Does a Finite Field with Four Elements Actually Exist? dear StackExchange community, so once again I'm confused with this problem concerning finite fields. The given field has four elements with the addition table as
$$\begin{array}{c|c|c|} 
+ & 0 & 1 & x & x + 1 \\ \hline
0 & 0 & 1 & x & x + 1 \\ \hline
1 & 1 & 0 & x + 1 & x \\ \hline
x & x & x + 1 & 0 & 1 \\ \hline
x + 1 & x + 1 & x & 1 & 0 \\ \hline
\end{array}$$
and multiplication table as
$$\begin{array}{c|c|c|} 
\times & 0 & 1 & x & x + 1 \\ \hline
0 & 0 & 0 & 0 & 0 \\ \hline
1 & 0 & 1 & x & x + 1 \\ \hline
x & 0 & x & x + 1 & 1 \\ \hline
x + 1 & 0 & x + 1 & 1 & x \\ \hline
\end{array}$$
Now the question I'm supposed to answer is "Does a finite field with four elements actually exist?". I know the answer to this question must be yes (via looking at multiple threads here) and I could argue the existence by proving every field axiom, but I don't think that is the wanted solution (proving distribution law would also take forever).
I'm just very confused about the question itself. My course is still very elementary and I don't understand the meaning of a field to exist. Maybe the question is poorly worded? I just don't know. Any help would be nice and thank you in advance.
 A: I think you are supposed to verify the axioms.  The ones for addition and multiplication are easy-you can just point to the groups that they represent.  As you say, distributivity is the hard one.  Nominally, given that addition is comutative you have $24$ cases to check, four multiplies times six sums.  Half of them are trivial because they are multiplying by $0$ or $1$.  That only leaves $12$, which isn't so many.  I would do them and declare victory.
A: A field with $4$ elements is easily built: consider the quotient field $\;\mathbf Z/2\mathbf Z[X]/(X^2+X+1)$. It is a field because the polynomial $X^2+X+1$ is irreducible over $\mathbf Z/2\mathbf Z$, as it is a quadratic polynomial with no root inAlso, as a $\mathbf Z/2\mathbf Z$-vector space, it has dimension $2$, hence its cardinality is $4$. 
The tables you provide reflect the laws on this quotient. For instance
$x(x+1)=x^2+x\equiv -1$ since $x^2+x+1\equiv0$, and $-1\equiv1$ in 
 $\mathbf Z/2\mathbf Z$. 
A: Simply observe that addition and multiplication as defined in the table is equivalent to addition and multiplication of integer polynomials in $x$ modulo $2$ and modulo $x^2+x+1$, and then observe that they commute in a certain sense.
$
\def\zz{\mathbb{Z}}
$
Namely, for any $f,g \in \zz[x]$ define:
  $f \oplus g = (f+g) \bmod 2 \bmod (x^2+x+1)$.
  $f \odot g = (f·g) \bmod 2 \bmod (x^2+x+1)$.
Then prove that for any $f,g,m \in \zz[x]$ we have:
  $f \bmod m \bmod m = f \bmod m$.
  $(f+g) \bmod m = ( f \bmod m + g \bmod m ) \bmod m$.
  $(f·g) \bmod m = ( ( f \bmod m ) · ( g \bmod m ) ) \bmod m$.
Together these can be easily used to show (for you to do!) that associativity and commutativity and distributivity of $\oplus,\odot$ on the set of polynomials $F = \{0,1,x,x+1\}$ is equivalent to associativity and commutativity and distributivity of $+,·$ on integer polynomials, which is very easy to understand intuitively and prove.
A: If you're asked whether something exists, and you want to assert that it does, an effective way is to probably just construct it!
What I mean by this is that you have been asked whether a certain thing exists (a field with $4$ elements), and you want to assert that it does. What you can do is say yes, and prove that what you are asserting is a field with $4$ elements actually is one. 
The meaning of "does a field exist?", is exactly what it says. They want you to prove it by giving them an example of one, or disprove it by providing a proof to the contrary. 
