3
$\begingroup$

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

  • 0
    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.2018-02-02

1 Answers 1

8

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.