# Proof only one finite field with p elements

I am not sure if it is possible with my experience to prove something like this but I would be interested in how such a proof would work. I am an undergraduate and have had a Modern Algebra course so my understanding may be limited.

Essentially, I would like to prove that given two finite fields $F_p$ and $F'_p$ each with $\textit{p}$ elements that there exists an isomorphism from one field onto the other.

I am thinking I would need to define some kind of addition and multiplication table for one of the fields that satisfied the conditions for being a field. Then I would imagine you could say something like $$f(x+y) = a_{x+y} = a_x + a_y$$ by the way we defined the addition for example. I understand this is very naive but I am just not really sure how to start.

Thank you!!

-
You say that $F_p$ and $F_p'$ are finite fields, but then go on to say that you need to define addition and multiplication tables. A field's addition and multiplication tables are fixed and are part of its definition. – Michael Albanese Sep 30 '12 at 3:40
Is your $p$ a prime number? – Berci Sep 30 '12 at 12:32
Finite fields of a given order are unique up to isomorphism, so it is not in my opinion a bad idea to begin with the abstract idea of a field with $p$ elements, construct addition and multiplication operations in each of them, and then make a bijection. Try using $\mathbb{Z}_p$ for the additive group of one of the fields, to start. – Alexander Gruber Oct 5 '12 at 3:24

You can construct the isomorphism based on the fact that the multiplicative identity in the field must generate the field under the operation of addition if the field has order p.

The set of elements generated by 1 form a subgroup under addition. The order of that subgroup must divide the order of the entire additive group, p, and since the subgroup contains 0 and 1, it must have order p. That means each element in the field must be writable as a multiple of 1. You can then use that identification to construct your isomorphism.

-
Thank you, I will use this to help strengthen my understanding to create a proof that I am satisfied with. – Frudrururu Oct 1 '12 at 4:15

A finite field with $p$ elements must have characteristic $p$. That is, $p1=1 + 1 + \cdots +1=0$ where we summed $p$ terms, and $n1\neq 0$ for $1\leq n \leq p-1.$ Thus the elements $n1$, $n=0,1,\cdots, p-1$, must all be distinct and thus make the entire field.

So now suppose we have two fields with $p$ elements, call them $F$ and $F',$ and also suppose there exists some isomorphism $\phi$ between them. By definition of field isomorphism, we must have $\phi(1_F) = 1_{F'}.$ Then using $\phi(a+b) = \phi(a) + \phi(b)$ we get $\phi(n1_F) = n\phi(1_F) = n1_{F'}.$ This already determines $\phi$ as all elements of $F$ are of the form $n1_F,$ so if there is an isomorphism, it must be the one which sends $n1_F\mapsto n1_{F'}$. We can check that indeed, this is an isomorphism (it is injective and surjective and satisfies field homomorphism conditions).

Thus, two fields with $p$ elements are canonically isomorphic - they are isomorphic in a very natural way, there was no choice involved in making our isomorphism.

-
Your first statement is true if and only if $p$ is prime; $p$ could be a prime power $p = q^n$, in which case the characteristic of $F_p$ is $q$. – Michael Albanese Sep 30 '12 at 6:13
Thank you very much. I was not familiar with characteristic but I read a little about it and I find it very interesting. In time I will clear up all my questions regarding this and have a constructed proof of my own essentially following what you have given but made sure to where I understand what is going on to my satisfaction every step of the way. – Frudrururu Oct 1 '12 at 4:13