Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
@Mariano, how about turning these comments into an answer? – lhf Apr 15 '11 at 11:49

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 $R$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.