Multiplication rule to turn finite set into finite field. From the 3rd edition of the book "The Linear Algebra a Beginning Graduate Student Ought to Know" by Jonathan S. Golan, we find the following exercise under chapter 2:
:
It is a matter of checking computationally (for closure and for the existence of inverses) that an appropriate candidate for addition is the column-wise summation of two tuples - with the zero tuple "$(0,0,0,0)$" being a working $0$ for the field.
However, identifying a suitable candidate for multiplication seems quite difficult. I have tried variations of addition only to realise that they will all violate distributivity. Note that the natural definition of entry-wise multiplication leads to a violation of closure:
$$ (1,2,1,1)*(2,1,2,2) = (2,2,2,2) $$
And we note that the RHS is not in $K$. Do you have any ideas what addition and multiplication should be?
 A: Write each $4$-tuple $(a,b,c,d)$ as a $2\times 2$-matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$, then use the usual $+,\cdot$ of the matrix ring $M_2(\mathbb F_3)$. 
Edit. (March 5, 2022)
Let me respond to the comment of the OP
about my thinking process for this problem.
The following are some linear algebra facts.

* An $\mathbb F$-vector space $V$ is an algebraic
object defined with
additive and scaling structure, with scalars from
the field $\mathbb F$.


* If $V$ is an $n$-dimensional $\mathbb F$-space,
then, after choosing an $\mathbb F$-basis for $V$, elements $v\in V$ may be written
as length-$n$ columns vectors, $[v]\in \mathbb F^n$.
Linear transformations $\varphi\colon V\to V$ may be written
as $n\times n$ matrices $[\varphi]\in M_n(\mathbb F)$.


* If $\mathbb F$ is a subfield of a field
$\mathbb K$,
then $\mathbb K$ may be thought of as an $\mathbb F$-vector
space where the additive structure on the vector
space $\mathbb K$ is the field addition of $\mathbb K$,
and the scaling structure
comes from restricting the field multiplication
$\ast \colon \mathbb K\times \mathbb K\to \mathbb K$
so that the first factor comes from $\mathbb F$:
$\ast \colon \mathbb F\times \mathbb K\to \mathbb K$


* If $\mathbb F\subseteq \mathbb K$ is an extension of fields,
and $\mathbb K$ is $n$-dimensional as a vector space over
$\mathbb F$, then for $\alpha\in \mathbb K$,  scalar
multiplication by $\alpha$, say $m_{\alpha}(x)=\alpha\ast x$,
is an $\mathbb F$-linear transformation
from $\mathbb K$ to itself. Hence $[m_{\alpha}]$
(which I will just write as $[\alpha]$) is representable
as an $n\times n$ matrix over $\mathbb F$.
The set of all $n\times n$-matrices $[\alpha]$,
$\alpha\in \mathbb K$, is a subfield of the ring
$M_n(\mathbb F)$ and $\alpha\mapsto [\alpha]$
is an isomorphism of $\mathbb K$ onto this
subring of $M_n(\mathbb F)$.


*
If $\mathbb K$ is a degree-$n$ extension of $\mathbb F$,
and $f\in \mathbb F\subseteq \mathbb K$, then the matrix representation
of $f$ in $M_n(\mathbb F)$
is $[f] = f\ast I=\begin{bmatrix}
f & 0 & \cdots & 0\\
0 & f &  & 0\\
\vdots &  &  & \\
0 & 0 & \cdots & f\\
\end{bmatrix}$
where $I$ is the $n\times n$
identity matrix in $M_n(\mathbb F)$.
This fact is independent of the basis chosen for $\mathbb K$
over $\mathbb F$.


* If $\mathbb K = \mathbb F[\alpha]$ is a simple
extension of $\mathbb F$, the minimal
polynomial of $\alpha$ over $\mathbb F$ is
$x^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0$, and 
$(1,\alpha,\ldots,\alpha^{n-1})$ is chosen for an
$\mathbb F$-basis of $\mathbb K$, then the matrix
$[\alpha]$ will be the companion matrix of the minimal polynomial
of $\alpha$,
namely
$[\alpha]=\begin{bmatrix}
0 & 0 & 0 & \cdots & 0&-a_0\\
1 & 0 & 0 & \cdots & 0&-a_1\\
0 & 1 & 0 & \cdots & 0&-a_2\\
\vdots & & & &&\\
0 & 0 & 0 &  & 1&-a_{n-1}
\end{bmatrix}$.
The isomorphic copy of $\mathbb K$ in $M_n(\mathbb F)$
will be exactly the subring of
$M_n(\mathbb F)$ that is generated by the scalar matrices
$f\ast I$ and the companion matrix $[\alpha]$.

For example, suppose that
$\mathbb F=\mathbb R$ and $\mathbb K=\mathbb C=\mathbb R[i]$.
The minimal
polynomial of $i$ over $\mathbb R$ is $x^2+1$. The companion matrix
for this polynomial is $J=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$,
so the isomorphic copy of $\mathbb C$ in $M_2(\mathbb R)$
is the subring generated by the scalar matrices $a\ast I$, $a\in \mathbb R$,
and the companion matrix $J$. This is just the ring of matrices in $M_2(\mathbb R)$ of the
form $aI+bJ =
a\begin{bmatrix}
1&0\\0&1
\end{bmatrix}+
b\begin{bmatrix}
0&-1\\1&0
\end{bmatrix}=
\begin{bmatrix}
a&-b\\b&a
\end{bmatrix}$.

To get back to the thought process behind this specific
problem, we are asked to construct a field $\mathbb K$ out of
nine $4$-tuples defined over $\mathbb F=\mathbb F_3$.
Since $9=3^2$, $\mathbb K$ must be a degree $2$-extension
of $\mathbb F_3$ and hence is realizable as a subring
of $M_2(\mathbb F_3)$. Since $-1$ is not a square in $\mathbb F_3$,
the polynomial $x^2+1$ is irreducible over $\mathbb F_3$,
so $\mathbb K=\mathbb F_3[i]$ where $i$ is a symbol
for a root of $x^2+1$ over $\mathbb F_3$.
The construction proceeds as in the construction of
$\mathbb C$ to get that the $\mathbb K$ may be taken
to be the subring of $M_2(\mathbb F_3)$ consisting
of the matrices
$\begin{bmatrix}
a&-b\\b&a
\end{bmatrix}$, $a, b\in\mathbb F_3$.
If you know the linear algebra facts from above,
you view this representation as a reasonably standard
way to view a $9$-element field.
But, for some reason, the author of your book decided
to obscure the linear algebra aspect of this problem by writing out
the $2\times 2$-matrices as $4$-tuples.
The clues (for me) that these $4$-tuples were meant to represent $2\times 2$ matrices were these: (i) I was expecting to see $2\times 2$-matrices over $\mathbb F_3$,
and (ii) the $4$-tuples included $(0,0,0,0)$,  $(1,0,0,1)$,
and $(2,0,0,2)$,
which strongly suggested that the tuples represented the matrices
$\begin{bmatrix}
0&0\\0&0
\end{bmatrix}$,
$\begin{bmatrix}
1&0\\0&1
\end{bmatrix}$,
$\begin{bmatrix}
2&0\\0&2
\end{bmatrix}$.
The only thing left to decide was whether $(a,b,c,d)$
should represent $\begin{bmatrix}
a&b\\c&d
\end{bmatrix}$ or
$\begin{bmatrix}
a&c\\b&d
\end{bmatrix}$, but that turned out not to matter since the
set of $4$-tuples is invariant under interchanging
the roles of $b$ and $c$. That is,
$(a,b,c,d)$ is in the given set of nine $4$-tuples iff $(a,c,b,d)$ is.
