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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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    Why do you think this result holds?2012-02-19
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    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?2012-02-19
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    Ok so if R has an element of infinte order, g, then we can generate an infinite subgroup of R from $$. If R does not have an element of infinite order then we can write R as $$ and $A_p=\{a\inA|o(g)\mbox{is a power of p}\}$ so could we then make the group by taking the direct sum of every second one of these subgroups and so we would still have an infinite abelian group that would be a subgroup of R?- this is only for a group not a ring, is this in the right direction is is this just wrong?2012-02-20
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    That argument doesn't quite work, even for abelian groups. Prufer p-groups (http://en.m.wikipedia.org/w/index.php?title=Pr%C3%BCfer_group&oldid=470427135) 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 examples2012-02-24
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    It all depends on your definition of subring. I might be wrong but doesn't $\mathbb{Z}$ contain no such subring?2012-02-25
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    @fretty Does $\mathbb{Z}$ not contain infinitely many such subrngs of the form $$ for $n\in\mathbb{Z}$?2012-02-25
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    @hmmmm Some people demand that a subring contains $1$, in which case those structures are all ideals but not subrings (well for $n\neq \pm 1$).2012-02-25
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    @fretty Yeah that's why in my question I said that we did not need the identity, also i think the notation rng means ring without identity. However my statement is totally wrong regardless as shown below!2012-02-25
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    Oh, I missed that part :p sorry.2012-02-25

1 Answers 1

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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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    Can you explain why $R$ has no infinite subfields?2012-02-20
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    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 $T = R$.2012-02-20
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    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.2012-02-20
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    @Rob: Nice. That example looks good, so no need to say that you *believe* the claim is false. It is false.2012-02-24
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    @George: thanks. I have strengthened my counterclaim as you suggest!2012-02-25