In binary fields, does it matter which irreducible polynomial to be used in multiplication? I have been reading about binary fields, i.e. $GF(2^m)$ where $m \in \mathbb{Z}^+$. I think this is a pretty neat explanation of it: http://www.lirmm.fr/~giorgi/PDF/HanMenVan_ch2.pdf
I understand that multiplication in $GF(2^m)$ requires modulo the chosen irreducible polynomial $f(z)$. I wonder if it matters which irreducible polynomial to use. 
To put things in context, in the example 2.2 from that pdf, it didn't really state the reason for choosing $z^4 + z + 1$. There should be other possible irreducible polynomial right? Does it matter which one we choose?
Thank you very much.
Jason
 A: You may already know that finite fields exist with all prime power orders.  
It is not obvious but all fields of order $p^n$ for prime $p$ are isomorphic.  Hence a finite field is characterised up to isomorphism by its order.  There is some detail here: Finite field at Wikipedia
So, it may or may not matter depending on exactly what you want to achieve.  
I guess that $z^4 + z + 1$ was chosen as the simplest polynomial that had the necessary properties.  
By request, a little more on isomorphisms but first homomorphisms.  
With fields, and many other algebraic objects, a function from one to another which preserves the operation is called a homomorphism.  So, more specifically for fields, if $F$ and $G$ are fields (finite or not) and $f : F \rightarrow G$ is a function from $F$ to $G$ with these properties
$$\forall x, y \in F: f(x + y) = f(x) + f(y)$$
$$\forall x, y \in F: f(x . y) = f(x) . f(y)$$
then $f$ is called a homomorphism.  
If it is also injective (one to one) and surjective (onto) then it is called an isomorphism and we say that $F$ is isomorphic to $G$.  Their behaviour will be the same except that the elements will be labelled differently.  We tend to think of them as really the same.  
A simple example is with the complex numbers $\mathbb{C}$ the function $f(z) = \bar z$ is an isomorphism of $\mathbb{C}$ to itself (which is called an automorphism).
So, what I said above is that such a function exists between any two finite fields of the same size.  They may look different but they are in a sense the same.  
So, if you use a different polynomial to construct $GF(2^4)$ then it may look different but it will behave the same.  This is what I meant but it might or might not matter.  If you are exchanging data about the field with someone else then you will need to tell them how you constructed the field.  If you are just looking for its algebraic properties then it won't matter how you constructed it.  
Note that is applies only to finite fields not infinite ones.  There are plenty of infinite fields e.g. $\mathbb{Q}$, $\mathbb{A}$, $\mathbb{R}$, and $\mathbb{C}$ and the concept of isomorphism still applies but even if they are the same size (in the sense that is usually defined for infinite sets) then they are not necessarily isomorphic.  In a sense $\mathbb{Q}$ and $\mathbb{A}$ are the same size but not isomorphic.  Also $\mathbb{R}$ and $\mathbb{C}$ are, in the same sense, the same size but also not isomorphic.  However, the first pair are not the same size as the second (and certainly not isomorphic).  
