2
$\begingroup$

Let $R$ a set with $+$ and $\cdot $ s.t.

1) $(R,+)$ is a group (not necessarily comutative)

2) $\cdot $ is associative, and distributive for $+$, i.e. $a\cdot (b\cdot c)=(a\cdot b)\cdot c$, $a\cdot (b+c)=a\cdot b+a\cdot c$ and $(a+b)\cdot c=a\cdot c+b\cdot c$

3) there is $1\in R$ s.t. $1\cdot x=x\cdot 1=x$.

Show that $R$ is a ring.

I really don't know how to did it. I think that I have to show that $(R,+)$ is commutative, but I didn't success.

1 Answers 1

6

Hint:

Compute $$ (1 + 1) \cdot (a + b) $$ in two ways, using the two distributive laws.

  • 0
    we get $a+a+b+b=(a+b)+(a+b)$ but I don't get the trick... – 2017-02-28
  • 0
    Now exploit the fact that $(G, +)$ is a group, so that you can simplify. In other words, add $-a$ ***on the left*** and $-b$ ***on the right***. – 2017-02-28
  • 0
    Very nice. thanks – 2017-02-28