commutative algebra - Can you construct a field with 4 elements?

Can you construct a field with 4 elements? can you help me think of any examples?

Answer

Hint: Two of the elements have to be $0$ and $1$, and call the others $a$ and $b$. We know the mulitplicative group is cyclic of order $3$ (there is only one group to choose from), so $a*a=b, b*b=a, a*b=1$. Now you just have to fill in the addition table: how many choices are there for $1+a$? Then see what satisfies distributivity and you are there.

Added: it is probably easier to think about the choices for $1+1$