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Show that there is no commutative ring with the identity whose additive group is isomorphic to $\mathbb{Q}/\mathbb{Z}$.

Galois
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    Note that if we define $a b = 0$ for all $a,b \in \mathbb Q/\mathbb Z$, then we equip $\mathbb Q/\mathbb Z$ with a commutative ring structure *without* identity. Can you see how the case where you require an identity might be different? – Matt E Feb 08 '12 at 04:19
  • @Matt E I kind of see the issue that we may run into with the identity but not quite: So if we had an identity $1\in\mathbb{Q/Z}$, then we actually have $1\in\mathbb{Z}$, hence $1=0$ which is a problem for us here. Is this right? – Galois Feb 08 '12 at 04:30
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    A related question: http://math.stackexchange.com/questions/93409/does-every-abelian-group-admit-a-ring-structure – Jonas Meyer Feb 08 '12 at 04:31
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    Some of us do not like being ordered to do things, even by someone as illustrious as Galois. Would you mind rephrasing your post so it's more of a question and less of a demand? While you're at it, you could tell us how you came to be interested in the problem. – Gerry Myerson Feb 08 '12 at 04:32
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    @Galois: Elements of $\mathbb Q/\mathbb Z$ are equivalence classes of rationals, so no element of $\mathbb Q/\mathbb Z$ is in $\mathbb Z$. The "$1$" in a ring need not be like the $1\in\mathbb Z$. – Jonas Meyer Feb 08 '12 at 04:34
  • @Galois, your last comment does not make much sense. You should probably think a bit more about what you wrote and what Matt suggested! – Mariano Suárez-Álvarez Feb 08 '12 at 04:35
  • Oh I see, now it makes sense. So the issue comes up when we are defining multiplication. If we define a*b=0 then we get that a*a^{-1}=1=0 which is a problem for us. – Galois Feb 08 '12 at 04:49
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    @Galois: If you define $ab=0$ then there is no identity (and no "$a^{-1}$"), but your problem is to show that *there is no way* to define multiplication that turns $\mathbb Q/\mathbb Z$ into a commutative ring with identity (with addition being the addition of the group $\mathbb Q/\mathbb Z$). So this one example that doesn't work is not enough. – Jonas Meyer Feb 08 '12 at 04:52
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    Dear Galois, My point was just to give an example of a ring structure which *does* exist, but which *doesn't* satisfy your requirements, just to give you something to build your thoughts on. It wasn't supposed to be a precise guide to the answer. Given your response to the example, I would recommend that you brush up on basic concepts, such as what exactly is meant by a "ring with identity". Regards, – Matt E Feb 08 '12 at 05:11
  • May I know why this question is closed? It seems to me that it is not "off topic" at all and the context is clear, at least to those who know the basics of abstract algebra. – Zuriel Feb 02 '18 at 04:06

1 Answers1

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Here's how I would organize this: suppose I have a commutative ring $A$ and an isomorphism $f$ of abelian groups from $(A, +)$ (the additive group of $A$) to $\mathbf Q/\mathbf Z$. Assuming that $A$ has a unit element $1 \in A$, we can look at $f(1) \in \mathbf Q/\mathbf Z$. This element has finite order (why?), so there exists some natural number $n$ such that $n \cdot 1 = 0$. [To avoid confusion: this just means "add $1 \in A$ to itself $n$ times".]

Now, does this imply something about every element of $A$? I'll leave a slight gap here for now, but I think the other important thing to notice (and maybe you noticed it at the "why?") is that $\mathbf Q/\mathbf Z$ contains elements of every finite order.

Dylan Moreland
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