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- Feb 29, 2012

- 342

$$\text{Let } R \text{ be a ring with identity. Let }a \in R \text{ and suppose that exists an unique } a' \in R \text{ such that }a a' =1. \text{ Prove that } a'a=1.$$

My solution:

Since we have an identity, it has an inverse (itself), which means we can do

$$(a a')^{-1} = 1^{-1} = 1,$$

but $(a a')^{-1} = (a')^{-1} a^{-1} = 1$. From this, we can multiply once through the right by $a$ getting $(a')^{-1} a^{-1} a = 1 a = a$ and from that $(a')^{-1} = a$. Finally, multiplying through the left by $a'$ we get $a' (a')^{-1} = 1 = a' a$.

Am I correct? I am particularly uneasy about stating that $(aa')^{-1} = (a')^{-1} a^{-1}$.