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From the article: "Cauchy and Lamé hung their proofs on the tacit assumption that complex numbers, like real numbers, can be factored into a unique set of prime

by Chinjut 11y ago

From the article: "Cauchy and Lamé hung their proofs on the tacit assumption that complex numbers, like real numbers, can be factored into a unique set of primes. The real number 6, for example, always equals 2 x 3. Aside from reordering the factors (3 x 2), no other product will work. But to Cauchy and Lamé’s embarrassment, their German contemporary Ernst Kummer showed that certain complex numbers can be split into prime factors in more than one way." Well, there's also 6 = 1.5 x 4, etc. It's just in the context of the whole numbers (i.e., integers) that 6 has specifically the prime factorization 2 x 3. And depending on what your notion of a whole complex number is, they have unique prime factorization as well. For example, the Gaussian integers (complex numbers of the form a + b * i, where a and b are ordinary integers) have unique prime factorization. The Gaussian integers are the n = 4 case of "Consider the complex numbers generated by adding, subtracting, and multiplying copies of a primitive n-th root of unity"; Lamé's mistake, pointed out and built upon by Kummer, was in assuming that the same property of unique factorization into irreducible elements would continue to hold even if n here was replaced by any prime (this assumption fails for the first time at n = 23).

jackmaney 11y ago
Yes, the article did a vast oversimplification there. In particular, Cauchy, Lamé, and Kummer were working with rings of algebraic integers[1]. One simple example is the ring denoted Z[sqrt(-5)], which is the set of all complex numbers whose real part is an integer and whose imaginary part is an integer multiple of sqrt(5), ie Z[sqrt(-5)] = { a + b * sqrt(-5) | a,b are integers } This ring has a lot of structure in common with the integers, but it does not have unique factorization. In particular: 6 = 2 * 3 = (1 + sqrt(-5)) * (1 - sqrt(-5)) and each of 2, 3, 1 + sqrt(-5) and 1 - sqrt(-5) are irreducible in Z[sqrt(-5)] (ie they don't have any other factors other than 1 or -1). This is in opposition to what happens in the integers, namely unique factorization of any integer (other than 0, 1, or -1) into primes. The failed proof mentioned above hinged on the following assumption: if p is a prime and r_p = cos( 2 * pi / p) + i * sin( 2 * pi / p) is a primitive p^th root of unity, then the ring of algebraic integers Z[r_p] = {a_0 + a_1 * r_p + a_2 * (r_p)^2 + ... + a_(p-1) * (r_p)^(p-1) | a_0, a_1, ... , a_(p-1) are integers} has unique factorization[2]. It turns out that Z[r_p] has unique factorization if p is a regular prime[3]. However, this is not the case for non-regular primes. Thus, Fermat's Last Theorem wasn't settled in the 19th century. On the bright side, however, these failed attempts contributed to renewed interest in algebraic number theory. [1]: http://en.wikipedia.org/wiki/Algebraic_integer http://en.wikipedia.org/wiki/Algebraic_integer [2]: http://people.math.gatech.edu/~jrabinoff6/mathcamp/lectures.pdf http://people.math.gatech.edu/~jrabinoff6/mathcamp/lectures.... [3]: http://en.wikipedia.org/wiki/Regular_prime http://en.wikipedia.org/wiki/Regular_prime
thaumasiotes 11y ago
> Well, there's also 6 = 1.5 x 4, etc. It's just in the context of the whole numbers (i.e., integers) that 6 has specifically the prime factorization 2 x 3. Not quite. You can easily extend prime factorization to rational numbers by allowing negative as well as nonnegative exponents for primes. The prime factorization for 6 (and any other rational) will still be unique. If you instead decided to represent the factorization of a rational as a list of factors for the numerator and another list for the denominator, your factorizations would stop being unique... but the first way works just as well.
- Chinjut 11y ago
  What you say is correct, interpreting "primes" as meaning "the values 2, 3, 5, 7, 11, etc.". Every positive rational is essentially uniquely a product of finitely many of these values raised to integer powers. That having been said, what is it that makes these particular values "prime"? It's that they cannot be (nontrivially) decomposed into further factors. But in the rationals, they can be! 2 = 1.6 * 1.25, 3 = 1.6 * 1.875, etc. The inability to further factor 2, 3, 5, 7, 11, etc., which causes us to single them out as "prime" in the first place, is only on an account of factorization into whole numbers excluding such rational factorizations. So it's in that context that my comment was to be understood: the fact that 6 specifically has the irreducible factorization 2 x 3 is not universally true, but only true in particular contexts, of which the most common would be the integers. In the mentioned context of the reals, or even just the rationals, we could instead note that 6 (or anything, really) could continue being divided indefinitely, thus having no particular "prime factorization" in the sense of a factorization which cannot be further refined.

by Chinjut 11y ago

jackmaney 11y ago

Yes, the article did a vast oversimplification there. In particular, Cauchy, Lamé, and Kummer were working with rings of algebraic integers[1]. One simple example is the ring denoted Z[sqrt(-5)], which is the set of all complex numbers whose real part is an integer and whose imaginary part is an integer multiple of sqrt(5), ie Z[sqrt(-5)] = { a + b * sqrt(-5) | a,b are integers } This ring has a lot of structure in common with the integers, but it does not have unique factorization. In particular: 6 = 2 * 3 = (1 + sqrt(-5)) * (1 - sqrt(-5)) and each of 2, 3, 1 + sqrt(-5) and 1 - sqrt(-5) are irreducible in Z[sqrt(-5)] (ie they don't have any other factors other than 1 or -1). This is in opposition to what happens in the integers, namely unique factorization of any integer (other than 0, 1, or -1) into primes. The failed proof mentioned above hinged on the following assumption: if p is a prime and r_p = cos( 2 * pi / p) + i * sin( 2 * pi / p) is a primitive p^th root of unity, then the ring of algebraic integers Z[r_p] = {a_0 + a_1 * r_p + a_2 * (r_p)^2 + ... + a_(p-1) * (r_p)^(p-1) | a_0, a_1, ... , a_(p-1) are integers} has unique factorization[2]. It turns out that Z[r_p] has unique factorization if p is a regular prime[3]. However, this is not the case for non-regular primes. Thus, Fermat's Last Theorem wasn't settled in the 19th century. On the bright side, however, these failed attempts contributed to renewed interest in algebraic number theory. [1]: http://en.wikipedia.org/wiki/Algebraic_integer http://en.wikipedia.org/wiki/Algebraic_integer [2]: http://people.math.gatech.edu/~jrabinoff6/mathcamp/lectures.pdf http://people.math.gatech.edu/~jrabinoff6/mathcamp/lectures.... [3]: http://en.wikipedia.org/wiki/Regular_prime http://en.wikipedia.org/wiki/Regular_prime

thaumasiotes 11y ago

> Well, there's also 6 = 1.5 x 4, etc. It's just in the context of the whole numbers (i.e., integers) that 6 has specifically the prime factorization 2 x 3. Not quite. You can easily extend prime factorization to rational numbers by allowing negative as well as nonnegative exponents for primes. The prime factorization for 6 (and any other rational) will still be unique. If you instead decided to represent the factorization of a rational as a list of factors for the numerator and another list for the denominator, your factorizations would stop being unique... but the first way works just as well.

Chinjut 11y ago
What you say is correct, interpreting "primes" as meaning "the values 2, 3, 5, 7, 11, etc.". Every positive rational is essentially uniquely a product of finitely many of these values raised to integer powers. That having been said, what is it that makes these particular values "prime"? It's that they cannot be (nontrivially) decomposed into further factors. But in the rationals, they can be! 2 = 1.6 * 1.25, 3 = 1.6 * 1.875, etc. The inability to further factor 2, 3, 5, 7, 11, etc., which causes us to single them out as "prime" in the first place, is only on an account of factorization into whole numbers excluding such rational factorizations. So it's in that context that my comment was to be understood: the fact that 6 specifically has the irreducible factorization 2 x 3 is not universally true, but only true in particular contexts, of which the most common would be the integers. In the mentioned context of the reals, or even just the rationals, we could instead note that 6 (or anything, really) could continue being divided indefinitely, thus having no particular "prime factorization" in the sense of a factorization which cannot be further refined.