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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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    Oh, I missed that part :p sorry.2012-02-25

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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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    @George: thanks. I have strengthened my counterclaim as you suggest!2012-02-25