Do the distributive laws in a ring hold if the additive group is abelian and the multiplication is associative

Question

With the axiomatic approach to a ring, our textbook has three properties for something to be ring
$\mathcal{R}_1 : \langle R, + \rangle$ is an abelian group
$\mathcal{R}_2 :$ multiplication is associative.
$\mathcal{R}_3 :$ The left and right distributive laws hold. ($a(b+c)=ab+ac, (a+b)c=ac+bc$)
Now,
I know that $\mathcal{R}_3$ wouldn't exist if the first two automatically guaranteed it, but is there way to prove the distributive laws without computation? For example, and this is the problem I'm specifically trying to solve, is a subring in the matrix group of integers, M$_2(\mathbb{Z})$. It seems to me that if the addition is commutative and the multiplication is associative, the distributive laws hold. I know I am wrong, I just don't know why. And how do I then show that the laws hold, is it just computation?

Answer 1

the third proposition is the only one that connects addition and multiplication, so the first and second dont imply the third.

In order to come up with a simple counterexample we can consider the ring $\mathbb R$ and then twist the multiplication with a permutation.

What I mean by this is taking a bijection $f$ of the reals and definining $a\cdot b = f^{-1}(f(a)f(b))$ where the second multiplication is normal multiplication.

If you look at the multiplication by itself then it is all good because it is the same as normal multiplication but everything gets relabled, the problem is that now it wont fit with the sum.