12
$\begingroup$

I've recently been studying some elementary number theory, and I've frequently come across the fact that there are a fair number of results (the main one being the law of quadratic reciprocity) for which $2$ has to be treated as a special case, separately from the odd primes.

My question is why this is. I can see that there are reasons in each proof that the theorem doesn't hold for $2$ (a pretty common one is the fact that $2-1$ is not divisible by $2$), but I'm curious if there's an overarching theoretical reason. My best guess is that it might have to do with the fact that $\left ( \mathbb{Z}/2\mathbb{Z} \right )^\times$ is trivial, although that might just be a fancier way of saying that $2-1 = 1$.

I am, of course, open to the possibility that there are a variety of reasons that $2$ is an exceptional prime, but I'm curious what those are and if there's a best example, or a most common example.

$\endgroup$
6
  • 10
    $\begingroup$ 2 is the oddest prime of them all... :) $\endgroup$ – J. M. isn't a mathematician Sep 8 '11 at 19:09
  • 6
    $\begingroup$ I recommend checking out the Mathoverflow questions mathoverflow.net/questions/915/… and mathoverflow.net/questions/15141/why-is-2-so-odd. $\endgroup$ – Jonas Meyer Sep 8 '11 at 19:10
  • $\begingroup$ Did you mean to say $2-1$ divides $2$? For no $n$ (except $1$) does $n$ divide $n-1$ $\endgroup$ – Ross Millikan Sep 8 '11 at 19:11
  • 1
    $\begingroup$ Well, there's Optimus Prime. $\endgroup$ – Curiosity Sep 8 '11 at 19:22
  • 2
    $\begingroup$ I think there might be something to the idea that $2$ is an exceptional prime. But what I cannot understand is some people attempt to justify it by saying $2$ is the only even prime. Well, true; but isn't $17$ the only prime divisible by $17$? (I find Iasafro's answer more acceptable than this :-).) $\endgroup$ – Srivatsan Sep 8 '11 at 19:23
6
$\begingroup$

Many of the more meaningful exceptions seem to come about because $1\equiv -1$ modulo 2. So the reasoning "$x=-x$, therefore $x=0$" is not available in characteristic 2.

$\endgroup$
3
$\begingroup$

To pick up on your example, quadratic reciprocity has everything to do with roots of quadratic equations - and the presence of a 2 in the denominator of the quadratic formula $$ x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} $$ is what screws things up. Similarly, the correspondence between quadratic forms and bilinear forms involves dividing by 2, which screws things up at 2 in algebraic number theory, for example.

$\endgroup$
3
$\begingroup$

I will say that the reason why $2$ is special is that $2$ is the smallest prime in the natural numbers.

$\endgroup$
4
  • $\begingroup$ I downvoted this, because I found this a very strange reason before reading Buzzard's and Carnahan's answers on MO. Now that I want to take it back, the system is not allowing me to do so saying that my vote is locked in until the answer is edited. :( $\endgroup$ – Soarer Sep 8 '11 at 19:44
  • $\begingroup$ Don't worry, we all are humans ;) $\endgroup$ – Josué Tonelli-Cueto Sep 8 '11 at 19:52
  • $\begingroup$ @Soarer: I made a trivial edit to allow undownvoting. $\endgroup$ – Jonas Meyer Sep 8 '11 at 20:22
  • $\begingroup$ @Jonas: Done. $\hspace{1mm}$ $\endgroup$ – Soarer Sep 8 '11 at 20:59
1
$\begingroup$

I believe that another important source of exceptionality for the prime $2$ is that if $n$ is odd, then in fact $n^2\equiv1\bmod4$ while for every prime $p>2$ there are numbers $n$ such that $n^2\equiv1\bmod p$ but $n^2\not\equiv1\bmod p^2$ (i.e., there are no squares $\equiv3\bmod4$).

Whether this, philosophically, is a "manifestation of smallness" I'm not entirely sure.

$\endgroup$
1
$\begingroup$

At a lesser elementary level, when the characteristic of a field is $2$, and only in this case, the relation between the tensor product $V\otimes V$ and the symmetric product ${\rm Sym}_2(V)$ is less straightforward. This has important implications for the theory of quadratic forms (which is so basic in Number Theory).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.