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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.

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  • $\begingroup$ Yes: this is how you show it is unique. $\endgroup$
    – Crostul
    Commented May 24, 2016 at 23:10
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    $\begingroup$ Your proof is correct; although it could be made a lot less verbose by just saying $ i_1 = i_1 + i_2 = i_2 + i_1 = i_2 $. $\endgroup$
    – Ege Erdil
    Commented May 24, 2016 at 23:10
  • $\begingroup$ That's a little verbose, but essentially it. $\endgroup$ Commented May 24, 2016 at 23:10

1 Answer 1

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Yes, although you can trim it down a bit; for clarity's sake.

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.

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