Finite Field Arithmetic and Reed-Solomon Coding(research.swtch.com) |
Finite Field Arithmetic and Reed-Solomon Coding(research.swtch.com) |
http://web.eecs.utk.edu/~plank/plank/papers/CS-07-593/
And if you want an excellent description of how Reed-Solomon coding is efficiently implemented in Linux, try this:
http://en.wikipedia.org/wiki/Shamirs_Secret_Sharing
It lets you divide a message into (say) three parts, where any two parts are necessary and sufficient to reconstruct the message. Or into seven parts, where you need all the parts to reconstruct the message. Or whatever. The wikipedia page gives a very readable basic introduction to how it works.
Edit: it seems I conflated two different constructions of a field of order 256. Apologies. It's been a few semesters since abstract algebra >.<
Do you mean the numbers mod 256 that are co-prime with 256 form a field? Even that doesn't work because 1+3=4, so it's not closed under addition.
So I'm not sure what you mean.
Added in edit ...
Given any positive number n, the numbers from 1 to n-1 that are co-prime to n form a group with multiplication as the group operation, but that's still not a field, so I'm still confused as to what you might mean.
From wikipedia: "The finite fields are classified by size; there is exactly one finite field up to isomorphism of size p^k for each prime p and positive integer k." -http://en.wikipedia.org/wiki/Finite_field
As has been pointed out, Z/256Z has zero-divisors, so it's not a field.
"Powers of primes are also fields" doesn't mean anything; integers are integers, not fields. "Powers of primes are orders of finite fields" is correct; so there is a field of order 256, but it's not Z/256Z. (Rather, you get it by finding a degree-8 irreducible polynomial over Z/2Z, and adjoining a root of it. In fact, the article gives the specific polynomial x^8 + x^4 + x^3 + x + 1.)
1. Z/pZ is a field if and only if p is prime.
2. for every p prime and n>=1 there exists a unique (up to isomorphism) field, called Galois field (see any book of algebra for the proof).
3. you can build a field of p^n elements for every p and n>1, using polynomials over Z/pZ mod an irreducible polynomial of degree n, e.g. (see link) you can build F_{2^8} as polynomials with coefficients in Z_2 (i.e. bits) mod x^8 + x^4 + x^3 + x + 1. If you chose another irreducible polynomial, e.g. x^8 + x^4 + x^3 + x^2 + 1, then you get another representation of a field of 256 elements, but "structurally" they are the same (this should "explain" the expression "up to isomorphism")
If you read the post you don't need the summary because you just read the post, which btw, ends with a summary.
If you didn't read the post yet and know nothing about the field, your summary won't teach them anything, and more importantly it won't tell them anything about the post. It won't tell them it's a great, catchy, read, it might only scare them away with hard, math stuff.
If you didn't read the post yet and know a lot about the field, your summary won't teach them anything because there's nothing to teach. All it can do is create a false impression about what the articles is about, but looking at your other posts in this thread I see this is your intention anyway.
OpenSSL is useless for learning the basis of crypto (and the post is not about crypto, but whatever) because you need to understand the algorithms and the math aspect of the problem before you can follow and understand any terse and abstract implementation, and OpenSSL is also bad because it sacrifices lisibility for performance.
The integers mod 256 are closed under addition and multiplication. Distributivity obviously holds, and there is obviously an additive inverse. My question is about the multiplicative inverse.
What's the multiplicative inverse of 2?
To say that there exists a field of size 256 is a different matter, and I'd be interested in learning more.
Added in edit:
Quoting from the same wikipedia article:
Even though all fields of size p are isomorphic to Z/(pZ), for n ≥ 2 the ring Z/((p^n)Z) (the ring of integers modulo p^n) is not a field.
(Parentheses added for reduction of ambiguity)
Further edit: Not sure why this got a downvote, but I don't much care. Maybe people didn't realize that I wrote this before all the other answers streamed in. Still, now I've learned what may have been intended, and a little more besides, so I'm content.
What do you mean by 2? :-)
The typical way to describe the finite field of order 256 is to look at a set of polynomials like this,
c_7 * x^7 + c_6 * x^6 + ... + c_1 * x + c_0
where c_i is an element of Z/2Z, so it's either 0 or 1. Count how many such polynomials of degree <= 7 there are - since each coefficient is either 0 or 1, there are 2^8 = 256.
Ok. Now we can define adding these polynomials in the usual term by term way.
For multiplication, if you tried to just straight up multiply two polynomials of order 7, you might get a polynomial of order 14. So, here's the trick: pick a polynomial of degree 8 that cannot be factored as a product of polynomials of lesser degree. This is the polynomial equivalent of a prime number (prime numbers cannot be factored further either, right?).
Just as we can look at Z/pZ, we can look at the set of polynomials over Z/2Z modulo g, where g is that irreducible poly of order 8. It should not be any more obvious to you that this works than that it is that Z/pZ is well defined, forms a field, etc. But you can verify it in pretty much the same ways as you'd prove it for Z/pZ.
Anyway. So what do you mean by '2'? In this representation, 1+1 is 0.
Sometimes by '2' people mean x. That's because there's a natural coding of these polynomials as bits in a binary word. Since each coefficient is 0 or 1, you can say the ith bit is the coefficient of x^i. With that encoding, 2 is 0b10, is 1 * x^1 + 0, is x. So, the inverse of x is going to be x^254 mod g (where g is that irreducible generating polynomial).
: the set of numbers mod 256 form a field.
And the labelling is opaque, because while it's obvious what polynomial an integer gets mapped to, the multiplication depends on the particular g.
"Even though all fields of size p are isomorphic to Z/pZ, for n ≥ 2 the ring Z/(p^n)Z (the ring of integers modulo p^n) is not a field. The element p (mod p^n) is nonzero and has no multiplicative inverse."