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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 $R$... –  Mariano Suárez-Alvarez Apr 15 '11 at 1:50
    
@Mariano: Do we need to use the fact that there are no zero divisors? –  Yuval Filmus Apr 15 '11 at 1:54
    
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...) –  Mariano Suárez-Alvarez Apr 15 '11 at 2:07
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@Mariano, how about turning these comments into an answer? –  lhf Apr 15 '11 at 11:49

1 Answer 1

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 $R$.

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