Suppose $R$ is an integral domain and $R$ is algebraically closed. Prove that it then follows $R$ is a field.
How to prove that every algebraically closed integral domain is a field?
3
$\begingroup$
abstract-algebra
ring-theory
integral-domain
-
10Let $a$ be a non-zero element of your ring $R$. Then the polynomial $ax-1$ has a root in $R$... – 2011-04-15
-
0@Mariano: Do we need to use the fact that there are no zero divisors? – 2011-04-15
-
0If 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
-
2@Mariano, how about turning these comments into an answer? – 2011-04-15
1 Answers
2
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 $R$.