Show that is not possible to form a unity ring structure on $(\Bbb Q /\Bbb Z , + )$

Question

I'm trying to show exactly what the title says.

I tried to construct a multiplication from $\mathbb{Q}/\mathbb{Z}\times \mathbb{Q}/\mathbb{Z}\to \mathbb{Q}/\mathbb{Z}$ defined as $\hat{x} \ast\hat{y} =\widehat{xy}$ and find a contradiction. But that didn't get me anywhere.

Any help please ? Thank you !

Answer 1

The reason $\Bbb Q/\Bbb Z$ is not a ring with unit is that there cannot be a unit. Indeed, every element in $\Bbb Q/\Bbb Z$ has finite additive order, so if $\Bbb Q/\Bbb Z$ was a ring with unit, it would have finite characteristic $m$, and thus every element would have for additive order a divisor of $m$; but the orders are not bounded, because for every $n\in\Bbb N\setminus \{0\}$, $\overline{\left(\frac1n\right)}$ has order $n$.

Answer 2

In a unity ring, the additive zero $0+\Bbb Z$ is a multiplicative annihilator, and we have some $u:=\frac ab+\Bbb Z\ne 0+\Bbb Z$ that acts as multiplicative identity. From $\underbrace{\frac ab+\ldots+\frac ab}_b\equiv 0\pmod \Bbb Z$, we conclude $0=(\underbrace{u+\ldots u}_b)\cdot x=u\cdot( \underbrace{x+\ldots +x}_b)=\underbrace{x+\ldots +x}_b$ for all $x$. But that euqality certainly does not hold for $x=\frac1{b+1}+\Bbb Z$ - contradiction.

Answer 3

I tried to construct a multiplication from $\mathbb{Q}/\mathbb{Z}\times \mathbb{Q}/\mathbb{Z}\to \mathbb{Q}/\mathbb{Z}$ defined as $\hat{x} \ast\hat{y} =\widehat{xy}$ and find a contradiction. But that didn't get me anywhere.

By this, I think you mean that you tried the "obvious" operation of just multiplying coset representatives. Someone really should address why this did not work.

This is destined to fail as long as the bottom of the quotient is not an ideal. For example we have

$(\widehat{1/2})(\widehat{1/3})=\widehat{1/6}$ and

$(\widehat{3/2})(\widehat{1/3})=\widehat{1/2}$, but obviously

$\widehat{1/6}\neq \widehat{1/2}$ despite $\widehat{3/2}= \widehat{1/2}$, so multiplication isn't well-defined.

The trick observing that the ring cannot have identity (already in other solutions) is the slickest way of reasoning that it can't be made a ring with identity. Of course, without identity, you can always have the zero multiplication to make it into a rng.