Ye Olde Fundamental Theorem of Algebra
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1 Ye Olde Fundamental Theorem of Algebra James Parson Hood College April 14, 2012 James Parson (Hood College) Ye Olde FTA April 14, / 12
2 Do you know the FTA? The Fundamental Theorem of Algebra A non-constant polynomial with real coefficients has a degree-2 or degree-1 factor with real coefficients. James Parson (Hood College) Ye Olde FTA April 14, / 12
3 Do you know the FTA? The Fundamental Theorem of Algebra A non-constant polynomial with real coefficients has a degree-2 or degree-1 factor with real coefficients. Do you know this? If you know this theorem, shouldn t you be able to factor, say, x 6 + x + 1? James Parson (Hood College) Ye Olde FTA April 14, / 12
4 Do you know the FTA? The Fundamental Theorem of Algebra A non-constant polynomial with real coefficients has a degree-2 or degree-1 factor with real coefficients. Do you know this? If you know this theorem, shouldn t you be able to factor, say, x 6 + x + 1? I will discuss an old method for factoring, which was generalized to give one of the more-algebraic proofs of the FTA. James Parson (Hood College) Ye Olde FTA April 14, / 12
5 More on x 6 + x + 1 Here is a graph of y = x 6 + x + 1: James Parson (Hood College) Ye Olde FTA April 14, / 12
6 More on x 6 + x + 1 Here is a graph of y = x 6 + x + 1: No roots to get us started... James Parson (Hood College) Ye Olde FTA April 14, / 12
7 More on x 6 + x + 1 Here is a graph of y = x 6 + x + 1: No roots to get us started... We need a quadratic factor. James Parson (Hood College) Ye Olde FTA April 14, / 12
8 Descartes/Hudde method I will explain an olde method to construct a quadratic factor. James Parson (Hood College) Ye Olde FTA April 14, / 12
9 Descartes/Hudde method I will explain an olde method to construct a quadratic factor. First a bit of history: James Parson (Hood College) Ye Olde FTA April 14, / 12
10 Descartes/Hudde method I will explain an olde method to construct a quadratic factor. First a bit of history: 1637 Descartes publishes La Geometrie (100 pages): dictionary between geometry and algebra. James Parson (Hood College) Ye Olde FTA April 14, / 12
11 Descartes/Hudde method I will explain an olde method to construct a quadratic factor. First a bit of history: 1637 Descartes publishes La Geometrie (100 pages): dictionary between geometry and algebra Second Latin edition (van Schooten et al) (500 pages): extensive commentary and many examples. James Parson (Hood College) Ye Olde FTA April 14, / 12
12 Descartes/Hudde method I will explain an olde method to construct a quadratic factor. First a bit of history: 1637 Descartes publishes La Geometrie (100 pages): dictionary between geometry and algebra Second Latin edition (van Schooten et al) (500 pages): extensive commentary and many examples.newton read this edition a few years later. James Parson (Hood College) Ye Olde FTA April 14, / 12
13 Descartes/Hudde method I will explain an olde method to construct a quadratic factor. First a bit of history: 1637 Descartes publishes La Geometrie (100 pages): dictionary between geometry and algebra Second Latin edition (van Schooten et al) (500 pages): extensive commentary and many examples.newton read this edition a few years later. Part of the commentary was a letter by Jan Hudde, On the reduction of an equation. James Parson (Hood College) Ye Olde FTA April 14, / 12
14 Descartes/Hudde method I will explain an olde method to construct a quadratic factor. First a bit of history: 1637 Descartes publishes La Geometrie (100 pages): dictionary between geometry and algebra Second Latin edition (van Schooten et al) (500 pages): extensive commentary and many examples.newton read this edition a few years later. Part of the commentary was a letter by Jan Hudde, On the reduction of an equation. Hudde s letter contains 22 rules for reducing equations in one variable essentially for factoring polynomials. James Parson (Hood College) Ye Olde FTA April 14, / 12
15 Descartes/Hudde method: XIX Regula James Parson (Hood College) Ye Olde FTA April 14, / 12
16 Descartes/Hudde method: Rule 19 For x 6 + x + 1, here is what Hudde tells us to do: James Parson (Hood College) Ye Olde FTA April 14, / 12
17 Descartes/Hudde method: Rule 19 For x 6 + x + 1, here is what Hudde tells us to do: Divide by x 2 + yx + w, where y and w are unknowns. We want to choose real y and w to make the remainder 0. James Parson (Hood College) Ye Olde FTA April 14, / 12
18 Descartes/Hudde method: Rule 19 For x 6 + x + 1, here is what Hudde tells us to do: Divide by x 2 + yx + w, where y and w are unknowns. We want to choose real y and w to make the remainder 0. The remainder is x( y 5 + 4y 3 w 3yw 2 + 1) + ( y 4 w + 3y 2 w 2 w 3 + 1). James Parson (Hood College) Ye Olde FTA April 14, / 12
19 Descartes/Hudde method: Rule 19 For x 6 + x + 1, here is what Hudde tells us to do: Divide by x 2 + yx + w, where y and w are unknowns. We want to choose real y and w to make the remainder 0. The remainder is x( y 5 + 4y 3 w 3yw 2 + 1) + ( y 4 w + 3y 2 w 2 w 3 + 1). (I did this on a computer.) James Parson (Hood College) Ye Olde FTA April 14, / 12
20 Descartes/Hudde method: Rule 19 For x 6 + x + 1, here is what Hudde tells us to do: Divide by x 2 + yx + w, where y and w are unknowns. We want to choose real y and w to make the remainder 0. The remainder is x( y 5 + 4y 3 w 3yw 2 + 1) + ( y 4 w + 3y 2 w 2 w 3 + 1). (I did this on a computer.) To get remainder 0, we need to make both coefficients 0 by finding a solution (y, w) to this system of equations: y 5 + 4y 3 w 3yw = 0 y 4 w + 3y 2 w 2 w = 0. James Parson (Hood College) Ye Olde FTA April 14, / 12
21 Are we better off? James Parson (Hood College) Ye Olde FTA April 14, / 12
22 Are we better off? Some algebra shows that shows that y 5 + 4y 3 w 3yw = 0 y 4 w + 3y 2 w 2 w = 0 James Parson (Hood College) Ye Olde FTA April 14, / 12
23 Are we better off? Some algebra shows that shows that y 5 + 4y 3 w 3yw = 0 y 4 w + 3y 2 w 2 w = 0 is equivalent to w 15 w 12 2w 10 2w 9 + w 7 + 2w 6 +w 5 + w 4 + w 3 1 = 0 y = w 6 + w + 1. James Parson (Hood College) Ye Olde FTA April 14, / 12
24 Are we better off? Some algebra shows that shows that is equivalent to y 5 + 4y 3 w 3yw = 0 y 4 w + 3y 2 w 2 w = 0 w 15 w 12 2w 10 2w 9 + w 7 + 2w 6 +w 5 + w 4 + w 3 1 = 0 y = w 6 + w + 1. Hudde s algebra was ad hoc, but these days one can do it by computing a Groebner basis in the lexicographic order y > w, which is something like the row reduction you would use to solve a system of linear equations. James Parson (Hood College) Ye Olde FTA April 14, / 12
25 Are we better off? Some algebra shows that shows that is equivalent to y 5 + 4y 3 w 3yw = 0 y 4 w + 3y 2 w 2 w = 0 w 15 w 12 2w 10 2w 9 + w 7 + 2w 6 +w 5 + w 4 + w 3 1 = 0 y = w 6 + w + 1. Hudde s algebra was ad hoc, but these days one can do it by computing a Groebner basis in the lexicographic order y > w, which is something like the row reduction you would use to solve a system of linear equations.(i used a computer!) James Parson (Hood College) Ye Olde FTA April 14, / 12
26 Solving Hudde s system James Parson (Hood College) Ye Olde FTA April 14, / 12
27 Solving Hudde s system 1 First we solve w 15 w 12 2w 10 2w 9 + w 7 + 2w 6 + w 5 + w 4 + w 3 1 = 0 for w. James Parson (Hood College) Ye Olde FTA April 14, / 12
28 Solving Hudde s system 1 First we solve w 15 w 12 2w 10 2w 9 + w 7 + 2w 6 + w 5 + w 4 + w 3 1 = 0 for w.odd degree, and so must have a root by geometry (or IVT). James Parson (Hood College) Ye Olde FTA April 14, / 12
29 Solving Hudde s system 1 First we solve w 15 w 12 2w 10 2w 9 + w 7 + 2w 6 + w 5 + w 4 + w 3 1 = 0 for w.odd degree, and so must have a root by geometry (or IVT). In fact, there is a root between 0 and 1. James Parson (Hood College) Ye Olde FTA April 14, / 12
30 Solving Hudde s system 1 First we solve w 15 w 12 2w 10 2w 9 + w 7 + 2w 6 + w 5 + w 4 + w 3 1 = 0 for w.odd degree, and so must have a root by geometry (or IVT). In fact, there is a root between 0 and 1. A bit of bisection shows that w = is very nearly a solution. James Parson (Hood College) Ye Olde FTA April 14, / 12
31 Solving Hudde s system 1 First we solve w 15 w 12 2w 10 2w 9 + w 7 + 2w 6 + w 5 + w 4 + w 3 1 = 0 for w.odd degree, and so must have a root by geometry (or IVT). In fact, there is a root between 0 and 1. A bit of bisection shows that w = is very nearly a solution. 2 Next we take y = w 6 + w James Parson (Hood College) Ye Olde FTA April 14, / 12
32 Solving Hudde s system 1 First we solve w 15 w 12 2w 10 2w 9 + w 7 + 2w 6 + w 5 + w 4 + w 3 1 = 0 for w.odd degree, and so must have a root by geometry (or IVT). In fact, there is a root between 0 and 1. A bit of bisection shows that w = is very nearly a solution. 2 Next we take y = w 6 + w Thus x 2 + yx + w x x is nearly a divisor. James Parson (Hood College) Ye Olde FTA April 14, / 12
33 Solving Hudde s system 1 First we solve w 15 w 12 2w 10 2w 9 + w 7 + 2w 6 + w 5 + w 4 + w 3 1 = 0 for w.odd degree, and so must have a root by geometry (or IVT). In fact, there is a root between 0 and 1. A bit of bisection shows that w = is very nearly a solution. 2 Next we take y = w 6 + w Thus x 2 + yx + w x x is nearly a divisor. The remainder is x James Parson (Hood College) Ye Olde FTA April 14, / 12
34 Why do we end up with a degree-15 equation? James Parson (Hood College) Ye Olde FTA April 14, / 12
35 Why do we end up with a degree-15 equation? Descartes wrote (roughly) that an equation of degree n has n roots, some real and some imaginary. James Parson (Hood College) Ye Olde FTA April 14, / 12
36 Why do we end up with a degree-15 equation? Descartes wrote (roughly) that an equation of degree n has n roots, some real and some imaginary. (But imaginary does not mean complex until you prove the FTA.) James Parson (Hood College) Ye Olde FTA April 14, / 12
37 Why do we end up with a degree-15 equation? Descartes wrote (roughly) that an equation of degree n has n roots, some real and some imaginary. (But imaginary does not mean complex until you prove the FTA.) To get a quadratic factor (not necessarily with real coefficients), we choose two roots a, b: the product (x a)(x b) is a factor. James Parson (Hood College) Ye Olde FTA April 14, / 12
38 Why do we end up with a degree-15 equation? Descartes wrote (roughly) that an equation of degree n has n roots, some real and some imaginary. (But imaginary does not mean complex until you prove the FTA.) To get a quadratic factor (not necessarily with real coefficients), we choose two roots a, b: the product (x a)(x b) is a factor. In particular for x 6 + x + 1, there should be ( 6 2) = 15 possible quadratic factors, which provide 15 solutions ( real and imaginary ) to Hudde s system for y and w. James Parson (Hood College) Ye Olde FTA April 14, / 12
39 Why do we end up with a degree-15 equation? Descartes wrote (roughly) that an equation of degree n has n roots, some real and some imaginary. (But imaginary does not mean complex until you prove the FTA.) To get a quadratic factor (not necessarily with real coefficients), we choose two roots a, b: the product (x a)(x b) is a factor. In particular for x 6 + x + 1, there should be ( 6 2) = 15 possible quadratic factors, which provide 15 solutions ( real and imaginary ) to Hudde s system for y and w. These 15 solutions correspond to the 15 real and imaginary roots of our degree-15 equation for w. James Parson (Hood College) Ye Olde FTA April 14, / 12
40 Why do we end up with a degree-15 equation? Descartes wrote (roughly) that an equation of degree n has n roots, some real and some imaginary. (But imaginary does not mean complex until you prove the FTA.) To get a quadratic factor (not necessarily with real coefficients), we choose two roots a, b: the product (x a)(x b) is a factor. In particular for x 6 + x + 1, there should be ( 6 2) = 15 possible quadratic factors, which provide 15 solutions ( real and imaginary ) to Hudde s system for y and w. These 15 solutions correspond to the 15 real and imaginary roots of our degree-15 equation for w. Making these notions completely clear had to wait for Lagrange and Gauss. James Parson (Hood College) Ye Olde FTA April 14, / 12
41 Can we turn Hudde s method into a proof of FTA? James Parson (Hood College) Ye Olde FTA April 14, / 12
42 Can we turn Hudde s method into a proof of FTA? Oddly even case Try degree 10: James Parson (Hood College) Ye Olde FTA April 14, / 12
43 Can we turn Hudde s method into a proof of FTA? Oddly even case Try degree 10: the quadratic factor system should reduce to solving a single equation of degree ( 10 2 ) = 45, James Parson (Hood College) Ye Olde FTA April 14, / 12
44 Can we turn Hudde s method into a proof of FTA? Oddly even case Try degree 10: the quadratic factor system should reduce to solving a single equation of degree ( 10 2 ) = 45, which is odd. James Parson (Hood College) Ye Olde FTA April 14, / 12
45 Can we turn Hudde s method into a proof of FTA? Oddly even case Try degree 10: the quadratic factor system should reduce to solving a single equation of degree ( ) 10 2 = 45, which is odd. Odd-degree equations always have real solutions, and so we can construct a real quadratic factor of any degree-10 polynomial. James Parson (Hood College) Ye Olde FTA April 14, / 12
46 Can we turn Hudde s method into a proof of FTA? Oddly even case Try degree 10: the quadratic factor system should reduce to solving a single equation of degree ( ) 10 2 = 45, which is odd. Odd-degree equations always have real solutions, and so we can construct a real quadratic factor of any degree-10 polynomial. The argument works for any oddly even (2 mod 4) degree. James Parson (Hood College) Ye Olde FTA April 14, / 12
47 Can we turn Hudde s method into a proof of FTA? Oddly even case Try degree 10: the quadratic factor system should reduce to solving a single equation of degree ( ) 10 2 = 45, which is odd. Odd-degree equations always have real solutions, and so we can construct a real quadratic factor of any degree-10 polynomial. The argument works for any oddly even (2 mod 4) degree. Other cases Try degree 12: James Parson (Hood College) Ye Olde FTA April 14, / 12
48 Can we turn Hudde s method into a proof of FTA? Oddly even case Try degree 10: the quadratic factor system should reduce to solving a single equation of degree ( ) 10 2 = 45, which is odd. Odd-degree equations always have real solutions, and so we can construct a real quadratic factor of any degree-10 polynomial. The argument works for any oddly even (2 mod 4) degree. Other cases Try degree 12: the quadratic factor system should reduce to solving a single equation of degree ( 12 2 ) = 66, James Parson (Hood College) Ye Olde FTA April 14, / 12
49 Can we turn Hudde s method into a proof of FTA? Oddly even case Try degree 10: the quadratic factor system should reduce to solving a single equation of degree ( ) 10 2 = 45, which is odd. Odd-degree equations always have real solutions, and so we can construct a real quadratic factor of any degree-10 polynomial. The argument works for any oddly even (2 mod 4) degree. Other cases Try degree 12: the quadratic factor system should reduce to solving a single equation of degree ( 12 2 ) = 66,which is not odd, James Parson (Hood College) Ye Olde FTA April 14, / 12
50 Can we turn Hudde s method into a proof of FTA? Oddly even case Try degree 10: the quadratic factor system should reduce to solving a single equation of degree ( ) 10 2 = 45, which is odd. Odd-degree equations always have real solutions, and so we can construct a real quadratic factor of any degree-10 polynomial. The argument works for any oddly even (2 mod 4) degree. Other cases Try degree 12: the quadratic factor system should reduce to solving a single equation of degree ( ) 12 2 = 66,which is not odd,but 66 is oddly even, which case we resolved! James Parson (Hood College) Ye Olde FTA April 14, / 12
51 Can we turn Hudde s method into a proof of FTA? Oddly even case Try degree 10: the quadratic factor system should reduce to solving a single equation of degree ( ) 10 2 = 45, which is odd. Odd-degree equations always have real solutions, and so we can construct a real quadratic factor of any degree-10 polynomial. The argument works for any oddly even (2 mod 4) degree. Other cases Try degree 12: the quadratic factor system should reduce to solving a single equation of degree ( ) 12 2 = 66,which is not odd,but 66 is oddly even, which case we resolved! This step is the start of an induction argument due to Foncenex, a student of Lagrange (1759). James Parson (Hood College) Ye Olde FTA April 14, / 12
52 Subsequent history James Parson (Hood College) Ye Olde FTA April 14, / 12
53 Subsequent history After Foncenex, there were further clarifications to the argument in the 18th century see, especially, Note IX of Lagrange s Traité de la résolution des équations numérique de tous les degrés (1808), which seems to be where Galois learned the theory of equations. James Parson (Hood College) Ye Olde FTA April 14, / 12
54 Subsequent history After Foncenex, there were further clarifications to the argument in the 18th century see, especially, Note IX of Lagrange s Traité de la résolution des équations numérique de tous les degrés (1808), which seems to be where Galois learned the theory of equations. Gauss used the basic outline of Foncenex in his second proof of the FTA (1815). James Parson (Hood College) Ye Olde FTA April 14, / 12
55 Subsequent history After Foncenex, there were further clarifications to the argument in the 18th century see, especially, Note IX of Lagrange s Traité de la résolution des équations numérique de tous les degrés (1808), which seems to be where Galois learned the theory of equations. Gauss used the basic outline of Foncenex in his second proof of the FTA (1815). A version of this argument (employing a tricky of Laplace) is in van der Waerden s Moderne Algebra and in Dummit and Foote s Abstract Algebra. James Parson (Hood College) Ye Olde FTA April 14, / 12
56 Subsequent history After Foncenex, there were further clarifications to the argument in the 18th century see, especially, Note IX of Lagrange s Traité de la résolution des équations numérique de tous les degrés (1808), which seems to be where Galois learned the theory of equations. Gauss used the basic outline of Foncenex in his second proof of the FTA (1815). A version of this argument (employing a tricky of Laplace) is in van der Waerden s Moderne Algebra and in Dummit and Foote s Abstract Algebra. Emil Artin also provided a well-known variant using Sylow s theorem instead of Foncenex s induction. This variant is popular in graduate-level algebra textbooks. James Parson (Hood College) Ye Olde FTA April 14, / 12
57 Thanks! Thanks for coming! James Parson (Hood College) Ye Olde FTA April 14, / 12
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