Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $R$ is a finite ring (commutative ring with $1$) of characteristic $3$ and suppose that for every unit $u \in R\:,\ 1+u\ $ is also a unit or $0$. We need to show that $R$ is a field. Is this true if ${\rm char}(R) > 3$?

Here is what I attempted to do. $\:$ First of all, $\:$ I noticed that the statement is not true if $\ R\ $ is infinite ($ \mathbb F_3[x]$ is an example of an infinite ring which is not a field but it satisfies all the required properties). Now, in a finite ring, a non-zero element is either a unit or a $\:0\:$ divisor, so I tried to show that $R$ has no $\:0\:$ divisors. Clearly, $R$ has no nonzero nilpotent elements (if $x$ is nilpotent, then $1+x$ is a unit, but then $1+(1+x)$ and $1+(2+x)$ is either a unit or $\:0\:.\:$ Hence $x$ is either a unit or $\:0\:,\:$ and since $x$ is nilpotent, it can't be a unit, so we must have $x = 0$). But this does not solve the problem, since $R$ could have elements that are $\:0\:$ divisors but not nilpotent (for example, $\ (1,0)\ $ is a $\:0\:$ divisor in $\ \mathbb Z/3\:\mathbb Z \times \mathbb Z/3\:\mathbb Z\ $ but it is not nilpotent).

Another observation I made is that the set of units, together with $\:0\:$ forms a group under addition, so that $J = R^{*}$, together with $\:0\:$ is a subring of $R$. hence we may view $R$ as a $J$-module (and since $J$ is clearly a field, $R$ is a $J$-vector space).

Another thing I tried is to show that $R$ has no proper nontrivial ideals. Viewing $R$ and $J$ as abelian groups, I noticed that a non-trivial ideal of $R$ can contain at most one element from each coset of $J$ in $R$, because if an ideal contains two distinct elements from the same coset of $J$ in $R$, this ideal would have to contain their difference, hence it'd have to contain a unit, hence it would not be a proper ideal. But again, I don't see how this observation leads to a solution.

As for the last part, I suspect that this statement will remain true if ${\rm char}(R) > 3$. Since $1$ is a unit, it follows that $1,2,3,\ldots$ are all either units or $0$, which can only happen if ${\rm char}(R) = p$, a prime number (and then I suspect that $R$ will have to be a finite field), but again I do not see how to prove (or disprove) this.

By the way, this is not a homework problem. I am studying algebra on my own, and after thinking about it for a few days and making the observations I listed above, I still don't see how to finish the proof. I would appreciate your suggestions. Thank you in advance.

share|improve this question
    
Sorry, after many false starts, I think I have the proof. –  Thomas Andrews Apr 2 '11 at 23:12
2  
You could fix the title so that it becomes a true statement! –  Mariano Suárez-Alvarez Apr 3 '11 at 7:33
    
Dear Mariano, I fixed the title, so now it is a true statement. Thank you! –  algebra_fan Apr 3 '11 at 9:25
    
See also this question. –  Bill Dubuque Sep 5 at 16:45

3 Answers 3

up vote 12 down vote accepted

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 $x^2=x$.]

share|improve this answer
    
@Thomas: If you want $\equiv$ rather than $\cong$, the LaTeX code is \equiv. –  Arturo Magidin Apr 3 '11 at 3:03
    
C'est ingénieux! Ad hoc, where can one learn the inequality mentioned there? –  awllower Apr 3 '11 at 4:43
    
@awllower - Are you referring to ${{p-1} \choose i} \equiv (-!)^i$? –  Thomas Andrews Apr 3 '11 at 5:33
    
Dear Thomas, Thanks a lot for your solution! I really appreciate your help! –  algebra_fan Apr 3 '11 at 9:14
1  
@awllower - This fact can be seen either by writing ${{p-1}\choose{i}}=\frac{(p-1)!}{i!(p-1-i)!}$ and doing the arithmetic, or using that ${p \choose i}=0 \pmod p$ when $1\leq i \leq p-1$ and then using ${n \choose k}={{n-1}\choose{k-1}}+{{n-1}\choose{k}}$ –  Thomas Andrews Apr 3 '11 at 20:31

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

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

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

share|improve this answer
    
Ah, yes, that's a nice proof. Both this proof and my proof first find an idempotent non-unit $e$, but $(2e-1)^2=1$ is a nice simplification, compared to my $(e+1)^{p-1}=1$, and exposes much more clearly why the case $p=2$ is an exception. –  Thomas Andrews Apr 9 '11 at 15:36

HINT $\: $ Being finite, $\rm\:R\:$ is Artinian. $\rm\:R\:$ has trivial Jacobson radical since if $\rm\:j\:$ is in every max ideal then $\rm\: j-1\:$ is in no max ideal so it's a unit. So, by hypothesis, $\rm\: 1+(j-1)\ =\ j\:$ is a unit or $0$, thus $0$ (else unit $\rm\:j\in$ max ideal). So, by the structure theorem for Artinian rings (here essentially $\rm CRT$), $\rm\: R = R/J =$ product of $\rm\:n\:$ fields. $\rm\:n = 1\:,\:$ else e.g. $\rm\: (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 $\rm\:2\ne 0\:$ (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. $\rm (\mathbb Z/2)^{\:n}\:$ for $\rm\:n>1\:$ is not a field but, having $1$ as its only unit, satisfies $\rm\: 1+$ 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.

share|improve this answer
    
Dear Bill, Thanks a lot for your hints. I really appreciate your help. –  algebra_fan Apr 3 '11 at 9:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.