Finding the order of elements in a Galois Field Does there exist a Galois field GF(4)?
GF(4)={0,1,2,3};
If we take this Galois field, then the element '2' is not having any degree..?
So is it possible to construct GF(4) ?
 A: The simple answer to what appears to be your most basic question:  "is there a finite field with four elements?"
The answer is:  "yes"
The easiest way to construct it is to take
$$\Bbb F_2[x]/(x^2+x+1)$$
where $\Bbb F_2$ is the field with two elements, which you might denote by GF(2).
The polynomial $x^2+x+1$ has no roots, and so is irreducible, so that the quotient is a field.
Its elements are $0,1,x,x+1$ as anything of higher degree can be reduced since $x^2=x+1$.
The multiplication rule is simple, $x(x+1)=1$ since $x(x+1)=x^2+x=-1=1$ is the only one which could be ambiguous, everything else is simple.
The symbols $2,3$ are misleading to use because they make you think you're using things that behave like those numbers. You aren't. If you wanted to use those you'd be more thinking about the ring $\Bbb Z/4\Bbb Z$ of integers modulo $4$, which is not a field.
In particular, let's address the original question in its many possible interpretations:
($1$) Finding the degree of an element of GF($4$), which I will denote by $\Bbb F_4$.
To do this you need the degree of a minimal polynomial for $a\in\Bbb F_4$ over $\Bbb F_2$. There are two possibilities, either that degree is $1$ or it is $2$. You can tell the elements of degree $1$ if they do not satisfy $x^2+x+1=0$ and the degree $2$ elements do satisfy that relation.
($2$) The additive order. In this case, since all fields of size $2^n$ have characteristic $2$, so the answer is universally: "$2$."
($3$) The multiplicative order. In the final case you first need to discard $0$, which has no such thing. Clearly $1$ has order $1$. For the other two, we note that $x^3-1=(x-1)(x^2+x+1)$ so that the other two elements, which are not $1$, satisfy $x^3=1$, so their order divides $3$. Clearly their orders are not $1$, so it must be that the other two elements have order $3$.
A: The 4-element field has 4 elements but not $0,1,2,3$. The characterestics is 2, so in the field $3=1$ and $2=0$. The other two elements are the solutions of the equation $x^2+x+1=0$. Their order is $3$.
A: Note that a polynomial over the field with 2 elements has a binary representation (technically this should be called "$x$-adic notation", since you're writing the polynomials in base $x$). In some contexts, it is understood that when we write an integer, that we mean the polynomial that has the same binary representation: e.g. $2$ means $x$ and $3$ means $1+x$.
(in my opinion, this is usually a bad convention)
