How Math's Most Famous Proof Nearly Broke
nautil.us
nautil.us
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).
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
[2]: http://people.math.gatech.edu/~jrabinoff6/mathcamp/lectures....
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.
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.
[1] http://www.amazon.com/Mathematics-Its-History-Undergraduate-...
I was actually studying Maths at Cambridge when Wiles gave his seminars. People were buzzing with excitement after the first one.
Found it in a NOVA transcript [1]:
"At the beginning of September, I was sitting here at this desk, when suddenly, totally unexpectedly, I had this incredible revelation. It was the most—the most important moment of my working life. Nothing I ever do again will. . . I'm sorry."
Because these famous problems were solved, there is now increased demand for Math PhDs with that specialty.
How do you know if the proof is valid, or if they're just scratching each others' backs and pretending they're all brilliant? You might say "Wait, a fraud that big couldn't happen." Just like one of the biggest investment funds couldn't be a Ponzi Scam? (Madoff)
I.e., the few Mathematicians who know enough to be able to check the full proof might be doing the equivalent of this guy:
http://nymag.com/scienceofus/2015/05/how-a-grad-student-unco...
No one in bleeding edge science can fully grasp bleeding edge work outside their specialty, but high profIle stuff like this gets a lot of attention.
But then this gets published and other people could point their mistakes. And then it's not only about what the authors got wrong, but also what the reviewers let slip.
And while the number of mathematicians that can understand the proof is low, it is > 1
(Professional mathematician here)
Such a conspiracy could never happen. Mathematicians are, in general, scrupulously honest to a fault. There are exceptions of course, but such a scam would need everybody's cooperation.
Moreover, if you have tenure then you would have no incentive to participate in such a fraud (even to the extent of keeping quiet about it). Once you get tenure you are basically working for pride and for the sheer joy of solving problems, and maintaining a big lie would do nothing for either.