Galois Fields (GF) are finite mathematical structures used in cryptography where binary values are represented as polynomials with coefficients of 0 or 1, enabling efficient cryptographic operations through addition (XOR), multiplication, and division; operations are constrained within a specific field size (e.g., GF(2^3) has 8 elements) using a primitive polynomial that cannot be factorized, allowing results to be reduced back into the finite field when polynomial degrees exceed the field's maximum degree.
Galois Field Arithmetic and Cryptography: An Introduction
Added:okay so let's look at galwa fields and understand how they are used in cryptography so what we basically have is that we might have a vector so let's say we've got four three four six as a vector it is possible to represent that Vector as a as a polynomial so we might Define this as three plus four X plus six x squared and it's these polynomials that we perform our operations and it allows us to simplify our cryptographic operations so we represent our binary values or we can represent our binary values as a polynomial so for example if we have say 0 1 2 and 3 then we have zero zero zero one zero zero zero one one zero and one one this has a degree of two and we would Define that as a gawa field 2 to the power of two for this and we always have zero and this we can represent as X and this we can this we can represent as one this we can represent this X and this will be X plus one if we have a gawa field of three then we could go all the way up from zeros or zeros to all ones this will be a gawa field of 2 to the power of three or or have a degree of three this will go from zero then one X right up to x squared plus X plus one okay so in this way we can represent our binary values in terms of a polynomial value and then we can operate on these in this case we either have a zero or a one for our represent for our coefficients of the polynomial powers so if we Define that we have value such as this there are values of of a is the a coefficients there whether it be a zero value or a one value represented by the bit position that we have so the operations that we perform are performed as a multiplication and also as foreign with a binary adder we will take zero zero zero one one zero one one and we get 0 1 1 0.
when we multiply then we have zero zero zero or one so this is like an um an ad and this is like an an and or a multiply operation okay so let's look at how we then operate on our polynomial values so let's take a simple one and we'll take x squared plus X plus one so in this case this will be the value of one one one and we're going to add it to X plus one which is the value of zero one one and binary so with the result will be x squared plus X plus one Plus X plus one so we only take into account our polynomial values together when we're doing the operation so this will be X plus X plus X plus one plus X plus one if we bring together the X values then we have this this will cancel and this will cancel as we see and we'll end up with x squared so The Gallows gawa field addition of these two values ends up being this value and this value is obviously one zero zero now let's do a multiplication foreign value and it's x squared plus X plus one multiplied by X plus one so now we get x cubed plus x squared plus x squared plus X plus X plus one and here we see these and these will cancel because of this operation here so the value becomes X cubed plus one so one thing that we can see here that will come back to is that we now have a power which is greater than the values that we started with so we'll see what we can do with this later but that's the operation that we have and so this will be one 0 0 1 so the calculation of 1 1 times 0 1 1 will equal this value in binary yeah let's look at division and the way that we do division is that we will take the inverse of a value and then multiply it so if we wanted to take say x squared plus X plus one divided by X plus one we find the inverse polynomial of this value and then we multiply it together once we have that then we can we can compute just as we as we did there okay so we could do the calculation but we can actually find out that the inverse of this is actually x cubed plus x squared plus X from here what we can do is that we can then multiply this and this to be able to get the result okay so I won't do the whole calculation here but we'll end up with this value here and apprecially it looks a bit strange because we actually end up with a higher power than we have here but I'll explain how we can reduce that power down later if we want to do this in a long-handed way that it's possible for us to do the division as we would do with long division but let's keep our value there just now for that the operation that we now use to be able to reduce these back into the finite Fields or into the constrained fields that we have so for example gallius gawa field off to the power of it gives us 256 different values and we can't have any more than that from zero to two five five so we need to constrain our outputs to bring them back within the powers of the polynomials that we actually have so for this what we have is a primitive polynomial and a preventive polynomial is a rather like remote P operation that we have within finite fields and that no value can divide or we can't factorize this this prime number here so that is possible to perform a mod operation and make sure that our calculations still work It's A Primitive polynomial or a primitive cannot be factorized into other polynomials and allows our calculations to work where we can divide by our primitive to be able to reduce back into our finite field okay so let's see we're operating on a gawa field of to the power of three or eight so a non-primitive that we can use is X cubed plus X plus one so what we can do now is we can take are the values that we have and hopefully we'll be able to reduce the value down into this finite field so let's now take this value so we now have x to the power of 5 plus x to the power of 3 plus X and we're going to divide it by our primitive that's cubed plus X plus one so let's do this long-handed obviously we can do it in this way by finding the inverse of this but we'll just do it as we would do along the vision so that goes into that x squared times x to the three and we end up with x squared so those will cancel when we add them together those will cancel and we end up with x squared plus X now that doesn't go into that so this becomes our remainder so the result of this operation will then be this value here and we can see this is constrained with inside finite field now we try this one so in this case we had X cubed plus one and we're going to divide it by X cubed plus X plus one so that goes into that one time so we end up that's one plus X plus one so that cancels we'd end up with x and that will be an this one which is zero so the value will be X here so the result after we divided by the Primitive will be X the result of this operation here will be this so you can see that the Primitive value is the way that we can constrain the power of our polynomial so that we fit into the finite field thank you
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