Tidy up

docs/lectures/cryptography/06_finite_fields.md

@@ -1,12 +1,12 @@
 # Finite Field Arithmetic
 
 - A **finite field** is a set containing a finite number of elements
-  - This is sometimes called a *Galois Field*
+    - This is sometimes called a *Galois Field*
 - In a Galois field you can:
-  - Add
-  - Subtract
-  - Multiply
-  - Invert (divide)
+    - Add
+    - Subtract
+    - Multiply
+    - Invert (divide)
 - Fields are an extension of *groups* and related to *rings*
 
 ### Groups
@@ -14,9 +14,9 @@
 A group is a set of elements $G$ together with an operation $\circ$ that combines two elements of $G$
 
 > 1. The operation $\circ$ is **closed**
->    - i.e. for all $a,b \in G$ then $a\circ b=c\in G$
+>     - i.e. for all $a,b \in G$ then $a\circ b=c\in G$
 > 2. The operation is associative
->    - i.e. $a\circ(b\circ c) = (a\circ b)\circ c$ for all $a,b,c \in G$
+>     - i.e. $a\circ(b\circ c) = (a\circ b)\circ c$ for all $a,b,c \in G$
 > 3. There is an element $1\in G$ called a **neutral element** such that $a\circ 1 = 1\circ a = a$ for all $a\in G$
 > 4. For each $a \in G$ there exists an element $a^{-1}\in G$ called the **inverse** of $a$ such that $a\circ a^{-1} = a^{-1}\circ a = 1$
 > 5. A group $G$ is **abelian** (commutative) if $a\circ b = b \circ a$ for all $a,b\in G$
@@ -26,7 +26,7 @@ A group is a set of elements $G$ together with an operation $\circ$ that combine
 - The set of integers $\mathbb{Z}_m = \{0,1,...m-1\}$ with the operation addition modulo m form a group with the neutral element 0
 - Every element would have an inverse where $a + (-a) = 0$ mod m
 - This group would not form a group with multiplication, as not all elements would have an inverse
-  - We wouldn’t have an inverse, we would need $5\times \frac15=1$ however $\frac15 \notin \mathbb{Z}$ 
+    - We wouldn’t have an inverse; we would need $5\times \frac15=1$, but $\frac15 \notin \mathbb{Z}$
 
 ### Fields
 
@@ -35,12 +35,12 @@ A field $F$ is a set of elements with the following properties
 > 1. All elements of $F$ form an **additive group** with the group operation $+$ and the neutral element 0
 > 2. All elements of $F$ except 0 form a multiplicative group with the group operation $\times$ and the neutral element 1
 > 3. When the two group operations are mixed, the distributivity law holds.
->    - i.e. for all $a,b,c \in F, a\cdot(b+c) = (a\cdot b) + (a\cdot c)$
+>     - i.e. for all $a,b,c \in F, a\cdot(b+c) = (a\cdot b) + (a\cdot c)$
 
 ##### Example Field
 
 - The set of real numbers $\mathbb{R}$ is a field with neutral element 0 for addition and 1 for multiplication
-- Every real number $a$ has a additive inverse $-a$
+- Every real number $a$ has an additive inverse $-a$
 - Every non-zero number $a$ has a multiplicative inverse $\frac{1}{a}$
 
 ![1646405235.png](img/1646405235.png)
@@ -80,7 +80,7 @@ $a \cdot a^{-1} \equiv 1 \space (mod \space p)$
 
 - A modular inverse exists when $gcd(a,p) = 1$
 - Because $p$ is prime, every number has a multiplicative inverse
-  - $gcd(a,p) = 1, \forall a \neq0 \in GF(p)$
+    - $gcd(a,p) = 1, \forall a \neq0 \in GF(p)$
 - $a^{-1}$ can be calculated using the **extended Euclidean algorithm**
 
 #### Extension Fields
@@ -97,16 +97,16 @@ The coefficients of the polynomial are elements in $GF(2)$ the **sub-field**
 ##### Example $GF(2^3)$
 
 - The field $GF(2^3)$, sometimes called $GF(8)$ is an extension field containing elements of the form: $A(x) = a_2 x^2 + a_1x^1 + a_0$
-- Its often easier to simply write the coefficients $(a_2, a_1, a_0)$ e.g. 001 or 101
+- It's often easier to simply write the coefficients $(a_2, a_1, a_0)$ e.g. 001 or 101
 - $GF(2^3) = \{0, 1, x, x+1, x^2, x^2+1, x^2 + x, x^2 + x + 1\}$ 
-  - $|GF(2^3)| = 8$
+    - $|GF(2^3)| = 8$
 
 #### Arithmetic in $GF(2^3)$
 
 - Adding or subtracting two polynomials happens as expected, but adding the coefficients
-  - $A(x) = x^2 + x + 1$
-  - $B(x) = x^2 + 1$
-  - $A(x) + B(x) = (1+1)x^2 + (1)x + (1+1) = x$
+    - $A(x) = x^2 + x + 1$
+    - $B(x) = x^2 + 1$
+    - $A(x) + B(x) = (1+1)x^2 + (1)x + (1+1) = x$
 - mod 2 is simply `xor`
 - Addition and subtraction are identical
 
@@ -130,7 +130,7 @@ $$
 
 - Inversion is performed in a similar way to prime fields, we find:
 - $A(x) \cdot A^{-1}(x) \equiv 1 \space (mod \space P(x))$
-  - $A^{-1}(x)$ is calculated using the extended euclidean algorithm
+    - $A^{-1}(x)$ is calculated using the extended Euclidean algorithm
 
 ### AES’ Finite Field