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Suppose $R$ is an integral domain and $R$ is algebraically closed. Prove that it then follows $R$ is a field.

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    Let $a$ be a non-zero element of your ring $R$. Then the polynomial $ax-1$ has a root in $R$...2011-04-15
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    @Mariano: Do we need to use the fact that there are no zero divisors?2011-04-15
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    If for every $a\in R\setminus 0$ the polynomial there exists a $b\in R$ such that $ab=1$, then there are no divisors of zero. (I am assuming the ring is commutative...)2011-04-15
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    @Mariano, how about turning these comments into an answer?2011-04-15

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This is a community wiki answer intended to get this question off the unanswered list.


As Mariano mentions in the comments, the solution of $ax-1$ for $a\neq 0$ furnishes an inverse for $a$ in $R$.