# Prove that every commutative infinite rng $R$ has an infinite subrng $S$ s.t $S\neq R$

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.

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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
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
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
Oh, I missed that part :p sorry. – fretty Feb 25 '12 at 10:45

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 $A$.)

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Can you explain why $R$ has no infinite subfields? – the L Feb 20 '12 at 21:22
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