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MATH 480-01 (43128): Topics in JB-387, TuTh 6-7:50PM SYLLABUS Spring 2013

John Sarli JB-326 (909)537-5374 [email protected] TuTh 11AM-1PM, or by appointment

Text: V.S. Varadarajan, in Ancient and Modern Times (AMS Mathematical World Vol. 12, 1998) ISBN 0-8218-0989-X Prerequisites: MATH 252, MATH 329, MATH 345, MATH 355

This is a mathematics course structured around historical developments that produced our current understanding of algebraic equations. Rather than being comprehensive, we will focus on a few ideas dating from antiquity that will help explain the interpretation of solutions to algebraic equations in terms of roots of polynomials. This modern interpretation required the systematic development of from their early representation as geometric quantities to their formalization, more than two thousand years later, as structured algebraic systems. One of our objectives is to come to an understanding of why we teach algebra the way we do.

The above text will be used as a guide. We will not cover all of it but you should read as much of it as possible. I will supplement its topics with notes that will appear on my website www.math.csusb.edu/faculty/sarli/ along with this syllabus.

In order to maintain a seminar approach to this course (active participation through discussion) the grading will be based on just four components, 25% each: 1) First Written Project; 2) First Exam; 3) Second Written Project; 4) Final exam. As the course progresses I will make suggestions for suitable project topics, some of which will derive from exercises in the text. The exams, for which you may use your notes, will be require you to implement some of the basic mathematical techniques that we will develop in class. The date for the First Exam and the due date of the First Written Project will be announced in class and recorded on the website. Guidelines for project format will also be posted there.

After assessing your performance on each of the four components, course grades will be assigned as follows:

Some important dates: April 2, University closed April 12, Late add period ends April 19, Last day to drop w/o record

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May 2, First Exam; First Project due May 31, University closed June 6, Last day of class (Second Project Due) Thursday, June 13, Final Exam

Constructible Numbers

Before the invention of the line, real quantities were represented by geometric constructions, typically with straightedge and compass. More than an entire book of Euclid's Elements is devoted to geometric arithmetic consistent with his postulates for the development of plane geometry. For example, given a unit length one can construct a square whose side has this length. Then, by the Pythagorean Theorem, a diagonal of this figure has the property that the square of its length is . From our perspective we would say that is a constructible number. It is natural to ask just what numbers are constructible. This is made precise by stating axioms of constructibility consistent with the proof methods of Euclidean geometry. The purpose of these axioms is to determine which points in the plane are considered to be constructible. From these constructible points we can then define constructible numbers. For example, the distance between two constructible points is considered to be a constructible length (which we would associate with a non-negative ).

Any two distinct points may be chosen and designated constructible, and the distance between them taken as the unit length. The intersection of two constructible figures results in constructible points. The line or segment determined by two constructible points is a constructible figure. A circle with a constructible point as center and a constructible length as radius is a constructible figure.

The following theorems are easily derived from these four axioms:

The line parallel to a given constructible line and passing through a given constructible point not on the given line is a constructible figure.

The perpendicular bisector of a constructible line segment is a constructible figure.

The circle determined by three constructible points is a constructible figure. (If the three points are collinear then the resulting "circle" is clearly a constructible line.)

These three theorems were all Euclid needed to do arithmetic with constructible numbers. By arithmetic we mean the operations of . Though Euclid did not have access to the coordinate plane, we can use that setting to describe the arithmetic of constructible numbers succinctly. In fact, we can take it further by interpreting the ordered pair as the complex number . This formalism was not fully developed before the time of Gauss, but we will see that the arithmetic of constructible complex numbers simplifies many of the historical discoveries we will study.

To get started, let us take the points and as the two points in axiom . These correspond to the complex (real) numbers and . By axiom the line we call the real axis is constructible, and by axiom the circle we call the unit circle is also constructible. Then, by axiom the number is constructible (as is any integer), and by T the line we call the imaginary axis is constructible. Now

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suppose that and are constructible real numbers, that is, they have been constructed on the real axis.

Exercise. Show that is constructible, as are and .

Now suppose and . The numbers and are clearly constructible, by axiom . By T the circle through , , and is constructible. From Euclid (III.35) we have

where is the length of the segment from to the other intersection of the circle with the real axis. It follows that the product is a constructible number. Similarly, by constructing the circle through the constructible numbers , , and we see that is also a constructible number. We have shown:

The collection of constructible real numbers is closed under the arithmetic operations.

In other words, we can perform arithmetic within the set of constructible real numbers. A set of numbers with this property is called a , to use modern terminology.

Note also that the set of rational numbers is a field consisting entirely of constructible numbers.

Exercise. Use T to show that the complex number is constructible if and only if the real numbers and are constructible. Deduce that the set of all constructible complex numbers is a field and develop formulas for the sum, difference, product and quotient of two constructible complex numbers.

Geometers of Euclid's time did not have the concept of complex numbers but they discovered many constructions that had an impact on modern algebra. Here is one that allows us to find the reciprocal of any complex number although it was originally applied only to constructible points:

Consider a constructible circle centered at with radius and let be a constructible point inside the circle, that is, . The line through and is constructible as is the line through perpendicular to line . Since the points and where this perpendicular intersects the circle are constructible, the segment between any two of these four points is constructible. Also, the tangents to the circle at and are constructible (why?), and these two tangents meet on the line , hence at a constructible point .

In the early nineteenth century mathematicians began referring to as the inversion of in the given circle and interpreted this construction as the generalization of reflection in a line. Note that the right triangles , , and are similar to one another (and analogously with in place of ). This produces many relations among these constructible lengths, in particular

Thus, the product of the distances from the center of the circle to two points related by inversion is the square of the radius of the circle. The symmetrical relationship between and suggests there is an analogous construction if the constructible point is given to be outside the circle, that is, (see suggested project 5, below).

Perhaps because they did not have a theory of number systems as we understand it, the ancients were aware of some of the limitations of geometric arithmetic. In particular, they devoted a lot of mathematics to the attempt to resolve the following four questions regarding constructibility using only straightedge and compass:

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Q Is it possible to trisect an arbitrary angle? Q Is it possible to construct any regular polygon? Q Is it possible to construct a square with area equal to that of a given circle? Q Is it possible to construct a cube with volume twice that of a given cube?

Eventually it was proved that each of these questions has a negative answer, but the proofs required an understanding of roots of polynomials that was not available to the ancients. Questions of constructibility in general are answered by the following theorem, the result of work by Pierre Wantzel (1814-1848):

If a complex number is constructible then it is a root of an irreducible polynomial with integer coefficients whose degree is a power of .

So Q has a negative answer because, for example, constructing a cube with volume would require that be constructible. But is a factor of any polynomial that has as a root. For Q , consider the unit circle, whose area is . A square with side is not constructible because is transcendental, that is, not a root of any polynomial with integer coefficients. (This was proved by Ferdinand Lindemann (1852-1939) working from ideas that Charles Hermite (1822-1901) used to prove that the natural log base is transcendental.) As for Q , it can be shown that the above theorem has the following corollary:

Let be the number of integers less than or equal to that are relatively prime to . The regular -gon is constructible if and only if is a power of .

Thus a regular polygon with a prime number of sides is constructible if and only if that prime is of the form .

Suggestions for First Project:

1. Find the first five primes of the form . Prove that is prime only if is a power of . Let be a power of ; find the smallest such that is not a prime.

2. Find an angle such that is a root of the polynomial . Use Wantzel's theorem to show that the angle cannot be trisected with straightedge and compass.

3. Let be a central angle of the unit circle (centered at ) with initial ray the positive real axis and the terminal ray intersecting the circle at . Let and be collinear with such that is on the circle, is on the real axis, and has unit length. Draw such a figure and show that . Explain why this is not a constructible trisection of .

4. Let and be constructible positive real numbers. Show how to construct . Carry out the construction of a segment whose length is

.

5. Let the constructible point be outside the unit circle in the complex plane, that is, . Find a construction for , the inversion of in the unit circle. Let be the reflection of in the real line. Show that is constructible and that and are reciprocals of each other as complex numbers.

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Euclidean Algorithm

Book VII of Euclid's Elements is a departure from the earlier books in that it begins a three-book discussion of what we would now call elementary number theory. The discussion deals with properties of positive integers and their arithmetic through application of the theorems on proportions developed in the earlier books. This treatment exhibits a recognition of the natural numbers as, if not an actual number system, at least a collection if mathematical objects that have systematic properties independent of their representation as geometrical magnitudes. The most cited result from Book VII is the process for finding the greatest common divisor of two numbers and (a common divisor, usually chosen to be positive, that is a multiple of any common divisor), which we now commonly denote . This has come to be known as the Euclidean Algorithm, a process of repeated subtraction that we now associate with the division algorithm. Here is how it works using the example

Thus because is the last non-zero remainder in the process. We can represent this process abstractly as follows. Given and with , let and . Then the above process would take the form

As Euclid noticed, the process must always terminate after a finite number of steps, that is, , for some natural number . Euclid made special mention of the case when ; we would now say in this case that and are relatively prime. These books on number theory had great influence on the beginnings of algebra. For example, the work of Diophantus about seven centuries later dealt in a rather sophisticated manner with finding whole number solutions to problems involving arithmetic combinations of quantities. (This branch of algebra is now called Diophantine analysis and includes, for example, Fermat's conjecture that was finally settled in 1994 by Andrew Wiles.) As an example of a Diophantine problem imagine you have two flasks, one that holds and another that holds . Is it possible to use these flasks to measure exactly from a source of water into a bucket? To answer this, Euclid would look at his procedure and work backward:

Reversing the procedure this way reveals a lot about the structure of integers. First, it shows that for any two integers

for some integers . (The greatest common divisor does not depend on the sign of the given integers. In number theory we consider both and to be units, so if then is usually considered a greatest common divisor as well.) The integers and are not unique. The point is that if then there exist integers such that . On the other hand, if and are any integers then clearly divides , so any integer combination of and is a multiple of their greatest common divisor. (We say that the ideal generated by and is the principal ideal generated by .) This answers our original question: The two flasks do not suffice to

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measure out exactly because is not a multiple of . However, we can measure out exactly (and therefore any multiple of ) because : Pour measures of the flask into the bucket and then remove measures of the flask.

The iterative procedure for finding from has interesting properties that are studied in computer science. For a given number of steps, for example, as in the above problem, would be determined in general as follows:

Thus

are expressed entirely in terms of the quotients , by iteratively removing the remainders for .

Suggestions for First Project:

6. Find integers and such that . For integers and , find and such that in terms of the quotients in the Euclidean Algorithm assuming ; do the same assuming .

7. Let and suppose are integers such that . Explain how is related to and use this explanation to show that and are relatively prime if and are relatively prime. Show that if and only if

for any integer .

Euclidean Algorithm in

The study of proportions together with the theory of natural numbers eventually resulted in the formalization of the concept of fractions as numbers. Today we refer to the ring of integers and its associated field of fractions as . Solutions to Diophantine often force the consideration of fractional roots; for example, the attempt to "find a quantity whose cube multiplied by results in the original quantity" could be expressed in modern notation as

Because we now take algebraic notation for granted we would solve for by factoring:

Early on, the solution would typically be ignored, and then the "unique" solution might be phrased as "the desired quantity is exactly half of one unit quantity". The idea of a complete set of solutions

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would not become meaningful until the acceptance of and negative quantities as numbers in their own right. (Note that we usually teach negative numbers through the concept of "take away" and then only later formalize the idea of taking away a positive number as the addition of the negative of that number.) In honor of the Euclidean Algorithm algebraists refer to as a Euclidean domain, because for any integers with there are integers such that

where either or . Notice that the condition is redundant because we insisted . The condition is stated because there are many other Euclidean domains that have a natural ordering that is defined only for the non-zero elements. An important example is , the polynomials in a single variable with rational coefficients: for any polynomials with not the zero polynomial, there are polynomials such that

where either is the zero polynomial or the degree of is strictly less than the degree of ,

Any constant polynomial, except the zero polynomial, has degree zero. The degree of the zero polynomial is undefined. In , the degree of a polynomial plays the same part as the absolute value of a non-zero member of . What happens when we implement the Euclidean Algorithm in ? For example, let and . Then

The last non-zero remainder in this process is the constant polynomial , a polynomial of degree . Thus and have no common divisor in that is not a constant. (In fact, and is irreducible over .) In the non-zero constants are called units; they are the only polynomials that have reciprocals in .

As in , we write

and say that and are relatively prime. For any polynomials we have

and so any polynomial of the form

has the polynomial as a factor. Working the above example backward we find

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and so any member of can be written as , for some polynomials .

Suggestions for First Project:

8. Find

and write it in the form where neither nor is the zero polynomial.

Pythagorean Triples

The problem of finding triples of integers such that

has a long history, perhaps because the search for all such triples requires only consideration of the natural numbers. The Babylonians studied right triangles with side lengths hundreds of years before Pythagoras. Pythagoras pre-dated Euclid by about 250 years and produced geometric methods for producing certain triples. Euclid showed how to generate triples from any two natural numbers and interpreted his results geometrically. Diophantus lived about 500 years after Euclid and showed how to produce all Pythagorean triples. In doing so he introduced the rudiments of algebra though his notation was primitive and inefficient. (Much of the work Diophantus produced did not survive the so-called Dark Ages in Europe, and that which did was characterized by a lack of generality that left many of his methods somewhat opaque.) Since the influence of this rather elementary problem continues within modern algebraic geometry we should review this method of producing Pythagorean triples, using complex number notation. In other words, we will work within the Euclidean domain , a ring now referred to as the Gaussian integers because of the extensive contributions Gauss made to understanding their properties.

Let and be nonzero integers (there is no loss of generality in assuming they are both positive) and consider the Gaussian integer . Then

Since for any complex number we have and it follows that

In other words, is a Pythagorean triple. (Since could be positive or negative we can always work with as needed.) To an algebraist, two questions immediately arise:

Are all possible triples produced by this method? Is there a way to produce only those triples with no common divisor?

The second question would ask us to consider and to be the same triple provided is a nonzero integer. For example, we would now say that and are equivalent triples and that is the primitive representative of this

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equivalence class since all members are of the form . Assuming is primitive and that it is of the form , then and cannot both be even numbers (otherwise would all be even). Similarly, and cannot both be odd. So one must be even and the other odd. We say have opposite parity.

From now on assume is odd, and therefore is even. Note that and are both odd and that , of course, is even. Is the triple primitive? Not necessarily because could have a common divisor , whereby would have the common divisor . This suggests we impose the additional condition on that . Now the triple will be primitive because any common divisor of would need to be odd and therefore divisible by some prime . But then and so divides either or since is a prime. But if divides one of these then it divides the other (since we are assuming that divides ) and that contradicts the condition . To summarize:

Let , with of opposite parity and . Then

is a primitive Pythagorean triple.

We have not yet answered the first question, which we can now refine: Are all primitive Pythagorean triples of this form for some ?

Euclid did not fully address this, and there are differing opinions as to whether Diophantus produced a rigorous proof that the answer is affirmative since his system of notation did not allow for more than one unknown and did not admit non-positive solutions to equations. (Curiously, Diophantus seemed to have no problem with what we would call rational coefficients, and argued that they must be considered as numbers.) However, Euclid made the connection with angles in right triangles that we can easily see from this complex number representation: Given a Pythagorean triple consider the right triangle with legs and angle such that the ratio of opposite to adjacent is ; if one bisects to create a smaller right triangle then the ratio of leg opposite to to the adjacent leg is . In other words,

Notice this says that the opposite leg of the smaller triangle is

and so the ratio of to this leg is

Suggestions for First Project:

9. Assuming only that is a primitive Pythagorean triple, not necessarily of the form , show that must be odd. What does this imply about and ?

Show that any primitive triple is of the form . (This is Exercise 7 on page 15 of [Varadarajan].

Addendum: The Modular

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Since the Pythagorean triples are points on the cone

and so they are closely connected to rational points on the algebraic curve represented by this cone. The study of rational points on algebraic curves is called arithmetic algebraic geometry, which is foundational to modern number theory. An important tool in this subject that appears in many branches of mathematics and physics is the modular group , which consists of matrices of determinant with integer entries. Given

the matrix/vector multiplication

extends to matrix/vector multiplication on by identifying with the matrix

which preserves the quadratic form and so leaves the cone invariant. Further, if the triple then the triple is given by

This is the spin representation of . If can be shown that any coprime pair can be produced as for some

, whereby any Pythagorean triple on the cone can be produced as .

Note that and have opposite parity if are both odd and are both even; the matrices with this property form a subgroup of . Suggestions for First Project:

10. If show that provided . (Note that the determinant condition

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implies .)

Pell's Equation

A not too infrequent occurrence in mathematics is the attribution of credit to a mathematician for work that was developed by others. For example, Leonard Euler (1707-1783) named a particular class of Diophantine equations after the English mathematician John Pell (1610-1685), a contemporary of Fermat. Perhaps because of Euler's status in regard to solving quadratic Diophantine equations, the attribution stuck and is still used today for equations of the form

where the coefficient is a non-square positive integer and is a non-zero integer. In fact, for any given , the above equation is often denoted . Euler may have meant to credit Fermat but in any case, the techniques for solving were developed by mathematicians who flourished in India during the European Dark Ages. Brahmagupta (seventh century) knew that equations of this type have infinitely many integer solutions, and believed that integer solutions exist for any (Lagrange finally proved this in 1766). Brahmagupta did a great deal of work on quadratic equations in general, as did Aryabhata a century earlier. The contributions of these two Indian mathematician/astronomers advanced the theory of numbers by developing precise approximations of square and cube roots.

Perhaps the most significant advance by Brahmagupta was due to the way he viewed solutions to . He developed methods of generating new solutions from known solutions that foreshadowed the modern theory of algebraic groups. An algebraic group is a variety (points that solve a collection of polynomial equations in several variables) that is also a group, that is, points can be multiplied and have inverses. A simple example of an algebraic group is the unit circle

where we represent the points as complex numbers

The group product is just complex multiplication, so if and then

Also, because

Note that the rational points on the unit circle form a subgroup. For example, if and then

which allows us to interpret signed Pythagorean triples as a group by looking at their projections onto the unit circle. Brahmagupta did not have complex numbers to work with nor did he likely think of quadratic equations such as as curves. But these ideas occur to us naturally and we will see that his methods previewed ideas from algebraic geometry. The method that Brahmagupta perfected is based on the observation that two known solutions can be composed to create another solution, a process he called bhavana (the Sanskrit word for "production"). His motivation was probably an attempt to approximate square roots (see page 18 in{Varadarajan]). For a given , consider a solution to and a solution to

Defining

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we have

and we note that

Thus, is a solution to , which we might symbolize as

We can symbolize this result by saying that and imply . In particular, if then is another solution to , and if we see that the set of solutions to is closed under composition (compare this with the unit circle group, above). These observations represented a significant advance in algebraic thinking. We would naturally interpret this result in terms of analytic geometry; for example, if we could graph the hyperbola

Since solves it follows that

is also solution. But then is another solution, etc. The equations and were often considered together because

In modern language we would say that the algebraic curve is a group and is a subgroup of index , its other coset being .

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and

Brahmagupta's work prompted Fermat, Lagrange and many others to ask if all solutions could be generated from a small number of them. These questions anticipated the development of that began in earnest after the work of Galois and Abel. Here the work of many mathematicians came together to advance modern algebra. Notice that solutions can be considered as members of

which is an integral domain under ordinary multiplication:

This may have been how Brahmagupta arrived at the bhavana, but in any case we have a close connection with the Gaussian integers , which we would obtain if we allowed . (The related fields , where can be any integer, provide a rich collection of examples for understanding the the theory of polynomial roots that Galois created in 1830. For which the integral domain is actually a Euclidean domain is a difficult one and still the focus of much research in number theory.) With , notice that the equation becomes

which is a circle of radius if , and that the bhavana becomes the rule for multiplying complex numbers.

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What we now call representation theory began when algebraists such as Cayley and Hamilton experimented with using matrices to do arithmetic within integral domains. For example, given complex numbers and there product can be represented without using imaginary numbers:

Similarly, in the product can be represented by

Note that if and solve then these matrices have determinant .

Suggestions for Second Project:

11. For any given we have shown that the collection of points with integer coordinates on the curve is closed under composition. Show carefully that these points form a group, as follows: a) Show that the composition is associative, that is,

b) Find the identity element, that is, the point such that

for all .

c) For any solution , find its inverse, that is the point such that

where is the identity element.

By substituting arbitrary values and we obtain a solution to

where . This is the idea behind the solution technique called the chakravala that Brahmagupta and others developed for solving the fundamental equation

Starting with initial positive integers the chakravala generates values inductively that produce solutions to for determined values of . It was noticed that this process eventually generates and thus solves the fundamental equation . That this inductive process terminates after a finite number of steps was not proved. Rather, it was justified by noticing that the could be used to provide

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increasingly precise approximations to . This was made precise by Lagrange who proved in 1766 that

The and are strictly increasing sequences of positive integers that are the convergents to the continued fraction expansion of The chakravala always terminates after a finite number of steps in a solution . This solution to generates all positive integer solutions to under cyclic composition (with itself).

Over the centuries there were many ways to implement the chakravala, but the process can be summarized as a straightforward inductive algorithm using modern notation:

1) Let and choose . The we have with . 2) For any , determine such that

and let

Thus, it can be shown that at some stage of the process. As an example, we use this algorithm to solve

1) We have since , and . 2) Since , gives us or , but narrows it down to , and so

3) and give us , so

4) and give us , so

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Checking, we have

Note also that

which are increasingly better approximations to .

These are the first four convergents to the continued fraction for , for , which is the standard notation for the sequence of fractions indicated above. Lagrange showed that the convergents to the continued fraction for contain all positive integral solutions of ; for example, shows up as the convergent .

The inductive process behind the chakravala probably was motivated by the bhavana composition. Starting from one notices that

which solves with

so a reasonable next step would be to choose as close as possible to such that is divisible by . The process then continues based on similar considerations.

Suggestions for Second Project:

12. Find the fundamental solution to . Determine the simple continued fraction expansion of and find the convergent that is equal to .

The inductive process behind the chakravala probably was motivated by the bhavana composition. Starting from one notices that

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which solves with

so a reasonable next step would be to choose as close as possible to such that is divisible by . The process then continues based on similar considerations.

If is divisible by

so then is also divisible by

To choose as close as possible to

and .

First Exam Topics

This is a test where you should use your own notes and be able to:

Explain whether a point in the plane is constructible or not.

Apply the Euclidean Algorithm in and in and find corresponding combinations of two elements that produce the greatest common divisor.

Find the complex "square root" of given Pythagorean triple.

Apply the bhavana to integral solutions of Pell's equations and exhibit the combination as a solution to such an equation.

Implement the chakravala algorithm to find the fundamental solution to for a given .

First Project Guidelines

Select one of the suggested problems or discuss one of your own with me.

Start with a brief introductory paragraph explaining the problem and how it fits into the course.

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Explain your solution clearly, in full sentences.

Cite any references you use. If from an internet source, cite the actual published references.

Print out your final draft or send it to me as a PDF. If you do not have math software then write it up clearly in a "bluebook" using only one side of each page.

Early Algebra of Polynomial Equations

During the European Dark Ages, Hindu-Arabic mathematics continued to flourish. In particular, our modern system of decimal notation was essentially established about a thousand years ago, after a long development that dates back to the earliest Chinese notation and systems of writing in India at the time of Euclid. The work of Al-Khwarizmi in the ninth century was particularly influential in the establishment of Algebra as a discipline distinct from Geometry. Though he did not use modern algebraic notation it is clear that his treatment of algebraic equations in one variable emphasized solution methods that apply to all equations similar to a given one. Notation we would consider modern was not established until the work of Francois Viète in the sixteenth century. He was trained as a lawyer and performed duties for the government comparable to the English code decipherers in WWII, using algebraic methods few could comprehend. He wrote equations using Latin words that translate in a fairly straightforward manner into modern symbols. For example, he wrote

for

which he would solve as follows:

or

as he would write it, the first occurrence of what we would call the quadratic formula.

Even though the Hindus introduced negative numbers early on, prior to the sixteenth century the reluctance to using them as "coefficients" resulted in grouping equations by type into categories we would now consider equivalent. (Even Viète grouped quadratics by whether or not the radicands turned out positive, negative, or zero, which we often still do when teaching them for the first time.) For example, Al-Khwarizmi used verbal categories to distinguish the cases

His contemporaries did the same and many of them insisted on geometric foundations to algebraic solutions that should be based on Euclid. For example, Thabit ibn Qurra justified the solution of by drawing a square with vertices and saying that represents , represents , and represents , where is the vertex of a rectangle obtained by extending segment . He then shows that

where is the midpoint of . He also provided geometric "justifications" for the other two quadratic types given by Al-Khwarizmi.

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Another contemporary, Abu Kamil, worked freely with not only negative coefficients but also non-rational square roots. He would show these could similarly be given geometric interpretations but took the view that these were not required for justification. This viewpoint, working freely with the arithmetic of coefficients and solutions, encouraged mathematicians at the time to discover many identities that would make possible the work of later algebraists, such as these two due to Al-Karaji:

By the twelfth century, trade between the West and the East rekindled interest in Mathematics in Europe. Mathematicians became reacquainted with methods for solving quadratics and associating various interpretations with these solutions. The possibility of finding constructible quantities from given quantities started to be analyzed in terms of roots of polynomials. For example, the possibility of trisecting a given angle might be associated with the solutions of a cubic equation. The identities of Al-Karaji and others were realized as solutions of polynomial

equations of higher degree, at first by working backward; for example, noticing that solves

but that it apparently it also solves

raised questions about the relation between these two polynomials that would lead to the formal concept of factoring and the extension of the Euclidean algorithm to polynomials. The consideration of non-real roots would occur much later, but Fibonacci (Leonardo of Pisa) made use of such identities to describe quantities that were not necessarily constructible, yet they arose from simple inductive processes. The most remembered of these is the sequence attributed to him, the historical importance of which survives in the form of two modern concepts:

1) Recursion relations; 2) Exponential growth.

The Fibonacci sequence

is generated, to use modern terminology, by the recursion relation

which suggests the generalization

It is typically shown in a basic combinatorics course that any such sequence can be "solved" by expressing in the form

where are the roots of some quadratic polynomial. Clearly such relations make sense over the entire field of complex numbers (as well as other domains). For example, with we obtain

which would seem to fit no obvious pattern until we introduce a technique often used in solving differential equations: Assume for some non-zero to be determined. Then

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There will always be two complex values (possibly equal) that satisfy this quadratic relation. But then, for any constants we have

because

So satisfies the relation for . From the given conditions we have

and so, if ,

For the original Fibonacci sequence, so

so

Whereas, for our sequence with the complex numbers, , so

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Thus

for example,

Notice that can have non-real roots for real values of and , so the power representation of can require complex numbers even if all members of the sequence are real. Fibonacci did not contemplate complex representations, but he did notice in general that sequences of this type grow at a rate we would describe as exponential when considered as a function of the index.

Suggestions for Second Project:

13. a) Carry out in detail Thabit ibn Qurra's geometric interpretation of the solution to . b) Assuming and are positive real numbers and that denotes the real cube root of any real number , prove Al-Karaji's identity

and find a polynomial for which this quantity is a root.

14. Find a recursion relation with real coefficients for which and are non-real complex conjugates and express in the form for this sequence. Find a relation for which . How do you express in terms of powers in this case?

15. The polynomial , where are the roots. Solve the recursion relation

to find the general term in terms of and . The Cubic Equation Solutions to cubic equations of the form

and their variants were considered by mathematicians of Euclid's time and even extended to equations of higher degree by the time of Diophantus. Though non-integer solutions were taken seriously by Hindu and Arabic mathematicians a thousand years ago, the reluctance to admit coefficients that were not positive numbers persisted through Fibonacci's era. Fibonacci himself worked on cubic equations, but it was not until work of Italian algebraists in the fifteenth century that the concept of polynomial algebra was accepted, and with that acceptance the development of general, abstract methods took hold. The most significant contributions were made by the following names:

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Islamic algebra was first translated into Latin in the twelfth century. Along with this process came the intent to convert to the Hindu-Arabic system of numeration, including algorithms for computing with decimal notation. The school that promoted this conversion in Europe became known as the Italian abacists. Pacioli was among its leaders. He himself did not produce much original mathematics, but he was a rigorous archivist and was responsible for scoping out what was known and what was important to yet discover. In particular, he wrote in 1494 that he believed a general procedure for solving cubic equations was obtainable along the lines of the quadratic formula. He challenged the mathematical community with this task, despite the protestations of many algebraists that finding such a solution was as unlikely as carrying out the "impossible" geometric constructions of the ancients. Within the next twenty years, del Ferro produced an algebraic solution to equations of the form , but he kept this knowledge among his pupils and immediate colleagues. Nonetheless, as word began to circulate Tartaglia claimed knowledge of the general solution to equations of the form

which prompted one of del Ferro's students, Antonio Maria Fiore, to challenge Tartaglia to a contest. He posed several practical problems that reduced to solutions of cubics of the form , which Tartaglia generally was able to solve. For example, the equation

resulted from one of these problems. Today we would note that there is exactly one "real" solution

but at the time the "nature" of solutions to such equations was an idea yet to be made precise. In fact, the general absence of either an term or an term in these equations indicates some awareness of conditions that would result in a positive real soultion but which might be difficult to characterize in all cases. It should be remembered that the concept of factoring polynomials as freely as we factor integers was not well developed at this time. Rather, different "types" of equations were associated with solutions of one character or another. For example,

was seen to have as a solution, but, perhaps extraneously, and , were also observed as "solutions" by most. Note that

On the other hand,

is solved by , though it was noted by this time that solutions involving real radicals could be produced, namely :

Finally, "sophisticated" examples such as

were noted. Here, solves the equation and the other two "solutions" involve roots of negative numbers: :

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The legacy of mathematicians such as Al-Karaji gave the Italian abacists confidence in working with such expressions, but now the task was to make sense of them as numbers and to develop consistent algorithms for calculating with them. Bombelli made the breakthrough that resulted in a systematic description of complex numbers. He performed many computations with complex quantities by trusting in the power of algebraic rules; for example, he showed how to divide a real number by the complex number using the technique of multiplying both numbers by . He was thus able to discuss what came to be called "Cardano's Formula" for the cubic in great generality, introducing a designation for complex quantities equivalent to the modern notation, though wordier.

Suggestions for Second Project:

16. Any cubic polynomial with real coefficients has at least one real root. Assume are positive real numbers.

a) Show that has exactly one real solution and that . Suggestion: Start by writing

where is a quadratic polynomial in .

b) can have either three real solutions or exactly one real solution. If is the unique real solution show that and that . If all three solutions are real show that one is negative and two are positive.

Del Ferro's Formula For positive numbers and , the cubic equation

was known to have a single real solution, clearly positive. If there were no linear term the solution would simply be , the real cube root of . Possibly due to familiarity with Fibonacci's work on sequences as well as the identities Al-Karaji's such as

del Ferro assumed a solution of the form

for some that could be expressed in terms of and . He then worked backward, starting by cubing this form:

Thus, if this form really is a solution it must be the case that

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Since , del Ferro must have realized that and are required to have opposite signs for this to work. To solve for and , note that

Let

Then we can take

which yields

Probably because he preferred to work with positive radicands, del Ferro presented his formula in the equivalent form

Returning to the "challenge" problem , we have so

From this we see that Tartaglia's solution anticipated del Ferro's formula, whether he discovered it on his own or not.

The quantity is called the discriminant of the cubic. I was eventually seen that the sign of determines the nature of the solutions in a manner analogous to the discriminant of a quadratic equation. Clearly, if are both positive, as we have assumed thus far, we have and the equation has a unique real solution. After discussing Cardano's formula we will see that a positive discriminant always implies a unique real solution, regardless of the signs of and , and that a negative discriminant always implies three real solutions. What if ? Consider the example

for which we easily see that and are solutions. In fact,

and so is a "double" solution. From the graph

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we see that the curve is tangent to the -axis at . A vanishing discriminant will always indicate a "multiple" root.

Suggestions for Second Project:

17. Let be a constant. a) Expand and make a substitution of the form that removes the square term. Compute the discriminant for the polynomial in . b) Choose a real value for . Let and compute and . What happens on the graph when ?

Cardano's Formula

Suppose is a solution to the cubic equation

In order to obtain a formula without restrictions on the signs of the real numbers and , return to the original form we obtained in the derivation of del Ferro's solution:

This form, usually called Cardano's Formula, has the advantage of expressing symmetrically as a sum of terms

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Note that precisely when

which is only an issue when , in which case a non-negative discriminant means

For example, for the equation

Cardano's formula produces

The radicands of both terms are positive numbers and we can extract the real cube root of each to obtain a positive number that does in fact solve the equation. (Check this by expanding and combining terms using the rules of real radicals.) To understand the actual advances Cardano introduced we should summarize briefly what was understood at the time about cubic equations, using modern notation for clarity. 1) Any cubic equation can be reduced to the form by the substitution . Thus, adding to any solution of the reduced equation produces a solution of the original equation. However, this was generally only applied to real solutions since computations with square roots of negative numbers were not yet systematized. 2) The discriminant was understood to determine the nature of the solutions, in particular, the case implied a unique real root, a fact independent of whether is positive or negative. 3) If it was generally believed that the equation had three real solutions, but the interpretation of was not yet formalized to the point where all three could be computed systematically. 4) The case was not yet interpreted in terms of multiplicity of solutions, though it was used to compute a real solution from either del Ferro's or Cardano's expression for .

In the Ars Magna, Cardano made significant advances in the interpretation of solutions that would eventually lead to the development of the algebra of polynomials as we understand it. With regard to 1), he realized that considering the general cubic equation was important to understanding solutions other than the one produced by his formula, and that the substitution used to remove the term was just one of many that could be used to analyze solutions in general. For 2) and 3), he gave convincing arguments why produces a unique real solution and produces three real solutions, which to him implied the need for a larger number system within which one could consistently interpret so as to distinguish the cases. (Before Cardano's death Bombelli would develop systematic computations with what would come to be known as complex numbers.) He was one of the first to realize that cases where three real roots could be found by inspection indicated that all three must be deducible from his formula; for example

is solved by , , and . Here, , and removing the term yields

which must then have solutions , , and , as is easily verified. However, the formula gives

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an "irreducible case" in his terminology which yet must somehow represent all three real solutions.

As for 4), Cardano offered interpretations in order to explain how the discriminant could change from negative to positive by making slight changes in the coefficients. This approach influenced the development of the function concept, leading eventually to the continuum view of the real numbers and the important technique of studying small "perturbations" of solutions that became essential to the development of calculus.

Analysis of the Discriminant

Before explaining all solutions to the cubic in terms of Cardano's formula it is worth examining the details of how determines the nature of those solutions when and are real. To summarize: unique real solution three real solutions solutions with multiplicity

We will assume knowledge of factoring and basic calculus so that the analysis is not unnecessarily complicated. Let and assume . Then

If it follows that is a solution of

However,

and so divides . In fact,

because implies and , and factors as

We would say that is a root of multiplicity 2, a double root, of the polynomial . This analysis has shown that is a third root. Since but we cannot have , so this third root differs from the double root. Conversely, if then , whereby is a root of the polynomial , of multiplicity 3. Look again at the example

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which Cardano probably tried to interpret using his formula

Such considerations led him to surmise that perhaps a larger number system would permit various interpretations of .

Next suppose and consider again

If then for any real value of and so for precisely one real value of . Of course, in this case . For the rest of the analysis, then, we can assume . Now has two distinct critical points, at the real values

Since , the product

Thus, if then and have the same sign, and so there is precisely one value of such that ; whereas if then and have opposite signs, and so there are three distinct values of such that .

Complex Interpretation of Cardano's Formula

By considering the general form of the cubic equation

Cardano discovered many relations between the coefficients and the solutions that would, two centuries later, lead to Lagrange's theory of resolvents. In particular, for the case he was able to make these relations explicit. In hindsight, we can use factoring to understand the most basic of these relations. Suppose , , are the three real solutions. Then

In this form the coefficients are seen to be symmetric polynomials in the roots of the polynomial that defines the equation. Further, the degree of the symmetric polynomial plus the degree of the term is always equal to . The Fundamental Theorem of Algebra, first proved by Gauss (1777-1855), allows us to generalize this observation to polynomials of degree by factoring them into linear polynomials:

The FTA asserts the existence of complex numbers that allow this factorization. This justifies Cardano's speculation about cubics and

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supports Bombelli's development of a system of numbers with possibly non-real parts, but goes further in that the coefficients can now themselves be complex numbers, whereby

Here, is the elementary symmetric polynomial of degree . For example, if then

and

is the coefficient of the linear term. Note that . The FTA also provides a general definition for the discriminant that is symmetric in the roots of the polynomial:

This definition applies to any complex polynomial, but note what happens when we focus on cubics with real coefficients. For example, precisely when two of the roots are equal, as expected. However, if all three roots are real and distinct then clearly . Whereas, if there is exactly one real root then the other two are complex conjugates of each other. In this case let be the real root, let and let , the complex conjugate of . Then

Let with real. Then

This is just the reverse of the discriminant condition we have been using but it is simpler to define. Thus Lagrange and others introduced this definition, which for the polynomial becomes . Note that if the coefficients of the original polynomial are not real then need not be real. For example,

is solved by with . The interpretation of in the most general case was a key component in the eventual development of .

We still need to apply the theory of complex numbers to a consistent interpretation of Cardano's Formula. Recall that we formally expressed solutions to the equation as

whereby . Expanding and recombining we obtain

However, , and since we can assume we are not in the case where is a solution we conclude that

This was del Ferro's original observation, but now we can interpret this condition in terms of complex roots. Look again at the example

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for which Cardano's Formula yields

By the FTA, and each have three distinct values, so formally there are nine possible sums. But the condition requires that we choose the values so that

This restriction eliminates all but three of the sums. Specifically, write

Then

where is the cube root of in the first quadrant, , and again denotes the complex conjugate of any

number . Now . Thus if we choose then we must choose .

Similarly, implies and implies because . Which pairs produce the corresponding roots ? Since in each case we have we find

because ( was chosen in the first quadrant) and is the only positive root. Now and . To decide which is which, consider the polar form

so is close to because is close to . Thus , the angular argument of , is close to and so the argument of is close to . Similarly, the argument of is close to . Since it follows that . We conclude

Suggestions for Second Project:

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18. Find all solutions to

in two ways: a) Factor and find the roots. Compute the Lagrange discriminant and use it to obtain the Cardano discriminant . b) Remove the term by substitution. Then solve the cubic using Cardano's Formula. Show that the same solutions are obtained as in a) by pairing the cube roots correctly.

Quartic Equations

Lodovico Ferrari (1522-1565) was a student of Cardano. He found a general solution to fourth-degree polynomial equations, now called quartics but known then as biquadratics as a result of the method Ferrari introduced. Cardano included Ferrari's solution in the Ars Magna, listing twenty types of quartic equations. For the general polynomial

the term can be removed as usual by the substitution , so Ferrari worked with equations of the form

Having noted that such equations with no linear term could be treated by completing the square, Ferrari applied this strategy even when the linear term was present. To accomplish this he needed to introduce an auxiliary variable and use the identity

The right side can now be substituted into the original equation to obtain

Since the left side is a perfect square, the idea is to choose the auxiliary so that the right side is also a perfect square quadratic. This will occur if the discriminant of the quadratic is zero:

Thus, we can choose be any value that solves

Sometimes at least one real solution to this auxiliary cubic is evident without first removing the term. But in any case a particular solution can be found using Cardano's Formula. Since produces a perfect square quadratic, we have

Each of these quadratic equations has two solutions and so we obtain four solutions, counting multiplicity if necessary, to the original quartic. As an example, consider

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which has the auxiliary cubic

Cardano's Formula produces

resulting in three real values for , one of which is the positive value obtained by adding the cube root in the first quadrant to its complex conjugate. Taking this one for (because is then real) we have

and then, using the quadratic formula,

Of course, the exact value of is cumbersome to insert into these expressions for . However, it

is not difficult to see that the real part of for this choice of cube root is approximately , and therefore . Thus the

four values of that solve the quartic are approximately and , all four of which

approximations will produce when substituted into . In any case, it is easy to see using calculus that the function has no real zeros.

Suggestions for Second Project:

19. The equation

can be solved using DeMoivre's Theorem by noting that . Find the solutions as roots of unity and then solve the equation by removing the term and using Ferrari's method. (Note: The auxiliary cubic factors in this case, providing a rational without using Cardano's Formula.)

Lagrange Resolvents

Lagrange (1736-1813) understood the importance of working with all roots of a polynomial, which in general will be complex numbers, in order to understand their relation to the coefficients. If are the roots of a quartic his discriminant takes the form

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From Ferrari's solution we can take the roots of to be

and then we find

Since is a root of the auxiliary cubic we can use the relation

to rewrite in terms of the coefficients only, by continually replacing powers of that are greater than in the numerator until we reach the following multiple of the denominator:

For the above example we had and so , where is the Cardano discriminant of the auxiliary cubic. But is Lagrange's discriminant for the cubic, so is the Lagrange discriminant of the auxiliary cubic. This is always the case:

Let be the roots of a given quartic and let be the roots of the Ferrari auxiliary cubic. Then

Lagrange was interested in permutations of the roots of polynomials and how they combined with roots of unity. He proved several theorems about these combinations (which came to be known as Lagrange resolvents) that led to the development of group theory. For the general cubic

with roots , recall that

Lagrange defined the resolvents and to be

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Since is a primitive cube root of unity it follows that

and, after some tedious algebra,

But now, since

we can take

Then can be written entirely in terms of the coefficients of the cubic:

We can take to be any one of its cube roots, and then , whose cube will of course satisfy the third equation of . The equations of provide the roots of the cubic. In this way, Lagrange was able to reproduce Cardano's Formula with a consistent interpretation of the cube root terms.

Note that if then we can take to be any cube root of

For example, for the polynomial

for which

so we can take and obtain

as expected, since . Then the roots of the cubic are

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so .

Suggestions for Second Project:

20. For the general cubic , express the Lagrange discriminant in terms of the resolvents and only ( will not appear in the final form). Explain why is independent of the coefficient .

Resolvents for Quadratics and Quartics

The resolvents and for the cubic were derived from a general theory introduced by Lagrange to study polynomials of any degree. If are the roots of a polynomial of degree and is a primitive root of unity ( and every root of unity is for some ) then

is an example of a Lagrange resolvent. Others can be obtained by replacing with for . For the cubic, with , we have

as above. Thus, Lagrange was attempting to generalize the sum of the roots of a polynomial, which is (the coefficient of the trace term), in hopes that these other resolvents could be expressed in terms of the coefficients of the original polynomial. Note that when we must take , and then the resolvents are

Let . Then

However, , where is the constant term in the quadratic. Thus

and so the roots are

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and we obtain the quadratic formula in a similar manner to the way Lagrange obtained Cardano's Formula from the cubic resolvents.

Lagrange then applied his theory to the general quartic

with , a primitive fourth root of unity, forming the resolvents

Noticing that and , Lagrange found it easier to work with

because

and so

where

Lagrange now produced a resolvent cubic, similar but not identical to Ferrari's auxiliary cubic:

because

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These relations are tedious to find but Newton had developed powerful theorems for expressing symmetric polynomials of various degrees in terms of each other. After finding the roots of this cubic the are computed in terms of the coefficients of the quartic by taking square roots. Since it follows that

and so the square roots must be taken so as to satisfy this relation. As before, the roots of the quartic can be solved in terms of the to obtain

This method gives an interpretation to the resolvent cubic that previewed the Galois theory of polynomial roots. Note that if then this resolvent cubic becomes

The Lagrange discriminant of this cubic is identical to the Lagrange discriminant of the original quartic, without having to introduce the factor of , because

From the viewpoint of group theory, Lagrange was exploiting the fact that are each left invariant by a subgroup of order 4 of the full symmetric group, yet are permuted among each other by the full symmetric group of order 24.

Note also that the discriminant is easily expressed in terms of the because

and so

whereas in terms of the original resolvents

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Polynomials of Higher Degree

The method of Lagrange resolvents runs into difficulties for most polynomials of degree 5 or higher. Let be a primitive fifth root of unity. and suppose

Then the fundamental resolvents are

and so

For the cubic polynomial, the product of the resolvents was easily expressed in terms of the coefficients of the polynomial. For the quartic, the product of the original resolvents was not a symmetric polynomial in the roots but Lagrange was able to replace those resolvents with linear combinations of them so that the product became symmetric in the roots and hence expressible in terms of the coefficients of the quartic. For the quintic we have

where

which is not symmetric and therefore cannot be expressed in terms of . Further, there is no linear combination of the resolvents whose product is symmetric. This was an indication to Lagrange that it is unlikely that powers of the individual resolvents could be expressed as polynomials in the coefficients of the quintic. Thus, the roots of the general quintic might not be determinable by extraction of radicals.

In 1799, Paolo Ruffini (1765-1822) tried to prove the existence of quintics not solvable in radicals. In 1824, (1802-1829) managed to do this. Of course, there are quintics whose roots can be determined by radicals. For example,

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and so the five roots are obtained by solving the quadratic and the cubic. Suppose, then, that the quintic is irreducible over , which means that it does not factor into polynomials of lower degree with rational coefficients. Then the polynomial may or may not be solvable in radicals. As a solvable example, consider

which has no rational root but which obviously has as a root. (Here, is the real fifth root of .) The other four roots are where is a primitive fifth root of unity. Notice that but also in this case. However, , consistent with the formula for the roots in terms of the resolvents.

If is an integer not divisible by the square of any prime then is irreducible over the rational numbers but is solvable in radicals. These irreducible binomials have an important property: Any root is a rational function of any two of the roots. For example, if we let and for , then

are the other three roots. Evariste Galois (1811-1832) discovered the following theorem that characterized solvable polynomials of any degree:

An irreducible polynomial of prime degree is solvable by radicals if and only if all roots are rational functions of any two of them.

For quintics there is an immediate corollary:

An irreducible quintic with three real roots and two non-real roots is not solvable by radicals.

Thus, a necessary (though not sufficient) condition for an irreducible quintic, or for any irreducible polynomial of prime degree , to be solvable by radicals is that it have all real roots or a unique real root. Consider the polynomial function

which is easily seen using calculus to have exactly three real zeros:

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The polynomial does not factor over the rational numbers. The other two roots guaranteed by the Fundamental Theorem of Algebra are non-real complex conjugates. Since they cannot be produced as rational functions of the real roots this polynomial is not solvable by radicals.

Radical Extensions of the Rational Numbers

The theory that Galois developed went much deeper than the statement of the above theorem. It described how the roots of solvable polynomials could be obtained by extending the field of rational numbers . This amounted to a vast generalization of Lagrange's work on the permutation of roots and explained how these permutations determined whether or not a polynomial is irreducible. The Fundamental Theorem of Galois Theory establishes a correspondence between a particular group of permutations associated with a given irreducible polynomial and the extensions of where the roots of the polynomial are found. This theory requires a considerable amount of to describe, but certain extensions of , called radical extensions, are not difficult to define. Examples of such extensions arose in the solution of Pell's equation. If is an integer not divisible by the square of any prime then we can consider all numbers of the form

and denote this extension of by . Clearly the sum or product of any two such numbers is again such a number. For example,

However, the reciprocal of such a number is also in because

For this reason, is called a field extension of because addition, subtraction, multiplication and division can be performed within these numbers. In this context we write to emphasize this fact. If we wanted to factor

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this would require a field extension that contained the roots and . Such an extension is

. This extension does not contain the numbers and but if we create the extension and then extend by to create then we get all numbers of the form

which contains the smaller extensions , and . These are all examples of what Galois called radical extensions because they could be obtained from number fields by adjoining numbers which would be in the smaller field if a large enough integer power were taken. Galois proved that a root of an irreducible polynomial belongs to a radical extension of if and only if all of its roots belong to some radical extension. He then showed that this condition is equivalent to being able to solve the polynomial by radicals. The polynomial , then, has no root that belongs to a radical extension of the rational numbers. Nonetheless, we can take a root of this polynomial and use it to describe the field extension of where all of the roots belong. Starting with

we find that the reciprocal of can be written as a polynomial in . Similarly, it can be shown that all numbers in the extension are of the form

for some rational numbers . We say that the field is a vector space of dimension 5 over the field . In general, if is a root of an polynomial of degree that is irreducible over then is a vector space of dimension over the field . Complex numbers that occur as roots of polynomials with rational coefficients are called algebraic numbers. They comprise a countable set and are themselves a field, which we refer to as the algebraic closure of .

Final Exam Topics

This is a test where you should use your own notes and be able to:

Find a formula for in terms of the roots of a quadratic polynomial determined by the recursion

Use the discriminant to determine the nature of the roots of a cubic polynomial and find those roots using Cardano's Formula.

Find the roots of a quartic polynomial using Ferrari's biquadratic method.

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Find the roots of a quartic polynomial using Lagrange resolvents.

Express an element of as a polynomial in with rational coefficients.

Show that an irreducible quintic may not be solvable by radicals.

Eisenstein Criterion: The polynomial , where the are integers, is irreducible if there is a prime such that

Solving a quartic with resolvents: For the quartic , we have , so the Lagrange's resolvent cubic is

Using the resolvents

we have

Choosing

satisfies

Then

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after using the identities

Note that

(Using Ferrari's biquadratic method, the auxiliary cubic is

whose roots are . Selecting produces the two quadratics

Choosing the plus (minus) sign we obtain ( ) and its complex conjugate.)

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