First, suppose $i_1$ and $i_2$ are additive identity in ring R. From the definition of "additive identity" $a+i_1=a$ such that there is $a$ $\in$ $R$, including for $a=i_2$, so $i_2+i_1=i_2$. But $i_2$ is also an additive identity, so there $a \in R$ making $a+i_2=a$. Then $a=i_1$. Then $i_1+i_2=i_1$. Since this is a ring, addition is commutative, so $i_1+i_2=i_2+i_1$. Thus, $i_2+i_1=i_1$. Since $i_2+i_1=i_2$ and $i_2+i_1=i_1$, $i_2+i_1$ is the same element of $i_2$ and is the same as $i_1$, making $i_2=i_1$. Is this how you show it's unique? If not, can someone show me how to prove it's unique.
$\endgroup$ 31 Answer
$\begingroup$Yes, although you can trim it down a bit; for clarity's sake.
$\endgroup$If $i_n$ is an additive identity, then for any $a$ in the ring: $~a+i_n=a~$ and $i_n+a=a$, by definition and the commutative property of addition in abelian groups.
If any $i_1, i_2$ are both additive identities, then we have: $i_1+i_2=i_1$ and $i_1+i_2=i_2$.
Thus for any additive identities $i_1,i_2$, then $i_1=i_2$.
That is: The additive identity for a ring is unique.
$\Box$
