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The History of the Formulation of Ideal Theory

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The History of the Formulation of Ideal Theory The History of the Formulation of Ideal Theory Reeve Garrett November 28, 2017 1 Using complex numbers to solve Diophantine equations From the time of Diophantus (3rd century AD) to the present, the topic of Diophantine equations (that is, polynomial equations in 2 or more variables in which only integer solutions are sought after and studied) has been considered enormously important to the progress of mathematics. In fact, in the year 1900, David Hilbert designated the construction of an algorithm to determine the existence of integer solutions to a general Diophantine equation as one of his \Millenium Problems"; in 1970, the combined work (spanning 21 years) of Martin Davis, Yuri Matiyasevich, Hilary Putnam and Julia Robinson showed that no such algorithm exists. One such equation that proved to be of interest to mathematicians for centuries was the \Bachet equa- tion": x2 +k = y3, named after the 17th century mathematician who studied it. The general solution (for all values of k) eluded mathematicians until 1968, when Alan Baker presented the framework for constructing a general solution. However, before this full solution, Euler made some headway with some specific examples in the 18th century, specifically by the utilization of complex numbers. Example 1.1 Consider the equation x2 + 2 = y3. (5; 3) and (−5; 3) are easy to find solutions, but it's 2 not obviousp whetherp or not there are others or whatp they might be. Euler realized by factoring x + 2 as (x + 2i)(x − 2i) and then using the facts that Z[ 2i] is a UFD andp the factorsp given are relatively prime that the solutions above are the only solutions,p namelyp because (x + 2i)(x − 2i) being a cube forces each of these factors to be a cube (i.e. (x + 2i) = (a + b 2i)3 for some integers a and b), from which we deduce our solutions to x2 + 2 = y3 by equating coefficients. Moreover, we see from the resulting system of equations that (5; 3) and (−5; 3) are the ONLY solutions. With this insight, Euler then realized that the proof for Fermat's Last Theorem would necessarily require similar methods. For example, as he realized in the year 1753 with x3 + y3 = z3, by considering the equation in Z[ζ3] (where ζ3 is a primitive cube root of 1), which is a UFD, one obtains a contradiction by assuming the solvability of the equation. Indeed, if there were a triple (x; y; z) satisfying this equation (z > 0), then 3 3 3 2 3 x + y = z would force (x + y)(x + ζ3y)(x + ζ3 y) = z , meaning each of the 3 factors on the left hand side 2 would have to be cubes of numbers that look like a + bζ3 + cζ3 . Then, using these, we could find a triple (a; b; c) (where 0 < c < z) solving the equation. Iterating, we get an infinite decreasing sequence of positive integers z > c > ··· , which is absurd. Fermat gave a valid proof that his last theorem holds for n = 4. It is also apparent that if n = pm where p is an odd prime, then xn + yn = zn would hold if and only if (xm)p + (ym)p = (zm)p. Therefore, it would suffice to prove Fermat's Last Theorem for odd primes p. Therefore, we restrict to this setting, if p p p 2 p−1 p we write x + y = z as (x + y)(x + yζp)(x + yζp ) ··· (x + yζp ) = z , assuming Z[ζp] is a UFD, we'd get a contradiction in the same was as p = 3, thus proving Fermat's Last Theorem. This is the purported proof that Lam´epresented in 1847 which was disproved by Ernst Kummer, who showed that Z[ζ23] is NOT a UFD. In fact, Z[ζp] is NEVER a UFD if p ≥ 23. 2 3 22 Recall that the norm of an element f(ζ23) 2 Z[ζ23] is the product N(f(ζ23)) = f(ζ23)f(ζ23)f(ζ23) ··· f(ζ23 ), which is an ordinary integer. With the help of Jacobi in this work, he found that the norm of any element 2 2 f(ζ23) 2 Z[ζ23] is an integer of the form (x + 23y )=4, and this means that no such f(ζ23) can have norm 47 because 4·47 = 188 can't be written in the form x2 +23y2 where x and y are integers. This led him to realize 21 21 that in Z[ζ23], since N(1 − ζ23 + ζ23 ) = 47 · 139, 1 − ζ23 + ζ23 is irreducible (if you could factor it, something 1 would have norm 47) and not prime (it divides 47 · 139 but can't divide 47 or 39 because it doesn't divide 22 139 21 their norms, 47 and 47 ). Thus, we've written 47 · 39 = N(1 − ζ23 + ζ23 ) as a product of 22 irreducible 23−n −n 10 13 8 15 7 16 2 elements. Also, notice that since ζ23 = ζ23 , we have that N(ζ23 + ζ23 + ζ23 + ζ23 + ζ23 + ζ23 ) = 47 10 13 8 15 4 19 2 is a product of 11 irreducible elements, each squared, and N(ζ23 + ζ23 + ζ23 + ζ23 + ζ23 + ζ23 = 139 is a product of 11 irreducible elements, each squared. Therefore, by taking half the factors from each, we can write ±47 · 139 as a product of 22 irreducible factors in a DIFFERENT way than we did before. At the time, Kummer had been studying higher reciprocity laws and was seeing the connection between these and the study of cyclotomic field extensions. One might ask what reciprocity laws have to do with cyclotomic extensions. Before we give a sense of Kummer's work on that question, it's important to con- sider the context: namely the findings and methods of Gauss when working with quadratic forms. In his revolutionary number theory text Disquisitiones Arithmeticae (published in 1798, when he was only 21 years old; the book was revolutionary in its rigor and in how it set the foundations for number theory to come), Gauss proved the law of quadratic reciprocity and various results about binary quadratic forms (that is, polynomials of the form f(x; y) = ax2 + bxy + cy2 2 Z[x; y]) by introducing the Gaussian integers Z[i], and he prophetically proclaimed that ordinary arithmetic with natural numbers cannot be used to establish a general theory for higher reciprocity laws and that \such a theory demands that the domain of higher arithmetic be endlessly enlarged." Now, let's return to Kummer. Recall that the Legendre symbol for two distinct odd primes p and q p is denoted by q and is 1 if p is a square modulo q and −1 if not. The law of quadratic reciprocity states that 8 q p < if p ≡ 1 or q ≡ 1 mod 4 = p q q :− p otherwise or equivalently p q = (−1)(p−1)(q−1)=4: q p ∼ p Let q ≡ 1 mod 4. In general, Gal( (ζq)= ) = q−1 and (ζq) contains ( q). Have σp denote the p Q Q Z Q Q σp( q) p q p p p−1 (p−1)=2 Frobenius map a 7! a . q = ( q) ≡ q mod p, so this is 1 if and only if p = 1 by Euler's a (p−1)=2 criterion, which states p ≡ a mod p for any prime p and any natural number a coprime to p. p The case q ≡ 3 mod 4 is solved similarly by considering Q( −q) instead, which gets us (−1)(p−1)=2 instead. With ideas like this in mind, and with the formulation of his \ideal numbers," Kummer was able to extend this kind of reasoning to higher reciprocity laws (i.e. determining when p ≡ an mod q for some integer a). 2 Kummer's Ideal Numbers in the Cyclotomic Integers Case 21 Let's return to the case of Z[ζ23] and f(ζ23) = 1−ζ23 +ζ23 . What I didn't mention is that we CAN'T find any prime factors for 47 within Z[ζ23]! Indeed, if some g(ζ23) were a prime factor of 47, its norm would have to 22 divide N(47) = 47 AND g(ζ23) would have to divide 47·139 = N(f(ζ23)), which would mean g(ζ23) divides one of the 22 factors of N(f(ζ23)), say h(ζ23) since it's prime, which would then mean N(g(ζ23))jN(h(ζ23)), forcing N(g(ζ23)) = 47; which is impossible! Kummer then thought: what if we introduced \ideal prime numbers" outside the given number system Z[ζ23] that could result in unique factorization into products of primes? To see how this works, let's continue with this example. Ideally, since N(f) = 47 · 139, we would want to decompose f into a product of a prime with norm 47 and a prime with norm 139. In taking these norms, we would want no duplication and all factors in this norm to be prime, which would mean we'd decompose 47 as a product of 22 distinct ideal primes (our ideal prime and its \conjugates," one for each multiplicand in N(f)), only one of which also divides f. N(f(ζ23)) Let P be the hypothetical prime divisor that divides both 47 and f(ζ23).Then, let Ψf (ζ23) := , i.e. f(ζ23) the product of the other multiplicands of N(f) besides f itself. Necessarily, Ψf is not divisible by P , but it 2 is divisible by all other \ideal prime factors" of 47. This means that fΨf is divisible by 47 if and only if f is divisible by P .
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