A finite ring is a field if its units $\cup\ \{0\}$ comprise a field of characteristic $\ne 2$

Solution 1:

The map $f: x\rightarrow {x^3}$ is a homomorphism from $R$ to $R$. The kernel of $f$ is $\{0\}$ because $x^3=0$ implies $(x-1)^3=-1$, which means that $x-1$ is a unit, hence $x$ is a unit or zero, hence that $x=0$. So $f$ is 1-1, and, since $R$ is finite, it is thus an automorphism of $R$.

Then there must be an $n>0$ such that $f^n$ is the identity, that is: $x^{3^n}=x$ for all $x\in{R}$

Assume $r$ is not a unit, and let $s=r^{3^n-1}+1$. A quick computation shows that $s^2=1$, so $s$ must be a unit. But that means that $s+2$ must be a unit or zero, and $s+2$ is a power of $r$, a non-unit, so $r^{3^n-1}=0$ and $r=r^{3^n} = 0$.

The same argument works for arbitrary prime $p$. First you prove that $x \rightarrow x^p$ is an ring automorphism, find n such that $r^{p^n}=r$, and then using $s=r^{p^n-1} +1$, prove that $s^{p-1}=1$. A little harder to prove, requiring to know that

$${{p-1}\choose{i}} \equiv (-1)^i \pmod{p}$$

[That doesn't work when $p=2$, of course, and the theorem is not true for p=2. In that case, you can take the ring $\{0,1,x,x+1\}$ with the rule $x^2=x$.]

Solution 2:

Hint $\:R\:$ is finite so Artinian. $\,R\,$ has trivial Jacobson radical: if $\,j\,$ is in every max ideal then $\, j\!-\!1\,$ is in no max ideal so it's a unit. So, by hypothesis, $\, 1+(j\!-\!1)\ =\ j\,$ is a unit or $0$, thus $0\,$ (else unit $\:j\in$ max ideal). By the structure theorem for Artinian rings (here essentially CRT), $\: R = R/J =$ product of $\:n\:$ fields. $\:n = 1,\:$ else e.g. $\: (1,1) + (-1,1)\, =\, (0,2)\ $ so $\: 1+$ unit $\ne$ unit or $0$.

Note that the proof doesn't invoke $\,{\rm char}\ R = 3\:$ but only $\:2\ne 0\:$ (in the final line), so it works for $\:{\rm char}\ R \ne 2,\:$ as you surmised. But if fails for $\:{\rm char}\ R = 2,\:$ e.g. $ (\mathbb Z/2)^{n}$ for $\:n>1\:$ is not a field but, having $1$ as its only unit, satisfies $\: 1+$ unit $\: =\: $ unit or $0.$

For a simpler way see the method in my ring-theoretic generalization of Euclid's proof to fewunit rings, i.e. an infinite ring has infinitely many maximal ideas if it has fewer units than elements.

If desired, you can specialize this proof to a more elementary proof for your specific case.

See also this post on structure theories for nilpotent-free finite dimensional algebras over fields.

Solution 3:

Below is a complete elementary proof of the more general result that you conjectured.

Theorem $\ $ Finite ring $ \,R\,\supset\,\mathbb Z/p\ $ is a field, if prime $ \,p > 2\,$ and unit $ \,u\in R\, \Rightarrow\, 1\!+\!u\,$ unit or $\,0$

Proof $ \ \ R\,$ satisfies $ \ x^{q} =\, x,\ \ q = p^n,\,$ since, $ $ as Thomas showed, the hypotheses imply $ \ f(x) = x^p\ $ is a permutation on the finite set $ \,R,\,$ so it has finite order $ \, f^{n}\!= 1\,.\,$ For $ \,r \in R\ $ let $ \,e = r^{q-1}.\, $ Then $ \,e^2\!-e\ =\ r^{\,q-2} (r^{q}\!-r)\ =\ 0.\, $ So $ \,(2e\!-\!1)^2\! = 4\,(e^2\!-e)+1 = 1\,$ so $\, 1+(2e\!-\!1)\, =\, 2\,e\,$ unit or $ \,0.\,$ $ \, 2^{-1}\in\mathbb Z/p\, \Rightarrow\, e $ unit or $ \,0.\,\ e$ unit $ \Rightarrow r\,$ unit; $ \ e=0\, \Rightarrow\, r = r^{q}\! = re = 0,\, $ i.e. $ \,r\in R\,$ is a unit or $\,0\,.$