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Multiplication is just repeated addition, so we can derive multiplication from addition.

Addition over integers is a monoid. Addition and multiplication over integers form a Ring, so there's one example where this is possible, but addition and multiplication have more properties, for example addition is commutative. I'm not sure if it's possible to derive the identity element of multiplication from the Monoid (in the general ring-sense of multiplication and addition).

Is it possible to derive a Ring from a Monoid in general?

Filip Haglund
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    Multiplication is only repeated addition if you are multiplying by an integer. How would you explain $x*x$ in $\mathbb R[x]$ as repeated addition? You should read this post and maybe this link that appears there. – rschwieb Aug 25 '17 at 13:04

2 Answers2

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It depends on what you want to do.

For any monoid $M$ there is alway the monoid ring $\Bbb Z[M]$. This is the left adjoint to the forgetful functor from rings to monoids.

If you want to take a commutative monoid as ground for the underlying group of the ring you probably want to take the of its Grothendieck group before. But this could be any abelian group. And if the ring is required to have a multiplicative identity element this can not always be achieved

user60589
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Good luck explaining $\sqrt{2}\cdot \sqrt{3}$ with addition alone...

So no, you can't just derive the multiplication from the addition. To give you a smaller example, consider the following two rings:

  1. $R = \mathbb{F}_2[x]/\langle x^2\rangle = \{0,1,x,x+1\}$
  2. $S = \mathbb{F}_2[x]/\langle x^2+x+1\rangle = \{0,1,x,x+1\} \cong \mathbb{F}_4$

In both cases, you have the exact same addition, so you can't tell the difference between these two rings while only looking at addition. But they are of course different, as the second one is a field and the first one is not.

Dirk
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