑔𝑔 An important consequence of the is the Factor Theorem:

Factor Theorem: Let ( ) [ ] with > 0, and let .

Then ( ) = 0 if and only𝑝𝑝 𝑡𝑡 if∈ (𝐾𝐾 𝑡𝑡 )| (𝜕𝜕)𝑝𝑝 in [ ]. 𝛼𝛼 ∈ 𝐾𝐾 𝑝𝑝 𝛼𝛼 𝑡𝑡 − 𝛼𝛼 𝑝𝑝 𝑡𝑡 𝐾𝐾 𝑡𝑡 Question 12: Prove the Factor Theorem. Hint: Apply the Division Algorithm with = ( ) and = .

𝑔𝑔 𝑝𝑝 𝑡𝑡 𝑓𝑓 𝑡𝑡 − 𝛼𝛼 The Factor Theorem is the reason that FTA implies that every non-constant polynomial over factor completely into linear factors.

ℂ Definition: Let , [ ]. We say that [ ] is a highest common factor (hcf) or (gcd) of and if 𝑓𝑓 𝑔𝑔 ∈ 𝐾𝐾 𝑡𝑡 𝑑𝑑 ∈ 𝐾𝐾 𝑡𝑡 • | and | 𝑓𝑓 𝑔𝑔 • 𝑑𝑑( |𝑓𝑓 and 𝑑𝑑 |𝑔𝑔 ) | We𝑒𝑒 use𝑓𝑓 the𝑒𝑒 notation𝑔𝑔 ⟹ 𝑒𝑒 hcf𝑑𝑑 ( , ), or gcd( , ), or sometimes simply ( , ). 𝑓𝑓 𝑔𝑔 𝑓𝑓 𝑔𝑔 𝑓𝑓 𝑔𝑔 hcf( , ) is not unique, but it is “almost unique”: it is unique up to a constant multiple. We will usually say “the hcf of and ” with the assumption that this is understood.𝑓𝑓 𝑔𝑔 𝑓𝑓 𝑔𝑔

Exactly as it happens in , the Division Algorithm is the basis for another algorithm that produces hcf( , ) for nonzero polynomials and over . It is called the .ℤ It is based on repeated applications of the Division Algorithm and a very simple𝑓𝑓 but𝑔𝑔 important fact (that also works𝑓𝑓 just𝑔𝑔 like 𝐾𝐾in ):

If | and | in [ ], then |( + ) , [ ] ℤ 𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑐𝑐 𝐾𝐾 𝑡𝑡 𝑎𝑎 𝑝𝑝𝑝𝑝 𝑞𝑞𝑐𝑐 ∀𝑝𝑝 𝑞𝑞 ∈ 𝐾𝐾 𝑡𝑡 The Euclidean Algorithm not only produces hcf( , ), but also allows you to find a combination of and that equals hcf( , ). That is, it allows you to explicitly find polynomials and as in the following lemma𝑓𝑓 𝑔𝑔: 𝑓𝑓 𝑔𝑔 𝑓𝑓 𝑔𝑔 𝑎𝑎 𝑏𝑏 Lemma: Let , [ ] be nonzero polynomials and let = hcf( , ). Then , [ ] such that = + . 𝑓𝑓 𝑔𝑔 ∈ 𝐾𝐾 𝑡𝑡 𝑑𝑑 𝑓𝑓 𝑔𝑔 ∃𝑎𝑎 𝑏𝑏 ∈ 𝐾𝐾 𝑡𝑡 𝑑𝑑 𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏 Irreducible Polynomials and Unique

Definition: Let be a ring and let [ ] be a nonconstant polynomial. We say that is irreducible in [ ], or irreducible over , if it cannot be written as a product = 𝑅𝑅of polynomials over𝑓𝑓 ∈ 𝑅𝑅with𝑡𝑡 < and < . Equivalently, if =𝑓𝑓 , then or 𝑅𝑅is 𝑡𝑡constant. 𝑅𝑅 𝑓𝑓 𝑔𝑔ℎ 𝑅𝑅 𝜕𝜕𝑔𝑔 𝜕𝜕𝑓𝑓 𝜕𝜕ℎ 𝜕𝜕𝑓𝑓 Of𝑓𝑓 course,𝑔𝑔ℎ if can𝑔𝑔 beℎ written as a product of two polynomials over , both of lower degree, then we say that is reducible over . 𝑓𝑓 𝑅𝑅 𝑓𝑓 𝑅𝑅 Notes: • Let and [ ]. Obviously, if is reducible over , then it is reducible over . So if is irreducible over , then it is irreducible over . 𝑅𝑅1 ⊆ 𝑅𝑅2 𝑓𝑓 ∈ 𝑅𝑅1 𝑡𝑡 𝑓𝑓 𝑅𝑅1 The converse is𝑅𝑅 false,2 of𝑓𝑓 course. For example,𝑅𝑅 2 2 is irreducible over or𝑅𝑅1 but reducible over . [But see Gauss’s Lemma below2 for an important exception: if a polynomial is irreducible over , then it is also𝑡𝑡 irreducible− over ]. ℤ ℚ ℝ • Polynomials of degree 1 are alwaysℤ irreducible. ℚ • If [ ] has degree 2 or 3, then it is clear from the Factor Theorem that is reducible over (a field) if and only if it has a zero in . Equivalently, if has no zeros𝑓𝑓 ∈in 𝐾𝐾 ,𝑡𝑡 then if is irreducible over . Of course, this is not true if > 3. 𝑓𝑓 𝐾𝐾 𝐾𝐾 𝑓𝑓 𝐾𝐾 𝐾𝐾 𝜕𝜕𝑓𝑓 Definition: , [ ] are coprime if hcf( , ) = 1.

We also say 𝑓𝑓that𝑔𝑔 ∈ 𝐾𝐾is prime𝑡𝑡 to . 𝑓𝑓 𝑔𝑔 𝑓𝑓 𝑔𝑔 Lemma: Let be irreducible in [ ]. If | , then | or | .

(This lemma says𝑓𝑓 that in [ ], an𝐾𝐾 irreduc𝑡𝑡 𝑓𝑓ible𝑔𝑔ℎ element𝑓𝑓 is𝑔𝑔 prime𝑓𝑓 ℎ. The converse is true in any integral domain). 𝐾𝐾 𝑡𝑡 The ring of polynomials over a field, [ ], has the important property of being a unique factorization domain (UFD), like : 𝐾𝐾 𝑡𝑡 ℤ

Theorem: Factorization into irreducible polynomials in [ ] is unique (up to the order of the factors and a constant multiple). 𝐾𝐾 𝑡𝑡

Irreducibility Criteria

It is useful to know if a polynomial is irreducible. Not only we do not need to bother looking for factors (an does not have nontrivial factors), but the zeros and the degree of an irreducible polynomial will play a crucial role later.

In general, deciding if a polynomial over a ring is irreducible is not easy (of course, there are exceptions. For example, since is algebraically closed (FTA), the only irreducible polynomials over are the linear polynomials). ℂ We will be mainly interested ℂin the irreducibility of polynomials over , and although there is no general method to solve this problem, we do have a few useful tricks that can help in some cases. ℚ

Gauss’s Lemma: Let [ ]. If is irreducible over , then it is also irreducible over . 𝑓𝑓 ∈ ℤ 𝑡𝑡 𝑓𝑓 ℤ ℚ This lemma is very useful because instead of having to consider factors with rational coefficients, we only need to consider factors having integer coefficients. This allows for divisibility arguments and other tricks. Eisenstein’s Criterion: Let ( ) = + + + [ ]. Suppose there is a prime such that 𝑛𝑛 𝑓𝑓 𝑡𝑡 𝑎𝑎0 𝑎𝑎1𝑡𝑡 ⋯ 𝑎𝑎𝑛𝑛𝑡𝑡 ∈ ℤ 𝑡𝑡 (1) 𝑝𝑝 (2) | for 0 < (3) 2 Then𝑝𝑝 ∤ is𝑎𝑎𝑛𝑛 irreducible over𝑝𝑝 𝑎𝑎 𝑖𝑖 . ≤ 𝑖𝑖 𝑛𝑛 𝑝𝑝 ∤ 𝑎𝑎0 𝑓𝑓 ℚ This criterion is proved using Gauss’s Lemma and divisibility properties of primes.

Examples: • 3, 3 2, 2 + 25 15 + 1000 + 45 are irreducible over . 2 8 4 3 2 • 𝑡𝑡Eisenstein’s− 𝑡𝑡 Criterion− 𝑡𝑡 could appl𝑡𝑡 −y for𝑡𝑡 more than𝑡𝑡 one prime. ℚ For example, 3 10 is irreducible over . 7 𝑥𝑥 − ℚ Question 13: Prove that + + + is irreducible over . 2 5 5 4 3 1 9 𝑡𝑡 3 𝑡𝑡 𝑡𝑡 3 ℚ The next lemma follows from a clever application of Eisenstein’s Criterion and the fact that if is prime, then | for 0 < < . 𝑝𝑝 𝑝𝑝 𝑝𝑝 � � 𝑟𝑟 𝑝𝑝 𝑟𝑟 Question 14: Prove that if is prime, then | for 0 < < . 𝑝𝑝 𝑝𝑝 𝑝𝑝 � � 𝑟𝑟 𝑝𝑝 𝑟𝑟 Lemma: If is prime, then ( ) = 1 + + + + is irreducible over . 2 𝑝𝑝−1 𝑝𝑝 𝑓𝑓 𝑡𝑡 𝑡𝑡 𝑡𝑡 ⋯ 𝑡𝑡 ℚ Question 15: (a) Prove that given any positive integer , there are irreducible polynomials in [ ] with degree larger than . 𝑛𝑛 ℚ(b)𝑡𝑡 By contrast, prove that if 𝑛𝑛 [ ] is irreducible over , then = 1 or 2. Hint for (b): Prove that if 𝑓𝑓 ∈ isℝ a 𝑡𝑡zero of , then so is itsℝ complex𝜕𝜕𝑓𝑓 conjugate . Deduce that the non-real zeros of come in conjugate pairs. 𝛼𝛼 ∈ ℂ 𝑓𝑓 α 𝑓𝑓 The last irreducibility criterion has to do with reducing the coefficients of a polynomial modulo .

𝑛𝑛 Reduction Modulo : There is a natural map [ ] [ ] that reduces the coefficients modulo . 𝒏𝒏 ℤ 𝑡𝑡 → ℤ𝑛𝑛 𝑡𝑡 For example, if = 𝑛𝑛3 + 7 + 6 + 5 + 11 [ ], then 4 3 2 • Reduction modulo𝑓𝑓 2 𝑡𝑡maps 𝑡𝑡 to =𝑡𝑡 +𝑡𝑡 + ∈+ℤ1 𝑡𝑡 [ ] 4 3 • Reduction modulo 5 maps 𝑓𝑓 to 𝑓𝑓 = 𝑡𝑡3 +𝑡𝑡 2 𝑡𝑡+ +∈ 1ℤ2 𝑡𝑡 [ ] 4 3 2 𝑓𝑓 𝑓𝑓 𝑡𝑡 𝑡𝑡 𝑡𝑡 ∈ ℤ5 𝑡𝑡 Suppose that does not divide the leading coefficient of , so that the leading term does not disappear upon reduction modulo , and so = . 𝑛𝑛 𝑓𝑓 If = , then from the laws of 𝑛𝑛 𝜕𝜕we𝑓𝑓 get𝜕𝜕 𝑓𝑓 = .

So,𝑓𝑓 if 𝑔𝑔 isℎ reducible over , then is reducible over . Therefore:𝑓𝑓 𝑔𝑔ℎ

𝑓𝑓 ℤ 𝑓𝑓 ℤ𝑛𝑛 If is irreducible over , then is irreducible over

𝑓𝑓(and therefore also overℤ𝑛𝑛 by𝑓𝑓 Gauss’s Lemma) ℤ ℚ This is very nice, because has only finitely many elements, so there are only finitely many possible factors of , and you can check them all. ℤ𝑛𝑛 The trick is to find a good value of𝑓𝑓 . In practice, you can usually choose it to be a prime . 𝑛𝑛 𝑝𝑝 Question 16 (Challenge): Let , , , and be odd integers. Must

( ) = + + +𝑎𝑎0 𝑎𝑎+1 𝑎𝑎2 𝑎𝑎be3 irreducible𝑎𝑎4 over ? 4 3 2 𝑓𝑓 𝑡𝑡 𝑎𝑎4𝑡𝑡 𝑎𝑎3𝑡𝑡 𝑎𝑎2𝑡𝑡 𝑎𝑎1𝑡𝑡 𝑎𝑎0 ℚ

Field Extensions

Definition: A is a pair of fields , such that .

We use the notation : . 𝐾𝐾 𝐿𝐿 𝐾𝐾 ⊆ 𝐿𝐿 𝐿𝐿 𝐾𝐾 Another way of saying this is to say that is a subfield of . A subfield is a subset of a field that is itself a field, and therefore it is closed under field operations: addition, multiplication, additive and multiplicative𝐾𝐾 inverses.𝐿𝐿 The fields that we will work with are all subfields of . Note that any subfield of must contain 0 and 1, and therefore (since it is closed under field operations) it must contain . ℂ ℂ

ℚ Definition: Let be a field and a subset. The subfield of generated by is the intersection of all the subfields of that contain . 𝐾𝐾 𝑋𝑋 ⊆ 𝐾𝐾 𝐾𝐾 𝑋𝑋 Equivalently, it is the (unique) smallest subfield𝐾𝐾 of containing𝑋𝑋 . If contains a nonzero element, this is also equivalent𝐾𝐾 to the set 𝑋𝑋of all elements of that can be obtained from elements of by a finite sequence of field operations (addition,𝑋𝑋 multiplication, additive inverses and multiplicative inverses). 𝐾𝐾 𝑋𝑋

We use the notation ( ) for the subfield of generated by .

ℚ 𝑋𝑋 ℂ 𝑋𝑋 Definition: Let : be a field extension and a subset of the large field. The subfield of generated by is denoted by ( ). 𝐿𝐿 𝐾𝐾 𝑌𝑌 ⊆ 𝐿𝐿 When = 𝐿𝐿{ } or = { ,𝐾𝐾…∪, 𝑌𝑌 }, we write (𝐾𝐾)𝑌𝑌 and ( , … , ) respectively, instead of ({ }) or ({ , … , }). 𝑌𝑌 𝛼𝛼 𝑌𝑌 𝛼𝛼1 𝛼𝛼𝑛𝑛 𝐾𝐾 𝛼𝛼 𝐾𝐾 𝛼𝛼1 𝛼𝛼𝑛𝑛 𝐾𝐾 𝛼𝛼 𝐾𝐾 𝛼𝛼1 𝛼𝛼𝑛𝑛

Examples: • ( ) = { + , }

• ℚ 𝑖𝑖 2 =𝑝𝑝{ +𝑞𝑞𝑞𝑞 ∣ 𝑝𝑝2𝑞𝑞 ∈,ℚ }

• ℚ(�√) =� 𝑝𝑝 𝑞𝑞√ ∣ 𝑝𝑝 𝑞𝑞 ∈ ℚ / • Letℝ 𝑖𝑖 =ℂ2 . Then ( ) = { + + , , } 1 3 2 • ,𝛼𝛼 5 = { ∈+ℝ + ℚ5 +𝛼𝛼 5𝑝𝑝 ,𝑞𝑞𝑞𝑞, , 𝑟𝑟𝛼𝛼 ∣}.𝑝𝑝 𝑞𝑞 𝑟𝑟 ∈ ℚ

ℚ�𝑖𝑖 √ � 𝑝𝑝 𝑞𝑞𝑞𝑞 𝑟𝑟√ 𝑠𝑠𝑠𝑠√ ∣ 𝑝𝑝 𝑞𝑞 𝑟𝑟 𝑠𝑠 ∈ ℚ It takes some work to verify the last two examples. The third one is obvious. As for the first two,

Question 17: Prove that = { + , } and = { + 2 , } are in fact fields. 𝐿𝐿 𝑝𝑝 𝑞𝑞𝑞𝑞 ∣ 𝑝𝑝 𝑞𝑞 ∈ ℚ 𝑀𝑀 𝑝𝑝 𝑞𝑞√ ∣ 𝑝𝑝 𝑞𝑞 ∈ ℚ Hint: the only matter that is not completely straight forward is that these sets are closed under multiplicative inverses. That is, if 0 , then , and similarly for . −1 ≠ 𝑧𝑧 ∈ 𝐿𝐿 𝑧𝑧 ∈ 𝐿𝐿 𝑀𝑀 Definition: A simple extension is a field extension : such that = ( ) for some . 𝐿𝐿 𝐾𝐾 𝐿𝐿 𝐾𝐾 𝛼𝛼 That is,𝛼𝛼 ∈ is𝐿𝐿 obtained from by adjoining a single element . Such an element is called a primitive element. 𝐿𝐿 𝐾𝐾 𝛼𝛼

Obviously, the extensions ( ), 2 and 2 / of that we saw in the examples above are simple extensions. 1 3 ℚ 𝑖𝑖 ℚ�√ � ℚ� � ℚ

What is not so obvious is that the extension , 5 : is also a simple extension. In fact, , 5 = + 5 . Can you prove this? ℚ�𝑖𝑖 √ � ℚ ℚ�𝑖𝑖 √ � ℚ�𝑖𝑖 √ � Definition: A field extension : is finitely generated if there exist finitely many elements , … , such that = ( , … , ). 𝐿𝐿 𝐾𝐾 𝛼𝛼1 𝛼𝛼𝑛𝑛 ∈ 𝐿𝐿 𝐿𝐿 𝐾𝐾 𝛼𝛼1 𝛼𝛼𝑛𝑛 It is clear that ( , ) = ( ) where = ( ), so an easy induction shows that any finitely generated field extension can be obtained by a finite sequence of 1 2 2 1 simple extensions.𝐾𝐾 𝛼𝛼 𝛼𝛼 𝐿𝐿 𝛼𝛼 𝐿𝐿 𝐾𝐾 𝛼𝛼

Algebraic and Transcendental Extensions

There is a crucial distinction between two kinds of simple extensions ( ): , which depends on whether or not satisfies a nonzero polynomial over : 𝐾𝐾 𝛼𝛼 𝐾𝐾 𝛼𝛼 𝐾𝐾 Definition: Let : be a field extension. An element is algebraic over if 0 ( ) [ ] such that ( ) = 0. Otherwise is transcendental over . 𝐿𝐿 𝐾𝐾 𝛼𝛼 ∈ 𝐿𝐿 𝐾𝐾 ∃If ≠is 𝑓𝑓algebraic𝑡𝑡 ∈ 𝐾𝐾 over𝑡𝑡 , we say𝑓𝑓 𝛼𝛼 that ( ): is a simple𝛼𝛼 .𝐾𝐾 If 𝛼𝛼 is transcendental 𝐾𝐾over , we say𝐾𝐾 that𝛼𝛼 𝐾𝐾( ): is a simple transcendental extension. 𝛼𝛼 𝐾𝐾 𝐾𝐾 𝛼𝛼 𝐾𝐾

Examples: • The complex numbers , 3 and = 2 / are algebraic over . They satisfy the nonzero polynomials + 1, 3 and1 3 2 (which have coefficients in ) 𝑖𝑖 𝛼𝛼 ∈ ℝ ℚ respectively. So ( ), √23 , and2 ( ) are simple3 algebraic extensions of . 𝑡𝑡 𝑡𝑡 − 𝑡𝑡 − ℚ • It is known thatℚ 𝑖𝑖andℚ �√ are� transcendentalℚ 𝛼𝛼 over . These facts are not easyℚ to prove. This means that, for example, is not a root of any (nonzero) polynomial with rational coefficients.𝑒𝑒 𝜋𝜋 So ( ): is a simple ℚtranscendental extension. 𝜋𝜋 ℚ 𝜋𝜋 ℚ Question 18: Prove that 3 + 5 : is a simple algebraic extension.

ℚ�√ √ � ℚ Notes: • When a complex number is algebraic (respectively, transcendental) over , we simply say that is algebraic (respectively, transcendental). That is, in this 𝛼𝛼 ∈ ℂ context, we drop the “over ”. For example, 3 + 5 is algebraic, and is transcendental.ℚ 𝛼𝛼 ℚ √ √ 𝜋𝜋 • It turns out that the set of all algebraic numbers forms a field. That means that this set includes 0 and 1 (why?) and is closed under field operations (addition, multiplication, additive and multiplicative inverses). The field of all algebraic numbers is denoted by . It is a subfield of .

𝔸𝔸 ℂ Definition: The field extension : is algebraic if every element of is algebraic over . 𝐿𝐿 𝐾𝐾 𝐿𝐿 𝐾𝐾 The Minimal Polynomial

Definition: A polynomial is monic if its leading coefficient is 1. So, a monic polynomial has form ( ) = + + + + . 𝑛𝑛 𝑛𝑛−1 Note that the product of two monic𝑓𝑓 polynomials𝑡𝑡 𝑡𝑡 𝑎𝑎 𝑛𝑛is− monic.1𝑡𝑡 Also,⋯ given𝑎𝑎1𝑡𝑡 a𝑎𝑎 nonzero0 polynomial over a field, you can multiply it by a (unique) constant (the multiplicative inverse of the leading coefficient) and obtain a monic polynomial.

Claim: Let : be a field extension and let be algebraic over . Then there exists a unique monic polynomial ( ) [ ] of smallest degree such that ( ) = 0. 𝐿𝐿 𝐾𝐾 𝛼𝛼 ∈ 𝐿𝐿 𝐾𝐾 𝑚𝑚 𝑡𝑡 ∈ 𝐾𝐾 𝑡𝑡 𝑚𝑚 𝛼𝛼 Question 19: Can you prove this claim? Hint: By well order, there exists a monic polynomial ( ) [ ] of smallest possible degree such that ( ) = 0. Suppose that ( ) is another such polynomial. Show that in fact = by considering∗𝑚𝑚 𝑡𝑡 ∈ 𝐾𝐾. 𝑡𝑡 𝑚𝑚 𝛼𝛼 ∗ 𝑚𝑚 𝑡𝑡 ∗ 𝑚𝑚 𝑚𝑚 𝑚𝑚 − 𝑚𝑚 Definition: Let : be a field extension and let be algebraic over . The unique monic polynomial ( ) [ ] of smallest degree such that ( ) = 0 is called the minimal𝐿𝐿 𝐾𝐾 polynomial of over . 𝛼𝛼 ∈ 𝐿𝐿 𝐾𝐾 𝑚𝑚 𝑡𝑡 ∈ 𝐾𝐾 𝑡𝑡 𝑚𝑚 𝛼𝛼 𝛼𝛼 𝐾𝐾 This polynomial is also called the irreducible polynomial for over and is denoted it by irr( , ). This is ok, because an alternative definition for it is as the unique monic irreducible polynomial ( ) [ ] such that 𝛼𝛼 ( ) =𝐾𝐾0: 𝛼𝛼 𝐾𝐾 𝑚𝑚 𝑡𝑡 ∈ 𝐾𝐾 𝑡𝑡 𝑚𝑚 𝛼𝛼 Lemma: Let : be a field extension and let be algebraic over . The minimal polynomial of over is irreducible over , and it divides every polynomial (𝐿𝐿 )𝐾𝐾 [ ] for which ( ) = 0. 𝛼𝛼 ∈ 𝐿𝐿 𝐾𝐾 𝛼𝛼 𝐾𝐾 𝐾𝐾 𝑓𝑓 𝑡𝑡 ∈ 𝐾𝐾 𝑡𝑡 𝑓𝑓 𝛼𝛼 Let ( ) be the minimal polynomial of over . Note of course that if in [ ], then ( ) = 0. Together with this lemma, we have that ( ) = 0. 𝑚𝑚 𝑡𝑡 𝛼𝛼 𝐾𝐾 𝑚𝑚 ∣ 𝑓𝑓 𝐾𝐾That𝑡𝑡 is, the𝑓𝑓 polynom𝛼𝛼 ials for which is a zero are precisely the𝑚𝑚 multiples∣ 𝑓𝑓 ⟺ 𝑓𝑓 of𝛼𝛼 ( ). 𝛼𝛼 𝑚𝑚 𝑡𝑡 Examples: • The minimal polynomial of over or over is + 1. 2 • The minimal polynomial of 𝑖𝑖 3 overℚ is ℝ3. However,𝑡𝑡 the minimal polynomial of 3 over is simply 3, since2 3 . √ ℚ 𝑡𝑡 − • More generally,√ if ℝ , then the𝑡𝑡 minimal− √ polynomial√ ∈ ℝ of over is . / th • Let = , the𝛼𝛼 5∈ 𝐾𝐾root of 1 that makes an angle of measure𝛼𝛼 2𝐾𝐾 /5𝑡𝑡 (radians)− 𝛼𝛼 with the positive2𝜋𝜋𝜋𝜋 5 -axis. Clearly satisfies (is a zero of) the polynomial 5 1𝜉𝜉 𝑒𝑒[ ]. Is this its minimal polynomial over ? No, because this𝜋𝜋 polynomial 5 is5 not irreducible 𝑥𝑥over . Clearly𝜉𝜉 1 is a zero, so 1 is a factor. It is easy to check 𝑡𝑡that− ∈ ℚ1 =𝑡𝑡 ( 1)( + + + + 1). Therefore,ℚ is a zero of + + + 5 + 1. This polynomialℚ4 is3 the 2minimal polynomial𝑡𝑡 − of over . 4 3 5 2 𝑡𝑡 − 𝑡𝑡 − 𝑡𝑡 𝑡𝑡 𝑡𝑡 𝑡𝑡 𝜉𝜉 𝑡𝑡 𝑡𝑡 𝑡𝑡 𝑡𝑡 𝜉𝜉5 ℚ Question 20: Prove the last statement. Vector Spaces

At this point we need to introduce some important concepts from linear algebra. You may already know a little about this subject if you studied systems of linear equations, matrices, and vectors.

Definition: A vector space over a field consists of a set , a field , and maps +: × (vector addition) and : × (scalar multiplication) such that 𝑉𝑉 𝐾𝐾 𝑉𝑉 𝐾𝐾 (a) 𝑉𝑉( , +𝑉𝑉)→ is 𝑉𝑉an abelian group, and ∗ 𝐾𝐾 𝑉𝑉 → 𝑉𝑉 (b) 𝑉𝑉 , , and , , , scalar multiplication satisfies ∀ 𝑘𝑘• 𝑘𝑘1(𝑘𝑘2 ∈) 𝐾𝐾= ( ∀ )𝑣𝑣 𝑣𝑣 1 𝑣𝑣2 ∈ 𝑉𝑉 • (𝑘𝑘1 𝑘𝑘+2𝑣𝑣 ) =𝑘𝑘1𝑘𝑘2 𝑣𝑣+ • 𝑘𝑘(1 +𝑘𝑘2 𝑣𝑣) = 𝑘𝑘1𝑣𝑣 + 𝑘𝑘2𝑣𝑣 • 𝑘𝑘1 𝑣𝑣=1 𝑣𝑣2 𝑘𝑘𝑣𝑣1 𝑘𝑘𝑣𝑣2 𝑣𝑣 𝑣𝑣 It is easy to check that if is a vector space over , then:

• 0 = 0 . Here 𝑉𝑉the 0 on the left is in and𝐾𝐾 the 0 on the right is in . • 0𝑣𝑣 = 0 ∀ 𝑣𝑣 ∈ 𝑉𝑉. Here 0 is in . 𝐾𝐾 𝑉𝑉 • (𝑘𝑘 ) =∀𝑘𝑘(∈ 𝐾𝐾) = ( ) 𝑉𝑉 , . −𝑘𝑘 𝑣𝑣 𝑘𝑘 −𝑣𝑣 − 𝑘𝑘𝑘𝑘 ∀𝑘𝑘 ∈ 𝐾𝐾 𝑣𝑣 ∈ 𝑉𝑉 Examples: • , the set of ordered -tuples ( , … , ) of real numbers, is a vector space over , where𝑛𝑛 addition of vectors is by components, ( , … , ) + ( , … , ) = 1 𝑛𝑛 ( ℝ + , … , + ), and𝑛𝑛 scalar 𝑎𝑎multiplication𝑎𝑎 is given by ( , … , ) = 1 𝑛𝑛 1 𝑛𝑛 (ℝ , … , ). We can replace with any field 𝑎𝑎. 𝑎𝑎 𝑏𝑏 𝑏𝑏 𝑎𝑎1 𝑏𝑏1 𝑎𝑎𝑛𝑛 𝑏𝑏𝑛𝑛 𝑘𝑘 𝑎𝑎1 𝑎𝑎𝑛𝑛 𝑘𝑘𝑎𝑎1 𝑘𝑘𝑎𝑎𝑛𝑛 ℝ 𝐾𝐾 • [ ], the ring of polynomials over a field , is a vector space over where addition of vectors is the ordinary addition of polynomials and scalar multiplication𝐾𝐾 𝑡𝑡 is also the ordinary multiplication𝐾𝐾 of polynomials, one𝐾𝐾 of them being just a constant .

𝑘𝑘 ∈ 𝐾𝐾

• If = × ( ) denotes the set of all × matrices with real entries, then is a vector space over . Addition is the usual matrix addition, and scalar 𝑚𝑚 𝑛𝑛 multiplication𝑀𝑀 𝑀𝑀 consistsℝ of multiplying every𝑚𝑚 entry𝑛𝑛 of a matrix by the scalar. We 𝑀𝑀can replace with any fieldℝ .

ℝ 𝐾𝐾 • Let be the set of all functions : . Then is a vector space over . Addition is the usual (pointwise) addition of functions, and scalar multiplication is given𝑉𝑉 by ( )( ) = ( ) for all𝑓𝑓 ℝ →.ℝ We could𝑉𝑉 also take to consist onlyℝ of the continuous functions . 𝑘𝑘𝑘𝑘 𝑥𝑥 𝑘𝑘𝑘𝑘 𝑥𝑥 𝑥𝑥 ∈ ℝ 𝑉𝑉 ℝ → ℝ Example: If : is a field extension, then is a vector space over . Addition of vectors and scalar multiplication are simply addition and multiplication in . 𝐿𝐿 𝐾𝐾 𝐿𝐿 𝐾𝐾 That is, if , , then + is simply addition in L. As for scalar 𝐿𝐿 multiplication, say and . Since , too, and is simply multiplication𝛼𝛼 𝛽𝛽 in∈ 𝐿𝐿. 𝛼𝛼 𝛽𝛽 𝑘𝑘 ∈ 𝐾𝐾 𝛼𝛼 ∈ 𝐿𝐿 𝐾𝐾 ⊆ 𝐿𝐿 𝑘𝑘 ∈ 𝐿𝐿 𝑘𝑘𝛼𝛼 This is the only kind𝐿𝐿 of vector space that we will use.

Let be a vector space over .

Definition:𝑉𝑉 A subset spans𝐾𝐾 if every can be written in the form = + + for some 0, , and . 𝑆𝑆 ⊆ 𝑉𝑉 𝑉𝑉 𝑣𝑣 ∈ 𝑉𝑉 The𝑣𝑣 sum𝑘𝑘1𝑣𝑣1 ⋯ 𝑘𝑘 𝑛𝑛is𝑣𝑣 𝑛𝑛called a linear𝑛𝑛 ≥ combination𝑘𝑘𝑖𝑖 ∈ 𝐾𝐾 of 𝑣𝑣𝑖𝑖,∈…𝑆𝑆, . 𝑛𝑛 So, spans∑𝑖𝑖 =1 if𝑘𝑘 𝑖𝑖every𝑣𝑣𝑖𝑖 vector in is a 𝑣𝑣1 of𝑣𝑣 𝑛𝑛vectors in . When𝑆𝑆 spans𝑉𝑉 , we also say that𝑉𝑉 the vectors of span . 𝑆𝑆 𝑆𝑆 𝑉𝑉 𝑆𝑆 𝑉𝑉 Definition: The vector space is finite dimensional if there exists a finite set that spans . 𝑉𝑉 𝑆𝑆 ⊆ 𝑉𝑉 𝑉𝑉 Definition: The vectors in a subset are linearly independent over if whenever + + = 0 with 1, , and , we must have = = = 0. Otherwise, they 𝑆𝑆are⊆ linearly𝑉𝑉 dependent over . 𝐾𝐾 𝑘𝑘1𝑣𝑣1 ⋯ 𝑘𝑘𝑛𝑛𝑣𝑣𝑛𝑛 𝑛𝑛 ≥ 𝑘𝑘𝑖𝑖 ∈ 𝐾𝐾 𝑣𝑣𝑖𝑖 ∈ 𝑆𝑆 𝑘𝑘1 ⋯ 𝑘𝑘𝑛𝑛 𝐾𝐾 So, a set of vectors is linearly independent over if the only way to express the vector 0 as a linear combination of these vectors is by having all the scalar coefficients equal to zero. 𝐾𝐾 If the vectors are linearly dependent, there is a linear combination of them which equals 0 having scalar coefficients that are not all zero.

Definition: A basis for over is a set whose vectors span and are linearly independent over . 𝑉𝑉 𝐾𝐾 𝐵𝐵 ⊆ 𝑉𝑉 𝑉𝑉 𝐾𝐾 Every vector of can be written uniquely as a linear combination of basis elements with scalar coefficients. 𝑉𝑉

Example: A basis for over is {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. It is easy to check that every element of 3 can be written uniquely as a linear combination of these three vectors, with realℝ coefficients.3 ℝ This is just like setting up coordinates. ℝ This example easily generalizes to in an obvious way. 𝑛𝑛 ℝ Finally, we come to the definition of the dimension of a vector space over a field:

Definition: Let be a vector space over . The dimension of over , denoted by dim (or simply by dim when is understood), is the number of elements in any basis for 𝑉𝑉 over . 𝐾𝐾 𝑉𝑉 𝐾𝐾 𝐾𝐾𝑉𝑉 𝑉𝑉 𝐾𝐾 𝑉𝑉 𝐾𝐾 Examples: • The dimension of over is . 𝑛𝑛 • The dimension of ℝ × ( ℝ) over𝑛𝑛 is . • dim = 2. A basis𝑀𝑀𝑚𝑚 for𝑛𝑛 𝐾𝐾 over 𝐾𝐾is {1𝑚𝑚, }𝑚𝑚. • However,ℝℂ is infinite-dimensionalℂ ℝ over𝑖𝑖 (as you are asked to prove below). ℂ ℚ Notes: • The fact that dim is well defined, i.e., that every vector space has a basis, and that any basis has the same number of elements, is proved in linear algebra 𝐾𝐾 courses. 𝑉𝑉 • dim is actually a , meaning it is the cardinality of a set (a basis). It could be finite or infinite. If it is finite, that is, if is finite dimensional, 𝐾𝐾 we write𝑉𝑉 dim < . 𝑉𝑉 • Every set 𝐾𝐾𝑉𝑉 that∞ spans contains a subset that is a basis for over . • Every set 𝑆𝑆 ⊆ 𝑉𝑉 whose vectors𝑉𝑉 are linearly independent𝐵𝐵 ⊆ 𝑆𝑆 over can be enlarged𝑉𝑉 𝐾𝐾 to a basis for over . By “enlarged” we mean , of course. 𝑆𝑆 ⊆ 𝑉𝑉 𝐾𝐾 𝐵𝐵 𝑉𝑉 𝐾𝐾 𝑆𝑆 ⊆ 𝐵𝐵 Question 21: Prove that there does not exist a finite basis for over .

Hint: Find an infinite set of complex numbers that are linearlyℂ independentℚ over . Use the fact that is transcendental. ℚ 𝜋𝜋

The Degree of a Field Extension

Finally, we are ready to introduce the key concept that we will use to find a useful property of constructible numbers. Also, now we start numbering some of our results, so that we can refer back to them in subsequent proofs.

Definition: The degree of the extension : , also called the degree of over , is the dimension of as a vector space over . 𝐿𝐿 𝐾𝐾 𝐿𝐿 𝐾𝐾 𝐿𝐿 𝐾𝐾 We denote the degree of : by [ : ]. So [ : ] = dim .

If [ : ] < , we say that𝐿𝐿 the𝐾𝐾 extension𝐿𝐿 𝐾𝐾 : 𝐿𝐿 is𝐾𝐾 finite. 𝐾𝐾𝐿𝐿 𝐿𝐿 𝐾𝐾 ∞ 𝐿𝐿 𝐾𝐾 Theorem 1: Let ( ): be a simple extension.

If it is transcendental,𝐾𝐾 𝛼𝛼 then𝐾𝐾 [ ( ): ] = . If it is algebraic, then [ ( )𝐾𝐾: 𝛼𝛼] =𝐾𝐾 , where∞ is the minimal polynomial of over . 𝐾𝐾 𝛼𝛼 𝐾𝐾 𝜕𝜕𝜕𝜕 𝑚𝑚 𝛼𝛼 𝐾𝐾 In fact, when is algebraic over , we have that {1, , … , } is a basis for ( ) over , where = . 𝑛𝑛−1 𝛼𝛼 𝐾𝐾 𝛼𝛼 𝛼𝛼 In𝐾𝐾 this𝛼𝛼 case,𝐾𝐾 [ ( ): 𝑛𝑛] = 𝜕𝜕𝑚𝑚 is also called the degree of over and is denoted by deg ( , ). 𝐾𝐾 𝛼𝛼 𝐾𝐾 𝜕𝜕𝜕𝜕 𝛼𝛼 𝐾𝐾 𝛼𝛼 𝐾𝐾 Tower Law: (a) If are fields, then [ : ] = [ : ][ : ].

(b) More𝐾𝐾 ⊆ generally,𝐿𝐿 ⊆ 𝑀𝑀 if 𝑀𝑀 𝐾𝐾 are𝑀𝑀 fields𝐿𝐿 𝐿𝐿 t𝐾𝐾hen [ :𝐾𝐾0 ⊆] =𝐾𝐾1[ ⊆ :⋯ ⊆ 𝐾𝐾][𝑛𝑛 : ] [ : ] 𝐾𝐾𝑛𝑛 𝐾𝐾0 𝐾𝐾𝑛𝑛 𝐾𝐾𝑛𝑛−1 𝐾𝐾𝑛𝑛−1 𝐾𝐾𝑛𝑛−2 ⋯ 𝐾𝐾1 𝐾𝐾0

This law holds regardless of whether the extensions are finite or infinite, although we will only work with finite extensions. Part (b) follows from part (a) by an easy induction, and part (a) follows from the fact (can you prove it?) that if = { } is a basis for over and = { } is a basis for over , then = { , } is a basis for over . 𝐴𝐴 𝑥𝑥𝑖𝑖 ∣ 𝑖𝑖 ∈ 𝐼𝐼 𝐿𝐿 𝐾𝐾 𝐵𝐵 𝑦𝑦𝑗𝑗 ∣ 𝑗𝑗 ∈ 𝐽𝐽 𝑀𝑀 𝐿𝐿 𝐶𝐶 𝑥𝑥𝑖𝑖𝑦𝑦𝑗𝑗 ∣ 𝑖𝑖 ∈ 𝐼𝐼 𝑗𝑗 ∈ 𝐽𝐽 𝑀𝑀 𝐾𝐾 It is not hard to see, using the Tower Law, Theorem 1, and the quadratic formula, that: Claim 1: Let : be a field extension of degree 2, where char 2 (this means that 2 = 1 + 1 0 in ). Then = for some . 𝐿𝐿 𝐾𝐾 𝐾𝐾 ≠ ≠ 𝐾𝐾 𝐿𝐿 𝐾𝐾�√𝛼𝛼� 𝛼𝛼 ∈ 𝐾𝐾 The Constructible Numbers Revisited

Recall that the set = { is constructible} consists of those complex numbers such that there is a finite sequence of ruler and compass constructions using R, C, , ℭ and 𝑧𝑧 ∈ thatℂ ∣ begins𝑧𝑧 with 0 and 1 and ends with . 𝑧𝑧 𝑃𝑃𝑙𝑙𝑙𝑙 𝑃𝑃𝑙𝑙𝑙𝑙 𝑃𝑃𝑐𝑐𝑐𝑐 𝑧𝑧 Theorem 2: The set of all constructible numbers is a field. Moreover,

(a) = + ifℭ and only if and . (Here , , of course) (b) 𝑧𝑧 𝑎𝑎 𝑖𝑖𝑖𝑖 ∈ ℭ . 𝑎𝑎 ∈ ℭ 𝑏𝑏 ∈ ℭ 𝑎𝑎 𝑏𝑏 ∈ ℝ

𝑧𝑧 ∈ ℭ ⟹ √𝑧𝑧 ∈ ℭ (Of course, every nonzero complex number has two square roots. Statement (b) says that if is constructible, then both of its square roots are constructible, since they are simply opposites, and is a field). 𝑧𝑧 ℭ Denote by ( , ) the line that passes through and (here ).

Denote by 𝐿𝐿(𝑧𝑧1, 𝑧𝑧)2 the circle with center and𝑧𝑧1 radius𝑧𝑧2 > 0.𝑧𝑧 1 ≠ 𝑧𝑧2 𝐶𝐶 𝑧𝑧 𝑟𝑟 𝑧𝑧 ∈ ℂ 𝑟𝑟 Proof: To show that is a field, we must show that it contains 0 and 1, it is closed under addition and multiplication, it contains the opposite of any of its elements, and it contains the multiplicativeℭ inverse of any of its nonzero elements. • By definition, 0 and 1 .

• Given 0 ∈, ℭ(0, ) and∈ ℭ (0, | |) intersect at ± . So . (Of course,≠ if𝑧𝑧 ∈=ℭ0,𝐿𝐿 then𝑧𝑧 =𝐶𝐶0 𝑧𝑧) 𝑧𝑧 −𝑧𝑧 ∈ ℭ • Given , 𝑧𝑧 , then the− “𝑧𝑧 ∈ ℭ law” tells us how to construct + when 0, and are not collinear: ( , | |) and ( , | |) have + as one of its points of𝑧𝑧 intersection.𝑤𝑤 ∈ ℭ When 0, and are collinear, it is even easier to construct𝑧𝑧 𝑤𝑤 + . So𝑧𝑧 +𝑤𝑤 . 𝐶𝐶 𝑧𝑧 𝑤𝑤 𝐶𝐶 𝑤𝑤 𝑧𝑧 𝑧𝑧 𝑤𝑤 𝑧𝑧 𝑤𝑤 𝑧𝑧 𝑤𝑤 𝑧𝑧 𝑤𝑤 ∈ ℭ So far, we have proved that is a group under addition (and 1 ).

ℭ ∈ ℭ • Next we prove statement (a): If = + , drop perpendiculars from to the - and -axes (which obviously can be constructed). We get and . Now (0, | |) intersects the𝑧𝑧 -axis𝑎𝑎 at𝑖𝑖𝑖𝑖 ∈, soℭ . 𝑧𝑧 𝑥𝑥 𝑦𝑦 𝑎𝑎 ∈ ℭ 𝑖𝑖𝑖𝑖 ∈ ℭ 𝐶𝐶 𝑖𝑖𝑏𝑏 Conversely,𝑥𝑥 if𝑏𝑏 𝑏𝑏 ∈ andℭ (where , ), then (0, | |) intersects the -axis at . So and therefore = + by what we already proved. 𝑎𝑎 ∈ ℭ 𝑏𝑏 ∈ ℭ 𝑎𝑎 𝑏𝑏 ∈ ℝ 𝐶𝐶 𝑏𝑏 𝑦𝑦 𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖 ∈ ℭ 𝑧𝑧 𝑎𝑎 𝑖𝑖𝑖𝑖 ∈ ℭ • To prove that is closed under multiplication and reciprocals, we will first prove an intermediate result: { > 0} is closed under multiplication and reciprocals. Theℭ following picture shows why this is true. From or , we construct or , thenℭ draw∩ 𝑥𝑥 appropriate∈ ℝ ∣ 𝑥𝑥 parallel lines, and then we reason with similar triangles. You provide the details. 𝑎𝑎 𝑏𝑏 𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖

It follows immediately that is a field.

Now we can show that is ℭclosed∩ ℝ under multiplication and reciprocals: Let = + andℭ = + . Then = ( ) + ( + ). Using statement (a), then the fact that is a field, and then statement (a) again,𝑧𝑧 it 𝑎𝑎follows𝑖𝑖𝑖𝑖 ∈ thatℭ 𝑤𝑤 . 𝑐𝑐 𝑖𝑖𝑖𝑖 ∈ ℭ 𝑧𝑧𝑧𝑧 𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑏𝑏 𝑖𝑖 𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏 ℭ ∩ ℝ Similarly, if 0, then𝑧𝑧𝑧𝑧 ∈=ℭ , and using the same reasoning, we see 1 𝑎𝑎 𝑏𝑏 2 2 2 2 that . 𝑧𝑧 ≠ 𝑧𝑧 𝑎𝑎 +𝑏𝑏 − 𝑖𝑖 𝑎𝑎 +𝑏𝑏 1 𝑧𝑧 ∈ ℭ Now we have completed the proof that is a field.

• All that remains is to prove statement ℭ(b): Let 0 . Write = with 0 < = | | . We need to construct = / . It is clear since𝑖𝑖 𝑖𝑖 that we can construct an angle of measure . ≠ 𝑧𝑧 ∈ ℭ 𝑧𝑧 𝑟𝑟𝑒𝑒 𝑟𝑟 𝑧𝑧 ∈ ℝ Since we 𝑖𝑖can𝑖𝑖 2 bisect it, we can construct an angle of measure /2. It is then also 𝑤𝑤clear √that𝑟𝑟𝑒𝑒 if we can construct 𝑧𝑧 ∈, thenℭ we can construct = and we will be𝜃𝜃 𝜃𝜃 done. You can check that it is possible to construct as in the following picture, and (using similar triangles) that√𝑟𝑟 the segment joining 1 to𝑤𝑤 has√𝑧𝑧 length . 𝛼𝛼 𝛼𝛼 √𝑟𝑟

Therefore and we are done.

√𝑧𝑧 ∈ ℭ ⎕

Definition: The Pythagorean closure py of is the smallest subfield with the property that . ℚ ℚ 𝐾𝐾 ⊆ ℂ 𝑧𝑧 ∈ 𝐾𝐾 ⟹ √𝑧𝑧 ∈ 𝐾𝐾 That is, py is the smallest subfield of that contains the square roots of all its elements. It is the intersection of all subfields of having that property. ℚ ℂ It can also be described as the subfield of obtainedℂ from 0 and 1 by using a finite sequence of field operations and square roots. ℂ

Theorem 3: = py.

ℭ ℚ Proof: From the last theorem, is a subfield of that contains the square roots of all its elements. By the definition of py as the smallest such field, we have py . ℭ ℂ ℚ py Theℚ ⊆ detailsℭ of the reverse inclusion are tedious, but the idea is simple. Any is obtained from 0 and 1 by a finite sequence of constructions using R, C, , and . Each constructedℭ point⊆ ℚ along the sequence is the result of intersecting𝑧𝑧 ∈ ℭ two lines, or a line and a circle, or two circles. And each such 𝑙𝑙𝑙𝑙 𝑙𝑙𝑙𝑙 𝑐𝑐𝑐𝑐 intersection𝑃𝑃 𝑃𝑃 is 𝑃𝑃a solution of a system of equations that can always be expressed using field operations and square roots starting from previously constructed points (numbers). Therefore py.

𝑧𝑧 ∈ ℚ ⎕ Theorem 4: if and only if there exists a finite tower

𝑧𝑧 ∈ ℭ = of subfields of such ℚthat 𝐾𝐾0 ⊆ 𝐾𝐾 and1 ⊆ [𝐾𝐾2 :⊆ ⋯ ]⊆=𝐾𝐾2𝑛𝑛− for1 ⊆ 1𝐾𝐾𝑛𝑛 . ℂ 𝑧𝑧 ∈ 𝐾𝐾𝑛𝑛 𝐾𝐾𝑖𝑖 𝐾𝐾𝑖𝑖−1 ≤ 𝑖𝑖 ≤ 𝑛𝑛 Proof: • First suppose that = is a tower of subfields of such that and [ : ] = 2 for 1 . We show by induction that 0 1 2 𝑛𝑛−1 𝑛𝑛 for 0 ℚ. Then𝐾𝐾 ⊆it follows𝐾𝐾 ⊆ 𝐾𝐾 that⊆ ⋯ ⊆ 𝐾𝐾, since⊆ 𝐾𝐾 . ℂ 𝑧𝑧 ∈ 𝐾𝐾𝑛𝑛 𝐾𝐾𝑖𝑖 𝐾𝐾𝑖𝑖−1 ≤ 𝑖𝑖 ≤ 𝑛𝑛 𝐾𝐾𝑖𝑖 ⊆ ℭ ≤ 𝑖𝑖 ≤ 𝑛𝑛 𝑧𝑧 ∈ ℭ 𝑧𝑧 ∈ 𝐾𝐾𝑛𝑛 ⊆ ℭ = because is a subfield of and every subfield of contains .

ℚSuppose𝐾𝐾0 ⊆ ℭ . Sinceℭ [ : ] = 2ℂ, Claim 1 above says thatℂ = ℚ for some . Since , . Therefore , since contains 𝐾𝐾𝑖𝑖−1 ⊆ ℭ 𝐾𝐾𝑖𝑖 𝐾𝐾𝑖𝑖−1 𝐾𝐾𝑖𝑖 𝐾𝐾𝑖𝑖−1�√𝛼𝛼� the square roots of all its elements (recall either statement (b) of Theorem 2 or 𝑖𝑖−1 𝑖𝑖−1 Theorem 𝛼𝛼3: ∈ 𝐾𝐾= py). Therefore𝐾𝐾 ⊆ ℭ =𝛼𝛼 ∈ ℭ . √ 𝛼𝛼▪ ∈ ℭ ℭ

ℭ ℚ 𝐾𝐾𝑖𝑖 𝐾𝐾𝑖𝑖−1�√𝛼𝛼� ⊆ ℭ • Conversely, suppose that . Since = py, can be obtained from 0 and 1 (equivalently, from ) by a finite sequence of field operations and square roots. Each step in this sequence either𝑧𝑧 ∈ ℭ can be performedℭ ℚ 𝑧𝑧 in the same field or requires adjoining a square rootℚ that produces a field extension of degree 2. Therefore, discarding the repeated fields, we get a tower = of subfields of such that and [ : ] = 2 for 1 . ▪ ℚ 𝐾𝐾0 ⊆ 𝐾𝐾1 ⊆ ⋯ ⊆ 𝐾𝐾𝑛𝑛−1 ⊆ 𝐾𝐾𝑛𝑛 ℂ 𝑧𝑧 ∈ 𝐾𝐾𝑛𝑛 𝐾𝐾𝑖𝑖 𝐾𝐾𝑖𝑖−1 ≤ 𝑖𝑖 ≤ 𝑛𝑛 ⎕ Finally, we are ready to state a useful necessary condition for a complex number to be constructible:

Theorem 5: If , then [ ( ): ] = deg( , ) = 2 for some 0. 𝑚𝑚 𝛼𝛼 ∈ ℭ ℚ 𝛼𝛼 ℚ 𝛼𝛼 ℚ 𝑚𝑚 ≥ Therefore, every constructible number is algebraic, and the degree of its minimal polynomial over is a power of 2, by Theorem 1.

ℚ Proof: Let . By Theorem 4, there exists a finite tower

𝛼𝛼 ∈ ℭ = of subfields of such ℚthat 𝐾𝐾0 ⊆ 𝐾𝐾 1an⊆d 𝐾𝐾[ 2 :⊆ ⋯ ⊆] =𝐾𝐾𝑛𝑛2− for1 ⊆ 1𝐾𝐾𝑛𝑛 . By the Tower Law,ℂ [ : ]𝛼𝛼=∈2𝐾𝐾𝑛𝑛. But we𝐾𝐾𝑖𝑖 also𝐾𝐾𝑖𝑖− 1have (≤ )𝑖𝑖 ≤ 𝑛𝑛 . Using the Tower Law again, 2 = [ : ]𝑛𝑛= [ : ( )][ ( ): ]. 𝑛𝑛 𝑛𝑛 𝑛𝑛𝐾𝐾 ℚ ℚ ⊆ ℚ 𝛼𝛼 ⊆ 𝐾𝐾 Therefore [ ( ): ] 2 ,𝐾𝐾 and𝑛𝑛 ℚ so [ 𝐾𝐾( 𝑛𝑛):ℚ ]𝛼𝛼= 2ℚ 𝛼𝛼for ℚsome 0 . 𝑛𝑛 𝑚𝑚 ℚ 𝛼𝛼 ℚ ∣ ℚ 𝛼𝛼 ℚ ≤ 𝑚𝑚 ≤ 𝑛𝑛 ⎕ Caution: Theorem 5 gives a necessary condition for a complex number to be constructible. But it does not give a sufficient condition: its converse is false. That is, if has degree 2 over , it does not follow that . 𝑚𝑚 𝛼𝛼 ∈ ℂ ℚ 𝛼𝛼 ∈ ℭ Question 22: Decide whether or not each of the following numbers can be constructed with ruler and compass:

(a) = 2 + 3 + 5 + 7 +

(b) 𝛼𝛼The real root� of� the polynomial√ 𝑖𝑖 ( ) = 5 + 10 + 4 + 6 11 3 𝑓𝑓 𝑡𝑡 𝑡𝑡 𝑡𝑡 𝑡𝑡 Impossibility Proofs

Finally, we are ready to prove the impossibility of doubling the cube, trisecting the angle, and squaring the circle. Keep in mind, despite how easy this will be to us now, that these problems were open for about 2000 years!

Claim 2: = 2 / (the real of 2) is irrational. 1 3 𝛼𝛼 ∈ ℝ Proof: Suppose for a contradiction that . Write = where , . Then 𝑘𝑘 2 = = = 2 3, so . But this is impossible𝛼𝛼 ∈ ℚ by unique𝛼𝛼 𝑙𝑙 prime factorization𝑘𝑘 𝑙𝑙 ∈ ℤ in 3 𝑘𝑘 3 3 , because the3 power of 2 that exactly divides is a multiple of 3, while the 𝛼𝛼 𝑙𝑙 𝑘𝑘 𝑙𝑙 power of 2 that exactly divides 2 is 1 mod3 3. Therefore . ℤ 3 𝑘𝑘 𝑙𝑙 ≡ 𝛼𝛼 ∉ ℚ ⎕

Theorem 6: The cube cannot be doubled by ruler and compass.

Proof: Doubling the cube is clearly equivalent to constructing = 2 / . 1 3 But the minimal polynomial of over is 2. Therefore [𝛼𝛼 ( ): ] ∈=ℝ3. 3 By Theorem 5, . 𝛼𝛼 ℚ 𝑡𝑡 − ℚ 𝛼𝛼 ℚ 𝛼𝛼 ∉ ℭ ⎕ Question 23: Why is doubling the cube equivalent to constructing = 2 / ? 1 3 𝛼𝛼 ∈ ℝ For the impossibility of squaring the circle, we must take for granted the fact that is transcendental (as we mentioned a while back, this is not easy to prove). 𝜋𝜋

Theorem 7: The circle cannot be squared by ruler and compass.

Proof: Squaring the circle is clearly equivalent to constructing . Suppose that . Then (since is a field) . But then Theorem 5 implies that is algebraic, a contradiction. √𝜋𝜋 √𝜋𝜋 ∈ ℭ ℭ 𝜋𝜋 ∈ ℭ 𝜋𝜋 ⎕ Question 24: Why is squaring the circle equivalent to constructing ?

√𝜋𝜋 Theorem 8: The angle cannot be trisected by ruler and compass.

Proof: It suffices to exhibit a single angle that cannot be trisected. Below we exhibit such an angle.

⎕ Of course, some angles can be trisected (like the straight angle and the right angle). But the theorem says that it is impossible to trisect a general, arbitrary angle.

Theorem 9: The angle of measure 2 /3 cannot be trisected by ruler and compass.

𝜋𝜋 Proof: Trisecting this angle is clearly equivalent to constructing = / . Now, if , then + . But we will show that = + . 2 π𝑖𝑖 9 𝜉𝜉9 𝑒𝑒 −1 / −1 Let𝜉𝜉9 ∈=ℭ = 𝜉𝜉9 = 𝜉𝜉9 ∈.ℭ It is easy to see that +𝛼𝛼 +𝜉𝜉91 =𝜉𝜉09 (by∉ ℭdirect calculation, or by3 symmetry2π𝑖𝑖 3 , or by noting that the 2minimal polynomial of over 3 9 is 𝑤𝑤+ 𝜉𝜉+ 1). ξ Therefore𝑒𝑒 + = 1. Now, 𝑤𝑤 𝑤𝑤 2 6 3 𝑤𝑤 ℚ 𝑡𝑡 𝑡𝑡 = (𝜉𝜉9 + 𝜉𝜉9 ) −= + 3 + 3 + 3 −1 3 3 −1 −3 Since = 1𝛼𝛼= 𝜉𝜉9 𝜉𝜉=9 , we𝜉𝜉 9get 𝜉𝜉9 𝜉𝜉9 𝜉𝜉9 −3 −3 −3 9 6 𝜉𝜉9 𝜉𝜉9 ⋅ 𝜉𝜉9=⋅ 𝜉𝜉9+ 3𝜉𝜉9 + 3 + = 3 1 3 3 −1 6 and so is a zero of the𝛼𝛼 polynomial𝜉𝜉9 𝜉𝜉 9 𝜉𝜉39 + 1𝜉𝜉9 [ ]𝛼𝛼. −But this polynomial is irreducible over (see below), so it is3 the minimal polynomial of over . Since its degree𝛼𝛼 is not a power of 2, 𝑡𝑡by− Theorem𝑡𝑡 ∈5.ℚ 𝑡𝑡 ℚ 𝛼𝛼 ℚ 𝛼𝛼 ∉ ℭ ⎕ The irreducibility of 3 + 1 over follows from Gauss’s Lemma, since it is easy to see that it has no3 zeros in and therefore it is irreducible over , by the Factor Theorem and the𝑡𝑡 − fact𝑡𝑡 that the polynomialℚ has degree 3. Even easier, it has no roots in by the “Rational Rootℤ s Theorem” from high school, whichℤ in this case tells us that the only possible rational roots are ±1 (clearly neither is a root). ℚ

The Constructible Regular Polygons

We have proved the impossibility of doubling the cube, trisecting the angle, and squaring the circle by ruler and compass. The other famous problem that the Greeks were unable to solve is to determine precisely which regular polygons can be constructed by ruler and compass.

The ancient Greeks knew how to construct the regular pentagon. Obviously, they also could construct the , the square, and the regular hexagon. In general, they were quite good at finding constructions when they are possible, but they were unable to prove the impossibility of other constructions. Also, they missed the construction of the regular 17-gon. This is understandable, as this construction is very elaborate. It was Gauss who found it (in 1796, when he was 19 years old). He went on to determine exactly which regular polygons can be constructed by ruler and compass. Here are some basic ideas:

• The regular -gon is constructible if and only if = / is constructible. 2π𝑖𝑖 𝑛𝑛 We proved this𝑛𝑛 near the beginning. 𝜉𝜉𝑛𝑛 𝑒𝑒

• Since we can bisect any angle, if we can construct the regular -gon, then we can construct the regular 2 -gon. 𝑛𝑛 𝑛𝑛 • If the regular -gon and the regular -gon are constructible and , are coprime, then the regular -gon is constructible. 𝑚𝑚 𝑛𝑛 𝑚𝑚 𝑛𝑛 This is true because, since𝑚𝑚 gcd𝑚𝑚 ( , ) = 1, there exist integers , such that + = 1. Therefore + = . 2𝜋𝜋 𝑚𝑚 𝑛𝑛2𝜋𝜋 2𝜋𝜋 𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑎𝑎 𝑏𝑏𝑏𝑏 𝑎𝑎 𝑛𝑛 𝑏𝑏 𝑚𝑚 𝑚𝑚𝑚𝑚 Question 25: Finish the proof that the regular -gon is constructible if the regular -gon and the regular -gon are both constructible and gcd( , ) = 1. 𝑚𝑚𝑚𝑚 𝑚𝑚 𝑛𝑛 𝑚𝑚 𝑛𝑛 The Greeks knew the last two bullets. Since they could construct the regular -gon for = 3, 4, 5, they knew that they could construct them for 𝑛𝑛 =𝑛𝑛3, 4, 5, 6, 8, 10, 12, 15, 16, 20, … 𝑛𝑛 The first missing case in this list is = 7. Can the regular be constructed with ruler and compass? No, it cannot: 𝑛𝑛

Theorem 10: The regular heptagon cannot be constructed by ruler and compass.

Proof: We show, equivalently, that = / . Since = 1, the minimal polynomial of over divides 1. 2𝜋𝜋𝜋𝜋 7 7 7 7 7 𝜉𝜉 𝑒𝑒 ∉ ℭ 𝜉𝜉 But 1 = (𝜉𝜉7 1)(ℚ + + 𝑡𝑡 −+ + + + 1) and the second factor is irreducible7 over (why?6 ). Therefore,5 4 this3 second2 factor is the minimal polynomial of 𝑡𝑡 over− . Its𝑡𝑡 −degree𝑡𝑡 is not𝑡𝑡 a power𝑡𝑡 𝑡𝑡 of 2, 𝑡𝑡so 𝑡𝑡 by Theorem 5. ℚ 𝜉𝜉7 ℚ 𝜉𝜉7 ∉ ℭ ⎕ By Theorem 9, the regular 9-gon cannot be constructed either. How about the cases = 11, 13, 14, 17, 18, 19, … ? Using the third bullet above, we can reduce the problem to the case when is a prime power. But this is still a difficult problem. 𝑛𝑛 Using marvelous ideas involving𝑛𝑛 the interplay between group theory and field extensions, Gauss was able to completely settle the question:

Theorem 11: Let > 2 be an integer. The regular -gon can be constructed with ruler and compass if and only if = 2 , where 0 and 0 are integers and the 𝑛𝑛 are distinct Fermat 𝑟𝑟primes. 𝑛𝑛 𝑛𝑛 𝑝𝑝1 ⋯ 𝑝𝑝𝑠𝑠 𝑟𝑟 ≥ 𝑠𝑠 ≥ 𝑝𝑝𝑖𝑖 A Fermat prime is a prime of form = 2 + 1. Fermat thought that = 2 + 1 𝑘𝑘 𝑘𝑘 is prime for all 0 ( = 3, = 5, 2= 17, = 257, = 65537 are indeed2 𝑘𝑘 prime). But he was wrong: Euler showed𝑝𝑝 that is not prime. In fact, 𝐹𝐹Fermat was 0 1 2 3 4 very wrong: as 𝑘𝑘of ≥2021,𝐹𝐹 the above𝐹𝐹 numbers𝐹𝐹 3, 5,𝐹𝐹 17, 257 and𝐹𝐹 65537 are the only 5 known Fermat primes, and heuristic arguments𝐹𝐹 suggest that these are very likely the only ones.

Question 26: Decide whether or not the regular -gon is constructible from ruler and compass for each of the following values of : 𝑛𝑛 (a) 25 (b) 51 (c) 768 (d) 771 (e) for an𝑛𝑛 odd prime (f) 2 for > 1 2 𝑘𝑘 𝑝𝑝 𝑝𝑝 𝑘𝑘