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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 !

Arnaud D.
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Eduard Valentin
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  • From the addition you get that $n \ast \frac{k}n = \underbrace{\frac{k}n+\ldots+\frac{k}n}_n = k = 0$ ie $\frac{k}n$ (any element) is a zero divisor. – reuns May 02 '17 at 16:12

3 Answers3

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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 $n$.

Arnaud D.
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  • Ok so, let me se if I got it, because $\mathbb{Q}/\mathbb{Z} $ has finite characterestic ,let it be m,and other elements should have as characteristic a divizor of m(by the Lagrange's theorem I think,correct me If I'm wrong ),but we get that for every n in $\mathbb{N}$ {0} , $\frac{1}{n}$ has order depending on n , right ? – Eduard Valentin May 02 '17 at 17:30
  • It's not exactly Lagrange's theorem, because $\Bbb Q/\Bbb Z$ is infinite. But every element would be such that $m\cdot x =0$, and thus you can prove that the order of every $x$ must divide $m$ (because the remainder of the division of $m$ by the order is smaller than the order, and also has the property that $r\cdot x=0$, hence it is zero). – Arnaud D. May 02 '17 at 19:11
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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.

Hagen von Eitzen
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  • Why is $(\underbrace{u+\ldots u}_b)\cdot x=u\cdot( \underbrace{x+\ldots +x}_b)=\underbrace{x+\ldots +x}_b$ for all $x$ ? – Eduard Valentin May 02 '17 at 17:43
  • @EduardValentin The point is the same as Arnaud's : every element has finite additive order, and if there is something that acts like the identity, it must be a bound to the additive orders of everything else. But the additive orders are, in fact, unbounded. – rschwieb May 02 '17 at 17:45
  • Now I understand completely ,thank you ! – Eduard Valentin May 02 '17 at 17:49
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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.

rschwieb
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  • Why is $ (\widehat{3/2}) = (\widehat{1/2}) $ ? aren't they 2 different cosets in the quotient group since $(\widehat{3/2}) = 3/2+\mathbb{Z}$ and $(\widehat{1/2}) = 1/2+\mathbb{Z}$ – Eduard Valentin May 02 '17 at 16:40
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    @EduardValentin $3/2-1/2=1\in \Bbb Z$, so it is the same coset. – Arnaud D. May 02 '17 at 16:43
  • I feel like just proving the multiplication I provided isn't well defined is not enough to prove my point right ? I feel like that's not corect. – Eduard Valentin May 02 '17 at 17:32
  • @EduardValentin Yes, that's right, but you apparently did not run into this problem while checking this candidate, and I thought it really should be addressed. If anything it should help you appreciate ideals more. I cannot really improve on the other two solutions for finishing the problem at hand. – rschwieb May 02 '17 at 17:43