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A ring $\langle R,+, \cdot\rangle $ is a set $R$ with two binary operations such that:

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

  2. Multiplication is associative.

  3. Left and right distributive laws hold.

Can someone give me an example where the left distributive law holds but the right one doesn't?

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Not related to the problem at hand, but: Your definition doesn't require a multiplicative identity? – Thomas Andrews Mar 30 '13 at 4:32
@ThomasAndrews Some authors do this. My first algebra book did not require identities. – Alex Becker Mar 30 '13 at 4:32
yes, one can always embed a ring without unit into one that does have one. – rotten Mar 30 '13 at 4:39
@rotten The main issue comes with defining ring homomorphisms (most authors require they send $1$ to $1$, for pretty good reasons). – Alex Becker Mar 30 '13 at 4:44

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.

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@ThomasAndrews Thanks, good point. – Alex Becker Mar 30 '13 at 4:47
Certainly a cleaner example then mine, and it fits a pattern - function composition is a problem for distributive laws on one side :) – Thomas Andrews Mar 30 '13 at 4:50
@AlexBecker why do you have two $b$'s on the RHS of the last equation? – ahorn Oct 6 '15 at 12:14

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 $a\times_{opp} b = b\times a$.

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