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.
Best Answer
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.