In the definition of a ring $R$, one has

  1. $a(b+c) = ab + ac$ and
  2. $(a+b)c = ac + bc$

for all $a,b,c\in R$

My question is (just out of curiosity) if one really needs both of these. I can't think of an example of something that is not a ring that only satisfies one of the sides of the distributive law. So can one prove that if $a(b+c) = ab + ac$ for all $a,b,c$, then $(a+b)c = ac + bc$ for all $a,b,c$.

Edit: I maybe should add that all rings in my definition have a unity $1$.

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    $\begingroup$ If the ring is a right ideal or a left ideal, it will only satisfy one of these given properties $\endgroup$ Nov 20, 2017 at 13:49
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    $\begingroup$ There is a related thing you might be interested in: near-rings which are only required to be distributive on one side. But also the definition allows the $+$ operation to be nonAbelian, so it's not what you asked for. $\endgroup$
    – rschwieb
    Nov 20, 2017 at 13:51
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    $\begingroup$ Related math.stackexchange.com/q/513223/29335 : and also, funnily enough, asked by a similarly nondescript username. Not really a duplicate though. $\endgroup$
    – rschwieb
    Nov 20, 2017 at 13:54
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    $\begingroup$ @rschwieb: Sorry, I really did try searching before asking. $\endgroup$
    – John Doe
    Nov 20, 2017 at 13:55
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    $\begingroup$ @JohnDoe It's true, the question does ask for a one-or two sided example, but I think it only got a "natural" example that isn't distributive on either side. That's why I only think it's related, not duplicate. So don't sweat it :) $\endgroup$
    – rschwieb
    Nov 20, 2017 at 13:56

1 Answer 1


Here is an example that fails precisely in left distributivity.

Consider $\mathbb{R}[X]$ - the polynomials with coefficients from $\mathbb{R}$ with the usual operation of pointwise addition (in fact, the ring of scalars is irrelevant here).

The tricky part is how we define multiplication: let $p \cdot q$ be the composition $p \circ q$. This multiplication is associative, and even has an identity, which is the identity polynomial $p(x)=x$.

Now, trivially $$(p_1 + p_2) \circ q = p_1 \circ q + p_2 \circ q,$$ but in general $$p \circ (q_1 + q_2) \color{red} \neq p \circ q_1 + p \circ q_2.$$

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    $\begingroup$ .. or condensed even further: Start from any abelian group $A$ and consider all maps $A\to A$ with pointwise addition (and with composition as multiplication). For all but the simplest few choices of $A$, this will be an example. $\endgroup$ Nov 20, 2017 at 18:00
  • $\begingroup$ If one is looking for a more "natural" ring where one doesn't have to define an "unusual multiplication", the ring of $n\times n$ matrices would do just fine (which is of course exactly your answer, as a matrix is nothing else than a linear mapping and the product of two matrices is the composition of the corresponding mappings). $\endgroup$
    – Hirshy
    Nov 20, 2017 at 21:45
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    $\begingroup$ @Hirshy But the usual matrix ring is distributive. $\endgroup$
    – lisyarus
    Nov 20, 2017 at 22:07
  • $\begingroup$ @lisyarus: It's distributive precisely because matrices are linear and linearity is essentially just right-distributivity. To get a counterexample of this form you need the mappings in you ring to be nonlinear. $\endgroup$ Nov 21, 2017 at 0:40
  • $\begingroup$ @R.. Of course, but this nonlinear mappings won't be matrices. $\endgroup$
    – lisyarus
    Nov 22, 2017 at 14:15

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