Introduction to Galois Fields

Summary

In this informative session, Bill Buchanan OBE delves into the fascinating world of Galois Fields and their critical application in cryptography. He illustrates how vectors can be expressed as polynomials, and through various operations, how these polynomials simplify cryptographic processes. The video explains binary values representation, polynomial operations, and highlights the importance of primitive polynomials in maintaining computations within finite fields. Addition, multiplication, and division of polynomials are demonstrated with clear examples, emphasizing the constraints of Galois fields, particularly for cryptographic applications.

Highlights

• Bill explains how vectors translate to polynomials for cryptography. 🤓
• Polynomials help perform cryptographic operations more efficiently. 🔍
• Binary representations correlate with polynomial expressions seamlessly. 💻
• Primitive polynomials maintain operations within finite field confines. 🔄
• Addition and multiplication with polynomials simplifies complex calculations. 🚀
• Exploring inverse polynomials aids in comprehending division processes. 🔄
• Utilizing Galois Fields aids in safeguarding cryptographic integrity. 🛡️

Key Takeaways

• Galois Fields simplify cryptographic operations by using polynomial representations. 🔐
• Vectors can be expressed as polynomials like 3 + 4X + 6X² for easier manipulation. 🧮
• Primitive polynomials are crucial for keeping calculations within finite fields. ✨
• Binary operations like addition and multiplication have polynomial equivalents. ➕✖️
• Division in Galois Fields involves finding an inverse polynomial and then multiplying. ➗
• The session illustrates how cryptographic constraints help maintain integrity. 🛡️
• Understanding Galois Fields enhances grasp of complex cryptographic processes. 🤓

Overview

Bill Buchanan OBE dives into the intricate field of Galois Fields, crucial for cryptography, by simplifying mathematical operations through polynomial expressions. Starting with vector representation, he shows how cryptography benefits from using polynomials to perform efficient operations.

The video seamlessly covers binary value translation into polynomial forms, allowing a better understanding of finite field constraints. Buchanan excels in breaking down complex ideas such as addition, multiplication, and division using polynomials, making intricate calculations more accessible.

Primitive polynomials are highlighted as vital for maintaining calculations within specific boundaries, ensuring cryptographic integrity. His discussion illustrates the importance of these concepts for securing cryptographic methods in an engaging and insightful session.

Chapters

• 00:00 - 00:30: Introduction to Galois Fields The chapter introduces Galois Fields and their application in cryptography. It starts with a basic example of a vector, using numbers like four, three, four, and six, to explain the concept.
• 00:30 - 01:00: Representation of Vectors as Polynomials The chapter discusses the concept of representing vectors as polynomials. It explains that a vector can be expressed in the form of a polynomial, such as "3 + 4x + 6x^2". This polynomial representation is used to perform various operations, particularly in simplifying cryptographic processes.
• 01:00 - 02:00: Binary Representation and Galois Fields The chapter explores the representation of binary values as polynomials. It begins with an example of representing binary values, specifically the numbers 0, 1, 2, and 3, in a polynomial form. The chapter further discusses the significance of the degree of these polynomials.
• 02:00 - 03:00: Operations on Polynomials This chapter introduces operations on polynomials within Galois fields. It begins with an explanation of a Galois field GF(2^2), where elements are represented as 0, 1, X, and X+1. The concept is extended to larger fields, such as a Galois field GF(3), indicating that similar rules and representations can be applied.
• 03:00 - 05:00: Polynomial Addition Example The chapter explains polynomial operations within Galois fields, specifically GF(2^3). It describes representing binary values as polynomials, illustrating how these polynomials work in terms of algebraic structures like Galois fields. The explanation includes how the fields are constructed, stating elements in the field include terms from zero up to x squared plus x plus one.
• 05:00 - 07:00: Polynomial Multiplication Example In this chapter titled 'Polynomial Multiplication Example', the transcript discusses the representation of polynomials using coefficients and powers. It focuses on operating with polynomials where coefficients are either zero or one, much like in binary systems. The chapter likely explores a specific example of polynomial multiplication by defining values for coefficients and illustrating how these are used in the multiplication process.
• 07:00 - 09:00: Polynomial Division Using Inverse This chapter explains polynomial division using inverse operations. The focus is on specific coefficients of 'a', whether zero or one, which are represented by respective bit positions in computations. The mathematical operations discussed involve multiplication and its inverse to perform effective polynomial division.
• 09:00 - 12:00: Constraining Values into Finite Fields The chapter 'Constraining Values into Finite Fields' discusses the concept of using binary adders. It illustrates this with an example of adding binary numbers: 0000 and 0001, resulting in 0110. The chapter also touches on binary multiplication, briefly explaining or implying how multiplication works within the constraints of binary operations. However, the provided transcript ends abruptly, leaving further details and the conclusion of the explanation unknown.
• 12:00 - 14:30: Using Primitive Polynomials The chapter delves into using primitive polynomials within mathematical operations, particularly focusing on additive and multiplicative processes. As an example, a basic polynomial x² + x + 1 is explored to demonstrate these operations.
• 14:30 - 16:00: Example of Division by Primitive Polynomial In this chapter, the process of dividing using a primitive polynomial is illustrated. The example provided breaks down a binary operation where 'one one one' is added to 'X plus one', represented in binary as 'zero one one.' The result of this calculation, x squared plus X plus one plus X plus one, is derived by following the principles of binary arithmetic.

Introduction to Galois Fields Transcription

• 00:00 - 00:30 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
• 00:30 - 01:00 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
• 01:00 - 01:30 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
• 01:30 - 02:00 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
• 02:00 - 02:30 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
• 02:30 - 03:00 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
• 03:00 - 03:30 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
• 03:30 - 04:00 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
• 04:00 - 04:30 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
• 04:30 - 05:00 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
• 05:00 - 05:30 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
• 05:30 - 06:00 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
• 06:00 - 06:30 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
• 06:30 - 07:00 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
• 07:00 - 07:30 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
• 07:30 - 08:00 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
• 08:00 - 08:30 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
• 08:30 - 09:00 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
• 09:00 - 09:30 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
• 09:30 - 10:00 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
• 10:00 - 10:30 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
• 10:30 - 11:00 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
• 11:00 - 11:30 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
• 11:30 - 12:00 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
• 12:00 - 12:30 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
• 12:30 - 13:00 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
• 13:00 - 13:30 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
• 13:30 - 14:00 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
• 14:00 - 14:30 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
• 14:30 - 15:00 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
• 15:00 - 15:30 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
• 15:30 - 16:00 of our polynomial so that we fit into the finite field thank you