Question on the definition of a ring. A ring $\langle R,+, \cdot\rangle $ is a set $R$ with two binary operations such that:


*

*$\langle R,+\rangle $ is an abelian group.

*Multiplication is associative.

*Left and right distributive laws hold.
Can someone give me an example where the left distributive law holds but the right one doesn't?
 A: Consider the set of affine maps from $\mathbb R$ to $\mathbb R$ under addition and composition. We have right-distributivity as for affine maps $ax+b,cx+d,ex+f$ we get
$$a(cx+d)+b+e(cx+d)+f=(a+e)(cx+d)+(b+f)$$
but not left-distributivity as
$$a(cx+d+ex+f)+b\neq a(cx+d)+b+a(ex+f)+b$$
when $b\neq 0$. We can make it left-distributive but not right-distributive by simply reversing the order of composition.
A: Consider the set of polynomials with integer coefficients, with addition defined as usual, but multiplication defined as composition:
$$(f\times g)(x) = f(g(x))$$
Then $(f+g)\times h = f\times h + g\times h$, but $h\times (f+g)\neq h\times f + h\times g$ in general.
For example, if $f(x)=x$ and $h(x)=x^2$ then $((f+f)\times h)(x) = 2x^2 = (f\times h)(x) + f\times h(x)$.  On the other hand, $h\times(f+f)(x) = 4x^2\neq 2x^2=h\times f(x) + h\times f(x)$.
i'm not sure if this is left- or right-distributive, but if $(R,+,\times)$ is right-distributive, you can create a left-distributive ring $(R,+,\times_{opp})$ with $a\times_{opp} b = b\times a$.
