## Fields of characteristic 2?

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Farabor
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### Fields of characteristic 2?

So, I'm a first semester grad student in mathematics, and in the course of my linear algebra class the professor messed up in the notes by not including the restriction that a bunch of the theorems for a vector space V over an arbitrary field F only held if the characteristic of the field is not 2. This has since been corrected, but it's raised the question in my mind....

Are there fields I'm already (or should be) aware of of characteristic 2? (Other than the trivial field of course). Is this an area of study? Are there interesting fields of char 2?

madaco
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### Re: Fields of characteristic 2?

The nimbers have characteristic 2.

If you store a nimber in binary, the sum is xor-ing the two, and multiplying is some complicated thing I don't understand.

I only know this because someone linked here linked me to them when I was looking for a field where addition was fast.
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Something Awesome
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### Re: Fields of characteristic 2?

Every finite field of characteristic 2 looks like a direct sum of copies of Z/2Z. The field of rational functions over Z/2Z turns out to be an infinite field of characteristic 2.

Usually, somewhere in the proof of a theorem, you'll get something like "2x = 0, so x = 0 (and then conclude something is linearly independent or maybe some map is injective)," which isn't valid in characteristic 2. Fields of positive characteristic can be interesting, but they're often very hard to deal with because we lack a lot of the basic tools.

jestingrabbit
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### Re: Fields of characteristic 2?

Farabor wrote:Are there fields I'm already (or should be) aware of of characteristic 2? (Other than the trivial field of course). Is this an area of study? Are there interesting fields of char 2?

The basic idea is that you start with the trivial field, F_2, and then you make the ring of polynomials with coefficients in F_2, F_2[x]. A ring has nice addition, and multiplication that is okay, but nothing is guaranteed to have a multiplicative inverse. The set of polynomials with coefficients from a field forms a ring.

And its a nice ring in a lot of ways. You can talk about p|q ie "p(x) dividing q(x)" iff p(x)*r(x) = q(x). And the GCD of polynomials is also a thing too, and you can get at it using Euclid's Algorithm.

https://en.wikipedia.org/wiki/Euclids_algorithm

Well, "So what?" you're probably asking. Well, if you have a polynomial, p(x), that has no factors in F_2[x], then F_2[x]/p = F_2[x] modulo p(x) is a ring (because its a factor of a ring) and you can use Bezout's identity to prove that if q(x) is in F_2[x]/p and q(x)!=0, then there are polynomials a(x) and b(x) st 1 = a(x)q(x) + b(x)p(x) ie a(x) is the multiplicative inverse of q(x).

Long story short: F_2[x]/p is a field if p has no factors.

An example: x^2 + x + 1 has no factors in F_2[x] (the only degree 2 polynomials with factors are x*x=x^2, x*(x+1) = x^2 + x and (x+1)*(x+1) = x^2 + 2x +1 = x^2 + 1). So, F_2[x]/x^2 + x + 1 is a field. What are its elements? They're of the form {a(x) + b(x)*(x^2 + x + 1)| b(x) in F_2[x]}. But we want nice representatives from these equivalence classes, like we do with Z/n*Z. Well, lets list elements until we get repeats. 0, 1, x, x+1, x^2... no wait, x^2 == x^2+ x^2 + x+ 1 = x+1. With that, and a little thought, you can see that the equivalence classes of {0, 1, x, x+1} is all the elements that we need. What about multiplicative inverses? Well, the first interesting case is x. So, lets do Euclid's algorithm with x^2+x+1 and x.

x^2+x+1 + x*(x+1) = 1.

Well, that was quick... x and x+1 are each other's multiplicative inverses. Normally when you write down the elements of this set you use alpha instead of x, and alpha is thought of as a root of x^2+x+1, like 2^(1/2) is a root of x^2 -2.

Another case that has some cryptographic applications atm is F_2[x]/(x^64+x^4+x^3+x+1). You can represent elements of this with 64 bit words in a computer.

Hope that helps.
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Farabor
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### Re: Fields of characteristic 2?

It does, quite a bit! Although I really need to bone down on my factor groups and ring ideals (We're doing factor groups now in Herstein's classic book, probably won't get to ring ideals until next semester at this pace)

Farabor
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### Re: Fields of characteristic 2?

So, I've progressed in my studies into the beginnings of multivariable analysis/linear algebra. I'm now curious as to how things like determinants act when the underlying field is of characteristic 2. Normally the determinant being 0 of a matrix if and only if the vectors are linearly independent comes from the alternating tensor property of the determinant (at least as far as I can understand)...but in fields of char 2, this is meaningless. So....can we have determinant 0 and yet have linear independence if the underlying field is one of these funky things?

(Oh, and on a side note, just recently saw in class the other fun results, where you can have an irreducible polynomial with multiple roots in the polynomial field over the rational functions mod 2. @whee!)

jestingrabbit
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### Re: Fields of characteristic 2?

If you look at section 3 here

http://www.math.pitt.edu/~annav/0290H/row_reduction.pdf

you can see a few claims. I invite you to prove them for a matrix over a finite field.

Put more directly, its my understanding that you can still use determinants just fine.
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Dark Avorian
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### Re: Fields of characteristic 2?

It requires a bit of checking, but for characteristic =/= 2, the alternating tensor property of the determinant is equivalent to the condition that if we have a duplicated entry we get 0. This condition is still meaningful in characteristic two, although obviously no longer equivalent. So we reformulate in terms of this and a lot still works.
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