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


*

*For each positive integer $n$, $A$ contains a unique subfield $F_n$ with $p^n$ elements.

*$A$ is the union of the $F_n$.

*$F_m \subseteq F_n$ iff $m$ divides $n$.

*The $F_n$ are the only finite subfields of $A$.

*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 $A$.)
