A finite field $(\mathbb F,+,\cdot)$ is a finite set $\mathbb F$ with two internal laws $+$ and $\cdot$, such that $(\mathbb F,+)$ is a commutative group with neutral noted $0$, and $(\mathbb F-\{0\},\cdot)$ is a commutative group with neutral noted $1$, and multiplication is distributive w.r.t. addition that is $\forall A,B,C\in\mathbb F$ it holds $A\cdot(B+C)=(A\cdot B)+(A\cdot C)$.
It can be shown that all finite fields with the same number of elements are isomorphic, that is we can map from one to another by a bijection $\mathcal F$ such that $\mathcal F(A+B)=\mathcal F(A)+\mathcal F(B)$ and $\mathcal F(A\cdot B)=\mathcal F(A)\cdot \mathcal F(B)$. We can thus talk about the finite field $\mathbb F$ with $q$ elements. It's often noted $\mathbb F_q$.
It can be shown that any finite field has a number $q$ of elements of the form $q=p^k$ for some prime $p$ and some $k\in\mathbb N^*$.
When $k=1$, the field $(\mathbb F_p,+,\cdot)$ is simply the ring of integers modulo $p$, that is $(\mathbb Z_p,+,\cdot)$.
For arbitrary $k\in\mathbb N^*$, we can think of the field $(\mathbb F_{p^k},+,\cdot)$ as the set of polynomials of degree up to $k-1$ and coefficients in $\mathbb Z_p$. That is, polynomials for an abstract variable $x$ with one coefficient in $\mathbb Z_p$ for each of the $k$ terms $x^i$ with $i\in\{0,1\ldots,k-1\}$. Addition in $\mathbb F_{p^k}$ is addition of polynomials. Multiplication in $\mathbb F_{p^k}$ is multiplication of polynomials followed by reduction modulo a particular reduction polynomial $R(x)$ of degree exactly $k$, and irreducible.
Equivalently, we can think of $\mathbb F_{p^k}$ as the set of $p^k$ tuples of $k$ elements of $\mathbb Z_p$, noted $(a_0,a_1\ldots,a_{k-1})$. Addition is defined by$$(a_0,a_1\ldots,a_{k-1})+(b_0,b_1\ldots,b_{k-1})=(a_0+b_0,a_1+b_1\ldots,a_{k-1}+b_{k-1})$$with the later additions carried in $\mathbb Z_p$, that is with reduction modulo $p$. If the tuple $A$ has $a_i=1$ and all the other terms $0$, and the tuple $B$ has $b_j=1$ and all the other terms $0$, then when $i+j<k$ the tuple $C$ for $A\cdot B$ has $c_{i+j}=1$ and all the other terms $0$. When $i+j=k$, the tuple $C$ for $A\cdot B$ is a constant tuple $(r_0,r_1,\ldots,r_{k-1})$ independent of $i$ and $j=k-i$. That tuple is such that the polynomial $R(x)=x^k-r_{k-1}\,x^{k-1}\ldots-r_1\,x-r_0$ with coefficients in $\mathbb Z_p$ is irreducible, implying $r_0\ne0$. This constant tuple $(r_0,r_1\ldots,r_{k-1})$, or equivalently the polynomial $R(x)$, combined with the previously stated rules and properties of $+$ and $\cdot$, fully defines multiplication, and it's neutral $(1,0\ldots,0)$.
We can compute the tuple $(c_0,c_1\ldots,c_{k-1})$ for $(a_0,a_1\ldots,a_{k-1})\cdot(b_0,b_1\ldots,b_{k-1})$ as follows:
$(c_0,c_1\ldots,c_{k-1}):=(0,0\ldots,0)$
for $i$ from $k-1$ down to $0$
- $m:=c_{k-1}$
- for $j$ from $k-1$ down to $1$
- $c_j:=m\cdot r_j+a_i\cdot b_j+c_{j-1}$
- $c_0:=m\cdot r_0+a_i\cdot b_0$
with the computations in the last two lines carried in $\mathbb Z_p$, that is modulo $p$.
