Galois Fields

Galois Fields

We have already looked at some examples of fields on the Field Axioms page. We will now look at another specific type of field known as a Galois Field. Before we do that though, we must first get a little bit acquainted to a special type of relation known as the modulus of an integer $a$ to a divisor $m$.

Definition: Let $a$ be an integer. Then we write $r = a \pmod m$ which says that $r$ is the remainder of $a$ when $a$ is divided $m$, that is for some positive integer $k$ we have that $a = km + r$.

For example, consider $7 \pmod 3$. We note that $1 = 7 \pmod 3$ since when $7$ is divided by $3$ we have a remainder of $1$, and so $7 = 2(3) + 1$. We are now ready to define what a Galois field is.

Definition: A Galois Field of the prime number $p$ denoted $GF(p)$ or $\mathbb{Z}_p$ is a set containing the nonnegative integers $0, 1, 2, ..., p$ whose operation of addition between any two elements $a, b \in GF(p)$ is defined by $a + b \pmod p$ and whose multiplication is defined as $a \cdot b \pmod p$.

The simplest Galois Field is $GF(2)$ whose addition and multiplication tables are given below.

$+$ 0 1
0 0 1
1 1 0
$\cdot$ 0 1
0 0 0
1 0 1

We can verify that all eleven axioms of a field hold for $GF(2)$. For example, commutativity of addition and multiplication can be observed on both tables since the tables have symmetric values upon the top-left to top-right diagonal.

Example 1

Write the addition table for $GF(5)$ and verify that an additive identity $0$ exists such that $\forall a \in GF(5)$ that $a + 0 = 0 + a = a$

The addition table for $GF(5)$ is as follows where the red column and row verifies that $0$ is the identity element of $GF(5)$.

$+$ 0 1 2 3 4
0 0 1 2 3 4
1 1 2 3 4 0
2 2 3 4 0 1
3 3 4 0 1 2
4 4 0 1 2 3

Example 2

Write down the multiplication table for $GF(3)$ and verify that multiplication is commutative.

The multiplication table for $GF(3)$ is as follows. The commutativity of multiplication is represented by the symmetry across the diagonal of this multiplication table, that is for all $a, b \in GF(3)$, $a \cdot b = b \cdot a$.

$\cdot$ 0 1 2
0 0 0 0
1 0 1 2
2 0 2 1

Example 3

Explain why $GF(4)$ does not form a field.

In this case we note that $p = 4$ is not a prime number. When we write the addition table for $GF(4)$ we get that:

$+$ 0 1 2 3
0 0 1 2 3
1 1 2 3 0
2 2 3 0 1
3 3 0 1 2

So nothing is wrong with addition. We can verify that all of the addition axioms hold. The reason why $GF(4)$ does not form a field actually lies within the operation of multiplication.

$\cdot$ 0 1 2 3
0 0 0 0 0
1 0 1 2 3
2 0 2 0 2
3 0 3 2 1

We first note that $1$ is the multiplicative identity. Suppose that we are given the element $2$ and want to find its multiplicative inverse $b$ such that $2b = 1$. If we look down the column/row containing $2$ we notice that no such element exists since $2b = 0$ or $2b = 2$ for all $b \in GF(4)$. Therefore not all elements in $GF(4)$ contain multiplicative inverses and so $GF(4)$ is not a field.

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