# Equivalence of $\mathbb{F}_p$ and $\mathbb{Z}/p\mathbb{Z}$

I feel like I have some fundamental misunderstanding and I'm not really sure how to phrase this question, but here's a first attempt. In Child's Concrete Introduction to Higher Algebra (ISBN: 978-0387745275), the author uses $\mathbb{F}_p$ and $\mathbb{Z}/p\mathbb{Z}$ interchangeably (especially when talking about polynomials with coefficients in some field $\mathbb{F}_p$). I understand that if $p$ is prime, then $\mathbb{Z}/p\mathbb{Z}$ is a field, but why is correct to say something like this:

Let $f(x),g(x) \in \mathbb{F}_3[x]$ and $f(x)=3x^2 + 2$. If $g(x)=2$, then $f(x)=g(x)$. Basically, why can I apply modulo...aren't there fields that aren't the set of congruence classes modulo $p$ (which is how I understand $\mathbb{Z}/p\mathbb{Z}$)? Is this some fundamental property of finite fields that I haven't read about yet?

• The point is that, up to isomorphism, there is only one field with a given prime number of elements. Apr 3, 2015 at 17:10
• There is only one field (up to isomorphism, as Andreas says!) of size $p$. This follows (non-trivially) from the statement that there is only one group of size $p$, which in turn follows from the theorem that the order of any given element of a group divides the order of the group (why?). For fields of size $p^n$ the statement is still true, but the theorem rapidly becomes non-trivial. Still, it's well-enough established that people speak of 'the' field $\mathbb{F}_{p^n}$. Apr 3, 2015 at 17:11

To flesh out my comment a bit, here's a proof of the result for $\mathbb{F}_p$, using the fact (which you should already know or should be able to easily prove) that there's only one group of order $p$ (and that group is commutative and generated by any of its elements):

We know that the additive group of our field of order $p$ must be $C_p$, and that any of the non-zero elements will generate it. Our field has a unit element for multiplication; we'll label this unit (of course) as $\bar1$. Now, since $\bar1$ generates the additive group, we can supply labels $\bar2=\bar1+\bar1$, $\bar3=\bar2+\bar1$, $\ldots$ for the rest of the elements of the group; note that they all obey $\overline{m+n}\equiv \bar{m}+\bar{n}\pmod p$, so that (at least additively) the bars are superfluous. From here, we just work by induction using the distributivity of multplication over addition: for all $n\in C_p$, $\bar2\times \bar n=(\bar1+\bar1)\times \bar n = \bar n+\bar n \equiv \overline{n+n}\equiv \overline{2\cdot n\pmod p}$ (where the first multiplication is the one in our field and the last multiplication is the canonical one over $\mathbb{Z}$); then $\bar3\times \bar n=(\bar2+\bar1)\times \bar n=\bar2\times \bar n + \bar1\times \bar n \equiv \overline{2\cdot n + n} \equiv \overline{3\cdot n\pmod p}$, etc. This provides a canonical isomorphism of our field's multiplication operation to multiplication mod $p$.

As Andreas and Steven said, there is only one field which $p$ elements, for $p$ prime, up to isomorphisms. This fact is part of a more general result:

If $p$ is a prime and $n\in \mathbb{N}$, there there exists exactly one field with $p^n$ elements, up to isomorphisms.

The symbol $\mathbb{F}_{p^n}$ denotes the field with $p^n$ elements. It is important to notice that:

• $\mathbb{Z}/p\mathbb{Z}$ is a field and has got $p$ elements, then it is isomorphic to $\mathbb{F}_p$.
• $\mathbb{Z}/p^n\mathbb{Z}$ is not a field if $n>1$.

Hence you are allowed to think about $\mathbb{Z}/p\mathbb{Z}$ when you read $\mathbb{F}_p$, but if $n>1$ then the field $\mathbb{F}_{p^n}$ cannot be identified with the set of congruence classes modulo $p^n$.

EDIT: my first answer of the following question was not completely correct.

The answer to "aren't there fields that aren't the set of congruence classes modulo $p$?" is "yes, every finite field is (isomorphic to) a set of congruence classes". As Pedro Tamaroff pointed out in his comments, every finite field can be realized as a quotient: $\mathbb{F}_{p^n}$ is isomorphic to $$\frac{\mathbb{Z}_p[X]}{(f_n)}$$ where $\mathbb{Z}_p$ is the field $\mathbb{Z}/p\mathbb{Z}$ and $f_n$ is an irreducible polynomial of degree $n$ with coefficients in $\mathbb{Z}/p\mathbb{Z}$. Indeed, the above quotient is a field which is an extension of $\mathbb{Z}_p$ of degree $n$, so it has got $p^n$ elements.

However, it is true that there are finite fields that cannot be realized as sets of integer congruence classes.

• Every finite field can be obtained as a set of congruente clases, just not of that specific set.
– Pedro
Apr 3, 2015 at 18:02
• @PedroTamaroff Are you sure about that? How do you build integer congruence classes that satisfy the field equations for $\mathbb{F}_4$, say? Apr 3, 2015 at 18:08
• @StevenStadnicki I never said integer congruence classes. You take the congruence classes of $\Bbb Z_2[X]$ under the congruence induced by the ideal $(X^2+X+1)$.
– Pedro
Apr 3, 2015 at 18:22
• @PedroTamaroff Fair enough - I prefer to use 'equivalence class' for those and reserve 'congruence class' for integer congruence, but it's certainly not inappropriate terminology. Apr 3, 2015 at 18:35