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.

Prove that every infinite commutative rng $R$ has an infinite subrng $S$ such that $R\neq S$. (Where the rng is not defined to have the identity as a member). Any help or hints of how to go about doing this would be great thanks, I thought I could use elements of infinite order in $\langle R,+\rangle$ but then I'm not sure that there is necessarily elements of infinite order in an infinite group.

Thanks for any help.

share|improve this question
1  
Why do you think this result holds? –  George Lowther Feb 19 '12 at 17:34
    
It is one of the practice questions for my course, so i suppose it could be wrong but I just assumed that is was not, do you think it is? –  hmmmm Feb 19 '12 at 18:10
1  
That argument doesn't quite work, even for abelian groups. Prufer p-groups (en.m.wikipedia.org/w/…) are infinite abelian groups with no infinite proper subgroups. However, fixing up your argument it does show that the Prufer p-groups are the only examples –  George Lowther Feb 24 '12 at 14:51
1  
It all depends on your definition of subring. I might be wrong but doesn't $\mathbb{Z}$ contain no such subring? –  fretty Feb 25 '12 at 10:22
1  
Oh, I missed that part :p sorry. –  fretty Feb 25 '12 at 10:45

1 Answer 1

up vote 5 down vote accepted

The claim is false. Take a prime $p$ and an algebraic closure $A$ of the field with $p$ elements. Then:

  1. For each positive integer $n$, $A$ contains a unique subfield $F_n$ with $p^n$ elements.
  2. $A$ is the union of the $F_n$.
  3. $F_m \subseteq F_n$ iff $m$ divides $n$.
  4. The $F_n$ are the only finite subfields of $A$.
  5. Any nontrivial subrng $S$ of $A$ is a field (if $0 \not= x \in S \subseteq A$, then, $x \in F_n$ for some $n$, so that $1 = x^{p^n-1} \in S$, since the multiplicative group of the finite field $F_n$ is cyclic).

Now let $R_i = F_{2^i}$ and $R= \bigcup_i R_i$. Then $R$ is infinite, and, by the above, the only subfields and hence the only subrngs of $R$ are $R$ itself and the finite subrngs $R_i$.

(hmmmm also asked for hints about how to go about the problem. The above example comes from trying to prove the claim, in the presumably easier case when $R$ is actually a ring. Any ring has at least one maximal ideal, $M$, say, and $R/M$ is then a field. If $M$ is infinite it is an infinite subrng, so we can assume it is finite. This suggests assuming $R$ actually is a field. If the field $R$ has characteristic $0$, then it has a subring isomorphic to $\mathbb{Z}$, so we can assume the characteristic is a prime $p$. Now an algebraic closure $A$ of the field with $p$ elements has a well-understood structure and it looks promising to try to disprove the claim by finding a counterexample inside $A$.)

share|improve this answer
    
Can you explain why $R$ has no infinite subfields? –  the L Feb 20 '12 at 21:22
1  
Let $T$ be a subfield of $R$ and let $T_i = T \cap R_i$. Then $T_i$ is a finite subfield of $R_i$ and hence is $R_j$ for some $j \le i$ (because every divisor of $2^i$ is a power of $2$). If $T = \bigcup_i T_i$ is infinite, every $R_j$ must be contained in some $T_i$, so $T = R$. –  Rob Arthan Feb 20 '12 at 22:24
    
Alternately, one can appeal to Galois theory. The Galois group of $R/\mathbb{F}_p$ turns out to be the (additive group of the) $2$-adic integers $\mathbb{Z}_2$ and every nontrivial subgroup of this group has finite index. –  Qiaochu Yuan Feb 20 '12 at 22:32
    
@Rob: Nice. That example looks good, so no need to say that you believe the claim is false. It is false. –  George Lowther Feb 24 '12 at 15:07
    
@George: thanks. I have strengthened my counterclaim as you suggest! –  Rob Arthan Feb 25 '12 at 10:16

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.