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Is it even possible? What consequences would this have if it is possible?

My attempt: Let us call this hypothetical universal identity $e$.

Fields require distributivity, right? $(a-a)a^{-1} = aa^{-1} - aa^{-1} = e-e = e$

But calculating without using distributivity $(a-a)a^{-1} = ea^{-1} = a^{-1}$

So any multiplicative inverse must be the identity. Then we can not have elements other than $e$ regardless of how we try and define addition?

Alex Wertheim
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mathreadler
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2 Answers2

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When $0$ is the additive identity for a ring, for any element $x$ in the ring, $x*0=x*(0+0)=x*0+x*0$. Subtracting $x*0$ from both sides of $x*0=x*0+x*0$ tells you that $x*0=0$.

If now $x$ is posited to be the multiplicative identity, it says that $x=x*0=0$. So such a ring is necessarily $\{0\}$.

As people have mentioned in the comments, the zero ring is excluded as a field by standard field axioms.

rschwieb
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  • And a ring is a field but without division i.e. multiplicative inverse are not required? – mathreadler Oct 10 '15 at 05:20
  • @mathreadler That is a fair description if a commutative ring, yes. Ordinarily, multiplication in general rings is not commutative. – rschwieb Oct 10 '15 at 11:32
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(1) Let $id_{+}$ = 0 = $id_{\times}$ in distributive system $\langle S, +, \times \rangle$. By distributivity, $(0)\times A = (0+0)\times A$, so $A = A+A$ for all $A$ in $S$.

(2) Also, given $A+A = A$ for all $A$ in a distributive system with $id_{+}$, $id_{\times}$, and $-id_{\times}$ exist, then $id_+ = 0 = id_{\times}$.

(3) Given $id_{+}$ = 0 = $id_{\times}$ without property $A+A = A$ for all $A$ in $S$, Implies that $\langle S, +, \times \rangle$ is a non-distributive system. Example: Let $S = \mathbb R$, with A\times B := AB + A+B$. $id_\times = 0$, $A+A = A$ only for $A=0$, But $\times$ does not distribute over real $+$.

Joy to You
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