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Let $F$ be a field and let $K$ be an associative $F$-algebra which is finitely generated over $F$. Suppose that there exists an element $y ∈ K$ satisfying the condition that for each $v ∈ K$ there exists an element $v' ∈ K$ satisfying $v'∙y = v$. Show that $v'$ must be unique.

This is an exercise from the book: The Linear Algebra a Beginning Graduate Student Should Know by Golan.

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Consider $\alpha:K\rightarrow K$ defined by $\alpha(x)=x∙y$. Now let $v,w\in K,c\in F$

$\Rightarrow\alpha(cv+w)=(cv+w)∙y=cv∙y+w∙y=c(v∙y)+w∙y=c\alpha(v)+\alpha(w)$,

and $\exists v'\in F\backepsilon v'∙y=v\Rightarrow\alpha(v')=v'∙y=v$

So $\alpha$ is an epic endomorphism and since $F$ is finitely generated $\alpha$ is an automorphism so $\alpha$ is monic and then if $\alpha(v')=v'∙y=v=v''∙y=\alpha(v'')\Rightarrow v'=v''$ showing that $v'$ is, in fact, unique. $\blacksquare$

Note: The condition that $K$ being "associative" is useless.

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