# What's so special about characteristic 2?

I've often read about things which do not work in a field with a characteristic $2$, mainly things which have to do with factoring, or similar things. I'm not exactly sure why, but the only example of such a field I could think of is $\mathbb{Z}/2\mathbb{Z}$, which itself is an interesting field because it contains only the identity elements for the two groups, and naturally, it is a cyclic field. Do these properties lead to the fact that many things don't work if the charateristic is $2$

Any examples of things which break in such a field are also welcome.

• mathoverflow.net/questions/915/… – Qiaochu Yuan Dec 13 '15 at 8:59
• Two is the oddest prime. It's even. – Asaf Karagila Dec 13 '15 at 9:00
• One I've always found interesting is that $2$ is the only prime $p$ for which we do not have $1 > 1/(p-1)$, and so the $p$-adic exponential fails to converge on $p\mathbb{Z}_p$ exactly when $p=2$. This is why $(\mathbb{Z}/p^n\mathbb{Z})^\times$ is cyclic for $p$ odd, but not for $p=2$. But the $2$-adic exponential converges on $4\mathbb{Z}_2$, which we can use to show that $(\mathbb{Z}/2^n\mathbb{Z})^\times$ has an index $2$ cyclic subgroup. – Slade Dec 14 '15 at 2:52
• As a user who (to exaggerate mildly) makes a living working in characteristic two I want to point out that it is not all so gloomy. In characteristic two you never make sign errors! The bright side of the point raised in Alan's (+1) answer. – Jyrki Lahtonen Dec 14 '15 at 7:30
• @JyrkiLahtonen: On the other hand, sometimes signs are the only thing one cares about (several quantities with multiplicative relations between them, where it is only the signs of the quantities that are of interest), and after transforming multiplicative notation to additive, this leads to linear algebra over $\Bbb Z/2\Bbb Z$. – Marc van Leeuwen Dec 14 '15 at 8:48

## 5 Answers

Two is the smallest (and as people sometimes say: the oddest) of all primes.

Just to take a contrived example, let's say you want to show

If the sum of two squares equals the square of the sum then one of the two is zero.

Well, that's easy, you just transform $a^2+b^2=(a+b)^2$ to obtain $0=2ab$; and as a product is $0$ only if one of the factors is zero, you conclude that $a=0$ or $b=0$. Done? No! We forgot the third factor. We should have said: $a=0$ or $b=0$ or $2=0$. And the latter is exactly what happens in characteristic $2$, i.e., in characteristic $2$ our claim does not (necessarily) hold.

To put it differently: In the attempt to arrive at $ab=0$ we had to divide by $2$, and as always when dividing we must make sure that we do not accidentally divide by zero. It happens ever so often that you have to divide by something. If you need to divide by $a-b$, say, you can circumvent the problem by adding a condition to your claim ("... provided $a\ne b$"). But sometimes you need to divide by an explicit constant such as $2$ in our example. In that case the condition to be added to the claim must be that the characteristic of the field is not a divisor of that constant.

The fact that it is often only characteristic $2$ that needs to be mentioned as exception might be called a consequence of the law of small numbers: It happens much more often that a factor $2$ pops up naturally than a factor $97$, say. That's why characteristic $2$ so often and characteristic $3$ sometimes plays a special role.

• Thanks, this is a good response! As one of the answers in the question Qiaochu Yuan posted above noted, it can be said that 2 and 3 are almost "vacuously" prime because of how small they are. – Juan Sebastian Lozano Dec 13 '15 at 9:17
• However, you could say the same about any prime characteristic and the corresponding prime powers, so this is not specific to 2. Perhaps the reason this particular fact is special is that squares (i.e. quadratic forms, or almost equivalently, inner products/bilinear forms) are special in a way that other powers are not. – Ryan Reich Dec 13 '15 at 22:39
• @RyanReich Yes, my lousy example was not meant to be generic, just an example of what can go wrong because one is simply used to divide by two without thinking – Hagen von Eitzen Dec 14 '15 at 16:54
• @HagenvonEitzen Okay, and not to push the point too much, but the problem is that it was generic, in that it did not explain why dividing by 2, specifically, was special, aside from "law of small numbers", which doesn't have much to do with the rest of your answer. – Ryan Reich Dec 14 '15 at 20:18

In a field of characteristic two, everything is its own additive inverse. so $x=-x$ for every element, and in a lot of proofs, we show something is $0$ by showing it is equal to its own additive inverse. This obviously fails in fields of characteristic 2.

Any finite field of characteristic 2 turns out to be the unique finite field of order $2^n$ for some $n$. Infinite fields exist as well, the simplest would be the field of rational polynomials over $\mathbb Z / \mathbb 2Z$

• The fact that "we show something is zero by showing it is its own additive inverse" is something that is so routine that I take it for granted, but yes, it makes complete sense why that would make things hard in such a field. – Juan Sebastian Lozano Dec 13 '15 at 9:04

One of many examples where characteristic $2$ makes things very hard is the classification of simple Lie algebras over algebraically closed fields. For characteristic zero the proof works very nicely. For prime characteristic $p$ things get much harder; also the result is more complicated. Nevertheless it has been achieved for all $p>5$; and partly for $p=5$ and $p=3$. Only $p=2$ seems to be hopeless. The reasons have been discussed here, e.g., the Killing form does not help very much then; the trace of the identity matrix may vanish etc.

There's a relatively low-level reason, partly mentioned by Hagen von Eitzen, that is $2$ is the smallest prime of them all, and hence appears in more places than others, even in the most abstract theorems. This kind of small special cases turns up everywhere because constants are generally small.

For example, consider the quadratic equation $ax^2 + bx + c = 0$ for $a,b,c \in F$ for some field $F$. One special case is $a = 0$, which we can easily deal with. Then we proceed by the usual complete-the-square technique to get $(2ax+b)^2 = b^2-4ac$. Suppose we let $\sqrt{d}$ denote some arbitrary square root of $d$ in $F$, we then get $2ax = -b \pm \sqrt{b^2-4ac}$ as usual. Since $a \ne 0$, we can divide both sides by $a$, but we're stuck with the $2$ unless $char(F) \ne 2$. Now how did the $2$ appear? It was a completely natural outcome of the identity $(s+t)^2 = s^2 + 2st + t^2$ involved here, simply because $1+1 = 2$. It so happens that $2$ is a prime, and so the characteristic that we need to exclude is $2$ itself to ensure that $2$ has a multiplicative inverse.

Similarly, when solving the cubic equation in an arbitrary field, one finds exactly the same issue with $3$. If we did it by Lagrange resolvents after obtaining the depressed cubic, we would have three roots $a,b,c$ such that $0 = a+b+c$, and we would let $x = a + ζ b + ζ^2 c$ and $y = a + ζ^2 b + ζ c$ where $ζ$ is a primitive cube root of unity. It turns out that $x^3,y^3$ are both in $F(\sqrt{D})$ where $D$ is the discriminant of the cubic, and so we can express them in terms of $a,b,c$ alone, provided $2$ has an inverse ($2$ appears in the computation 'by accident'). Then we add all three equations and divide by $3$ to get $a$, which of course needs $3$ to be invertible. So at least for this method we need the field to have characteristic neither $2$ nor $3$.

Normally, in a field, each element with a square root (other than zero) has two of them: x2-a2 = (x+a)(x-a), so both a and -a are roots. So by the pigeonhole principle, in a finite field (of odd characterisitic) half the nonzero elements have two square roots, and the other half have none. But in characteristic 2, a = -a, so all the elements have exactly one square root.
If you know the bare basics of elliptic curve cryptography, you might think this would make ECC impossible in fields of characteristic 2, but that isn't the case.
Some polynomial factoring algorithms over finite fields work more simply when the characterisitic is odd, just because 2 is a divisor of the number of nonzero elements (which is the order of the cyclic multiplicative group of the field). For characteristic 2, with size 22k, 3 is a divisor, and those factoring algorithms work with little modification. But with size 22k+1, 22k+1-1 might even be prime, requiring more major modifications for efficient factoring.