TITLE: What is an extension field? Covered differently in math & in cryptography.
QUESTION [7 upvotes]: In his book on Cryptography, Paar has this theorem

Theorem 4.3.1 A field with order m only exists if m is a prime
power, i.e., m = p^n, for some positive integer n and prime integer
p. p is called the characteristic of the finite field.

So here he says that the order has to be a prime power -
He also has this as examples

This theorem implies that there are, for instance, finite fields with 11 elements, or with 81 elements (since 81 = 3^4) or with 256 elements (since 256 = 2^8, and 2 is a prime).

So he explicitly says that you can have a field with 256 elements - the order of a finite field needs to be a prime power & not necessarily a prime itself.
He then goes on to talk about extension fields - he says that if the order of the field is not a prime then it's called as an extension field.

In AES the finite field contains 256 elements and is denoted as GF(2^8). This field was chosen because each of the field elements can be represented by one byte. For the S-Box and MixColumn transforms, AES treats every byte of the internal data path as an element of the field GF(2^8) and manipulates the data by performing arithmetic in this finite field.
However, if the order of a finite field is not prime, and 2^8 is clearly not a prime, the addition and multiplication operation cannot be represented by addition and multiplication of integers modulo 2^8. Such fields with m > 1 are called extension fields.

So as per this, I get the definition of an extension field as this - an extension field is any finite field where the order of the field is a prime power but not a prime itself.
However, when I look at books on abstract algebra, I see a totally different definition of extension fields which seem to be unconnected with what Paar says.
For e.g. from "Topics in Algebra" by Hernstein:

Let F be a field; a field K is said to be an extension of F if K
contains F. Equivalently, K is an extension of F if F is a subfield of K.

So are Extension fields described in Cryptography different from those described in Algebra? But is Paar's description wrong? Or are the 2 definitions equivalent in some way?

REPLY [2 votes]: Paar uses “extension field” to mean nontrivial extension field, that is, an field that is an extension field of some field other than itself. In other words, he calls a field an “extension field” a field that has a proper subfield.
(It's possible that he'd prefer a different definition in general, perhaps to restrict to a finitely generated extension, but that doesn't make a difference for finite fields.)
Any field $F$ has a smallest subfield, which is the subfield generated by the empty set. It's the intersection of all the subfields of $F$. A field must contains the neutral element for addition $\bar 0$ and the neutral element for multiplication $\bar 1$, and $\bar 1+\bar 1$, $\bar 1+\bar 1+\bar 1$, etc. There are two cases for what this smallest subfield can look like.

If there is a number $p \ge 1$ such that $p \bar 1 = \mathop{\underbrace{\bar 1+\bar 1+\bar 1+\ldots+\bar 1}}_\limits{\text{\(p\) terms}} = \bar 0$, take the smallest such $p$. $p$ must be prime, because if $p = q r$ with $p, q \ge 2$ such that $(q r) \bar 1 = \bar 0$ then $(q \bar 1) \cdot (r \bar 1) = \bar 0$ and therefore one of $q \bar 1$ or $r \bar 1$ is $\bar 0$. Addition and multiplication on $\{\bar 0, \bar 1, \bar 1+\bar 1, \bar 1+\bar 1+\bar 1, \ldots, (p-1)\bar 1\}$ coincides with the same operations in $\mathbb{Z}/p\mathbb{Z}$, and it follows that division also coincides and therefore $\{\bar 0, \bar 1, \bar 1+\bar 1, \ldots, (p-1)\bar 1\}$ is closed by division and is itself a field which is isomorphic to $\mathbb{Z}/p\mathbb{Z}$. The number $p$ must be prime, and is called the characteristic of the field.
Otherwise, $\bar 0$, $\bar 1$, $\bar 1+\bar 1$, $\bar 1+\bar 1+\bar 1$, etc. are all distinct. In this case, the smallest subfield is isomorphic to $\mathbb{Q}$ (the rationals), and the field is said to have characteristic 0. Note that this cannot happen for a finite field, since $\{\bar 0, \bar 1, \bar 1+\bar 1, \bar 1+\bar 1+\bar 1, \ldots\}$ is an infinite subset.

I omit the proofs, which should be easy to find. Paar should prove these statements for the case of finite fields as lemmas before the theorem you cite.
A nontrivial extension field is therefore a field which is neither $\mathbb{Z}/p\mathbb{Z}$ nor $\mathbb{Q}$ (up to isomorphism). For a finite fields, the number of elements must be a prime power $p^n$, where $p$ is the characteristic. If $n = 1$ the field is $\mathbb{Z}/p\mathbb{Z}$, which doesn't have a proper subfield (as we saw above, no proper subset containing $1$ is closed even by addition alone ). If $n \gt 1$ the field is a proper extension of $\mathbb{Z}/p\mathbb{Z}$.
The reason why the terminology “field extension” is used even though it's redundant with “subfield” is that it is very common in algebra (but not so common in the study of finite fields) to construct fields by starting from a base field and adding elements to it. Formally, what's really happening is to start from a “large” field $L$ and a “small” field $K$ which is a subfield of $L$, and then consider fields generated by some subset $S$ of $L$, written $K(S)$ (with $L$ implicit).