1
$\begingroup$

Let $R$ be a abelian group under a binary operation denoted by + and let its identity element be denoted by $0$ . If we define multiplication by $ab= 0$, for all $ a, b \in R$, then $R$ becomes a ring.

Is there any other way of defining multiplication on $R$ so that $R$ becomes a ring?

  • 1
    Have a look: http://math.stackexchange.com/questions/102761/define-two-differents-vector-space-structures-over-a-field-on-an-abelian-group?rq=1 – 2017-01-10
  • 0
    @Aweygan A ring need not to be a vector space. – 2017-01-10
  • 0
    @Awaygan & Ergul - yes! I made a mistake. I actually wanted to ask a question about rings, which I have posted now. – 2017-01-10
  • 1
    Check this answer: http://math.stackexchange.com/a/93412/133781 – 2017-01-10

1 Answers 1

3

For some choices of $R$ there are no other multiplications on $R$, even if you require that they satisfy no axioms besides distributing over addition. For instance, consider $R=\mathbb{Q}/\mathbb{Z}$ and suppose you have a product on $R$ that distributes over addition. Let $x,y\in\mathbb{Q}/\mathbb{Z}$ be arbitrary; say $x$ is represented by a rational number $\frac{a}{b}$ and $y$ is represented by a rational number $\frac{c}{d}$ (with $a,b,c,d\in\mathbb{Z}$). Let $x'$ be the element of $\mathbb{Q}/\mathbb{Z}$ represented by $\frac{a}{bd}$. Then $dx'=x$ (where $dx'$ just means a sum of $d$ copies of $x'$), so since our product distributes over addition we have $$xy=(dx')y=d(x'y)=x'(dy).$$ But $dy$ is represented by the rational number $d\frac{b}{d}=b$ which is an integer, so $dy=0$. Thus $xy=x'\cdot 0=0$.

(More generally, a similar argument applies to any abelian group $R$ which is divisible but in which every element has finite order.)

  • 0
    Thank you. The elements $x, y$ of the quotient group $R$ you have mentioned are cosets; is it ? – 2017-01-10
  • 0
    Yes, that's correct. – 2017-01-10