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Someone told me that it was possible to prove that 0 is unique without using the commutative property. I don't see how, and I constructed a multiplication/addition table with elements $0_1, 0_2$, and $1$, but it is a pain to check that the table is associative/distributive. Can anyone exhibit a proof of the uniqueness of 0 in a field using only Associativity of addition and multiplication $x + 0 = x$ and $x \cdot 1 = x$ existence of additive and multiplicative inverses distributive property.

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So what properties do you want to show of $0$ exactly? I understand that you want to assume that $1$ is a right-neutral element for multiplication (do you also want left?), existence of inverses (left and right, or just left/just right, and do they have to be the same?) the distributive properties is clear already though. Do you want to show that $0$ is the unique two-sided neutral element for addition? In this case you need none of the above, since if you want to show that the zero of a field is unique such that $0+x = x+0 = x$ for all $x \in K$, then if $0_1, 0_2$ are two candidates, $0_1 = 0_1 + 0_2 = 0_2$ (but I used the fact that $0_1$ is a left-neutral element and $0_2$ is a right-neutral element, which proves that the two-sided inverse is unique if it exists). For existence this is another story.

Another issue : you said you assumed existence of additive inverses, but how can you assume this if you are trying to show uniqueness of the neutral element? These inverses will be inverses to what? Say you have $x$ and its inverse is $-x$, $x+(-x)$ equals which neutral element, if you have not proven unicity yet?

I hope this leaves you with stuff to think about. Feel free to ask more questions in the comments.

Hope that helps,

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  • $\begingroup$ I would like to rule out the following scenario: There exists $n \ne 1$ neutral elements $0_1, 0_2, ... 0_n$ such that $x + 0_i = x$ for all $x$, but not necessarily that $0_i + x = x$. For all $x$, there exists $(-x)$ such that $x + (-x) = 0_i$ for some $i$ possibly dependent on $x$. $\endgroup$ – vukov Sep 10 '15 at 12:20
  • $\begingroup$ @vukov : I don't see the point in trying to prove such a thing, and the conditions you give me are kind of weak (I don't expect unicity since these $0_i$ are only right-identity elements). Are you in a particular situation where such an algebraic structure showed up and you are trying to prove something about its elements? $\endgroup$ – Patrick Da Silva Mar 9 '16 at 16:41

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