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

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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 8 '12 at 4:19
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A related question: math.stackexchange.com/questions/93409/… –  Jonas Meyer Feb 8 '12 at 4:31
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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. If it's homework, that's OK, but you should add the homework tag. –  Gerry Myerson Feb 8 '12 at 4:32
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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 8 '12 at 4:52
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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 8 '12 at 5:11

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up vote 7 down vote accepted

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.

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