Complete Cayley table for a field

Question

Let $M=\{0,1,a,b,c\}.$ Can I complete the Cayley tables so that $M$ is a field?

My thought was that $(M,+)$ must be an abelian group. This abelian group must be isomorphic to $(\mathbb Z_5,+).$ Correct? I managed to find that if $a=3, b=4$ and $c=2$ it works for addition.

But this does not work for the multiplication table. Does that mean that I can´t complete the Cayley table so that $M$ is a field?

Answer 1

I think you're right; this can't be a field of five elements.

If $a+c=0$ and $a\cdot a=c$, then $a=-c$, so $a\cdot a=-a$, so $a\cdot a+a=a(a+1)=0$,

so $a=0$, which doesn't make sense,

or $a=-1$, which also doesn't make sense because then $c=1$.

Answer 2

Although you can use ad-hoc arguments (as the other answer did) for specific cases, the general way to deal with this kind of question is in fact to prove that any two finite fields of the same size are isomorphic! The easiest proof is as follows. Take any finite field $F$. Then $\text{char}(F) = p$ for some prime $p$. Thus $F$ has a subfield isomorphic to$\def\ff{\mathbb{F}}$ $\ff_p$ and hence $F$ is a vector space over $\ff_p$, which implies that $\#(F) = p^k$ for some $k∈ℕ$. Since $(F^*,·)$ is a group where $F^*$ is the set of nonzero elements of $F$, we have $x^{p^k-1} = 1_F$ for every $x∈F^*$ by Lagrange's theorem. Thus every element of $F$ is a root of the polynomial $f = (x↦x^{p^k}-x)$ over $\ff_p$, and hence $f$ splits completely over $F$ since $f$ has at most $p^k$ distinct roots over any field. Therefore $F$ (being nothing more than the roots of $f$ in $F$) is a splitting field for $f$ over $\ff_p$. But splitting fields of the same polynomial over the same base field are unique up to isomorphism. Thus we are done.

Just for example, in this question, the field has size $5$ so it must be isomorphic to $\ff_5$. The first table allows us to easily find that the elements corresponding to $0,1,2,3,4$ are $0,1,c,a,b$ respectively since $\{a,c\} = \{2,3\}$ and $a$ is followed by $b$. But then clearly the second table is wrong because $3·4≠1$ in $\ff_5$. Although it may seem like more trouble here, it is easier in general since you know exactly what the finite field must look like.