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NIELS ABEL

NIELS ABEL. Francesca Davide MAT 604. Short Biography of Neil Abel’s Life. Born in Norway on August 5 th , 1802. Died from Tuberculosis on April 6 th , 1829 at the age of 26.

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NIELS ABEL

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  1. NIELS ABEL Francesca Davide MAT 604

  2. Short Biography of Neil Abel’s Life • Born in Norway on August 5th, 1802. Died from Tuberculosis on April 6th, 1829 at the age of 26. • Attended the Cathedral School, a prestigious college preparatory school in Norway at the age of 13. Here is where he worked with a professor privately on advanced mathematics. • Attended Royal Frederick University at the age of 19. • Already known as the most knowledgeable mathematician in Norway.

  3. Famous Mathematical Contributions • At the age of 16, proved the Binomial Theorem is valid for all numbers, not just rational numbers. • First person to formulate and solve an integral equation • Innovator in the field of elliptic functions (complex functions) • Discovered Abelian Functions • PROVED THAT IS IMPOSSIBLE TO SOLVE EQUATIONS ALGEBRICALLY WITH DEGREE 5.

  4. Definitions To Know • Solvable: A group that can be constructed from abelian groups using extensions. • Galois Group: Let L be an extension field of K. G is a group of transformations of L called Galois Group of L/K. • Automorphism: An isomorphism of a system of objects onto itself.

  5. ABEL-RUFFINI PROOF Goal: To show that there exists a quintic who Galois group is the entirety of S5. Claim: If p is quintic, and G=Gal(p) then G<S5 • S5is not solvable **This is important because if we show there exists a Galois group whose entirety is S5, then we are showing that the Galois group is NOT SOLVEABLE!** This implies that it satisfies the biconditional: If Gal(p) is not solvable  p is not solvable.

  6. What does this mean? We need to find a polynomial with degree 5 who Galois group is isomorphic to all of S5. Choose: (t5-6t+3) Claim that this polynomial has a Galois group that is isomorphic to S5. We know this polynomial is… -Irreducible (Eisenstein’s Criterion, p=3) - Separable: No 2 roots are equal - 2 real critical points that are both nondegenerate

  7. Let’s assume the roots will look like this: (t-α1) (t-α2) (t-α3) (t-α4) (t-α5) •  (t- α1)(t- α2)(t- α3)(t2-At+B) **Which creates an irreducible polynomial with degree 2 such that α4 and α5 are complex conjugates of each other** •  It is an automorphism of the roots over the Rationals. • Galois Group has order 2 • G has a cycle of 2

  8. Now according to Lagrange’s Theorem, G has a cycle of 5. This implies that our Galois Group has a cycle of 2, a cycle of 5 and is a transitive subgroup of S5.

  9. …Knowing that we can transpose any 2 adjacent symbols in G. • We can get any 2,3 4 or 5 cycle transpositions as a collection of 2 cycle combinations • WE JUST SHOWED THAT EVERY SINGLE ELEMENT OF S5 IS IN G. THUS THE GALOIS GROUP IS S5!!! • SINCE G=S5, G IS NOT SOLVABLE! • P IS NOT SOLVEABLE

  10. THUS QUINTIC EQUATIONS AND ANY OTHER EQUATIONS WITH DEGREE GREATER THAN 4 ARE INSOLVABLE OVER THE RATIONALS.

  11. Sources • O’Connor, J. June 1998. NielsHenrik Abel. Retrieved from: http://www-history.mcs.st-and.ac.uk/Biographies/Abel.html • Sorum, Erik. NielsHenrik Abel. Retrieved from: http://www.math.wichita.edu/history/men/abel.html • Wikipedia. July 2014. Niels Abel. Retrieved from: http://en.wikipedia.org/wiki/Niels_Henrik_Abel • Salmone, Matthew. May 2014. An Insolvable Quintic. Retrieved from: http://www.youtube.com/results?search_query=proof+that +quintic+equation+is+not+solvable

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