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<blockquote>
 elements of $GF(p)$ are integers, while elements of $GF(p^n)$ are polynomials. Why the difference?
</blockquote>

TL;DR: because integers modulo $p^n$ do not form a field.

<hr>

Here's a pragmatic construction of $GF(p^n)$ useful for applied cryptography, including AES which uses the field $GF(2^8)$ extensively.

$GF(p)$ for prime $p$ is the field of arithmetic modulo $m$ for modulus $m=p$. Arithmetic modulo an integer $m$ works as integer arithmetic except that all results are reduced to the range $[0,m)$ by subtraction of $q\,m$ for some appropriate integer $q$. Arithmetic modulo $m$ keeps the usual properties of addition and multiplication: their are <a href="https://en.wikipedia.org/wiki/Binary_operation#Terminology" rel="nofollow noreferrer">internal laws</a> with <a href="https://en.wikipedia.org/wiki/Associative_property" rel="nofollow noreferrer">associativity</a>, <a href="https://en.wikipedia.org/wiki/Commutative_property" rel="nofollow noreferrer">commutativity</a>, <a href="https://en.wikipedia.org/wiki/Distributive_property" rel="nofollow noreferrer">distributivity</a>, <a href="https://en.wikipedia.org/wiki/Neutral" rel="nofollow noreferrer">neutral</a> $0$ for addition and $1$ for multiplication, existence of an <a href="https://en.wikipedia.org/wiki/Additive_inverse" rel="nofollow noreferrer">opposite</a> (w.r.t. addition) for all elements. And, specifically when $m$ is prime, arithmetic modulo $m$ gains a property of rationals and reals that integers do not have: existence of an <a href="https://en.wikipedia.org/wiki/Inverse_element" rel="nofollow noreferrer">inverse/reciprocal</a> (w.r.t. multiplication) for all elements except $0$. That additional property makes arithmetic modulo $p$ a <a href="https://en.wikipedia.org/wiki/Field_(mathematics)" rel="nofollow noreferrer">field</a> of $p$ elements, when arithmetic modulo a composite $m$ is not¹.

When we want to make a field of $m$ elements and $m$ is not prime, we thus can't use arithmetic modulo $m$. It turns out that there is an alternative construction if and only if $m$ is of the form $p^n$ with $p$ prime. That's $GF(p^n)$.

One way to think an element $A$ of $GF(p^n)$ is as a <a href="https://en.wikipedia.org/wiki/Vector_(mathematics_and_physics)" rel="nofollow noreferrer">vector</a> $(a_0,a_1,\ldots,a_{n-1})$ of $n$ elements $a_i\in GF(p)$. At least that gives us the right count of elements $p^n$, and we can define a well-behaved addition as
$$A+B=(a_0+b_0\bmod p,a_1+b_1\bmod p,\ldots,a_{n-1}+b_{n-1}\bmod p)$$
with the all-zero vector the neutral. But somewhat we need to define multiplication in such a way that $A\,B$ is also a vector of $n$ elements in $GF(p)$, and all except one elements have an inverse².

That's where <a href="https://en.wikipedia.org/wiki/Polynomial" rel="nofollow noreferrer">polynomials</a> come to the rescue. We assimilate $A=(a_0,a_1,\ldots,a_{n-1})$ to the (<a href="https://en.wikipedia.org/wiki/Univariate" rel="nofollow noreferrer">univariate</a>) polynomial $a_0+a_1\,x^1+\ldots+a_{n-1}\,x^{n-1}$ of <a href="https://en.wikipedia.org/wiki/Degree_of_a_polynomial" rel="nofollow noreferrer">degree</a> less than $n$ and coefficients $a_i\in GF(p)$. Notice that the rules of addition of polynomials with coefficients in $GF(p)$ coincide with our earlier definition of addition.

Now we pick a polynomial $M=m_0+m_1\,x^1+\ldots+m_{n-1}\,x^{n-1}+m_n\,x^n$ of degree³ $n$ with coefficients $m_i\in GF(p)$, and define the product $A\,B$ in our would-be field as
$$\underbrace{A\,B}_{\text{in would-be field}}=C=\underbrace{A\,B\bmod M}_{{\text{in polynomials with}\\\text{coefficients in }GF(p)}}$$
The right side means that $C$ is the⁴ polynomial with coefficients $c_i\in GF(p)$, of degree less than the degree of $M$ (hence of degree less than $n$, hence representable as a vector of $n$ elements in $GF(p)$ as desired), such that there exists a polynomial $Q$ with coefficients $q_i\in GF(p)$ with
$$0=C+Q\,M-A\,B$$
per the normal rules of addition and multiplication of polynomials with coefficients in $GF(p)$. That is
$$0=\sum_{0\le i&lt;n}\left(\left(c_i+\sum_{0\le j&lt;n}\left(q_j\,m_{i-j}-a_j\,b_{i-j}\right)\right)x^i\right)$$
or equivalently when we get back to vector notation rather than polynomials
$$\forall i\in[0,n),\ \underbrace{0=c_i+\sum_{0\le j&lt;n}\left(q_j\,m_{i-j}-a_j\,b_{i-j}\right)}_{\text{in }GF(p)\text{, that is}\pmod p}$$
We've already seen that this construction of multiplication is an internal law. It's easy to show that it inherits from the field of arithmetic in polynomials with coefficients in $GF(p)$ the properties of associativity, commutativity, distributivity, and neutral the constant polynomial $1$, that is the vector $(1,0,\ldots,0)$. We have constructed a commutative <a href="https://en.wikipedia.org/wiki/Ring_(mathematics)" rel="nofollow noreferrer">ring</a> with $p^n$ elements.

It can be shown that if (and only if) polynomial $M$ is irreducible, each element has an inverse, which completes the field properties. And it can be shown that there exits an <a href="https://en.wikipedia.org/wiki/Isomorphism" rel="nofollow noreferrer">isomorphism</a> among the different fields that we obtain for different irreducible polynomials $M$, which is why mathematicians speak of the field $GF(p^n)$, rather than of the field $GF(p^n)$ obtained for a particular irreducible polynomial $M$, as practical cryptographers often do.

<hr>

<blockquote>
 Relation between extension fields and non-prime fields
</blockquote>

A field $G$ is an extension field of $F$ when $F$ (or $F'$ with a trivial mapping to $F$) is a subset of $G$ and is a field under the same addition and multiplication laws that make $G$ a field.

When we restrict $GF(p^n)$ constructed as above to the set of constant polynomials (equivalently, the set of vectors of $n$ elements in $GF(p)$ with all expect the first set to zero), we are back to a subset of $p$ elements that matches $GF(p)$.

Otherwise said, $GF(p^n)$ is an extension field of $GF(p)$.

<hr>

Notes:

¹ Arithmetic modulo a composite $m$ is only a commutative ring of $m$ elements, but not a field. Proof: if $m$ is composite, let $a$ be the smallest prime divisor of $m$. If $a$ had an inverse $x$ modulo $m$, we'd have $a\,x\equiv1\pmod m$, that is $\exists q\in\Bbb Z,\ a\,x=1+q\,m$. We can write $m=a\,b$ for some $b\in\Bbb Z$. Hence $a\,x=1+q\,a\,b$, hence $a\,(x-q\,b)=1$ in the ring of integers, which can't be since $a$ is prime. Hence $a$ has no inverse modulo $m$ when $m$ is composite. Since $a$ is not zero, one of the field requirement is not met.

² That's the hard part. In particular, the construction of the product as $A\,B=(a_0\,b_0\bmod p,a_1\,b_1\bmod p,\ldots,a_{n-1}\,b_{n-1}\bmod p)$ leaves multiple $A$ without an inverse: all those with at least one of their $a_i=0$ in $GF(p)$. 

³ That's exactly degree $n$, otherwise said $m_n\ne0$ in $GF(p)$.

⁴ The polynomial $C=A\,B\bmod M$ is uniquely defined by the relation $0=C+Q\,M-A\,B$. That $C$ can be (and in practice often is) obtained by computing the product $A\,B$ as a polynomial with coefficients in $GF(p)$ of degree less than $2\,n-1$, and progressively reducing that product's degree from at most $n+j$ to less than $n+j$, by computing $q_j$ and subtracting $q_j\,x^j\,M$, with $q_j=d_{n+j}/m_n$ computed in $GF(p^n)$, and $d_{n+j}$ is the high-order coefficient of the progressively reduced product. To simplify that computation, it is customarily used $m_n=1$, that is $M$ a <a href="https://en.wikipedia.org/wiki/Monic_polynomial" rel="nofollow noreferrer">monic polynomial</a>.

I was watching this class about AES in this <a href="https://www.youtube.com/watch?v=x1v2tX4_dkQ" rel="nofollow noreferrer">LINK</a> and I was trying to grasp the concept of prime fields, which is a finite field with prime order $p$.

The non-prime field part (order is $p^n$) is where it gets harder for me. The teacher jumps into "extension fields", and does not explain the link between it and non-prime fields. Also, he just affirms that elements of $GF(p)$ are integers, while elements of $GF(p^n)$ are polynomials.

Why the difference between the two?

Prime fields vs non-prime fields

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我在这个链接里看了这个关于AES的课，我试着理解素数场的概念，它是一个有限域，有质数阶的p。非素数字段部分(order是p^n)对我来说变得更难了。老师跳到“扩展字段”，没有解释它和非素数字段之间的联系。同时，他也肯定了GF(p)的元素是整数，而GF(p^n)的元素是多项式。为什么两者之间有区别？

问素数域与非素数域
EN

回答 1

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