Prove that dimension of $A$ is zero iff it is a field, where $A$ is an integral domain. 
Let $A$ be an integral domain. Show that $\dim(A)=0 \iff A$ is a field.

The backward implication is trivial.
For the forward implication, if we can show that $1 \in <a>$, where $a(\neq 0) \in A$. Then, we are done. However, I don't know how to show it. 
Any suggestions are appreciated.

Since maximal ideals are prime, "zero-dimensional" becomes the same thing as "prime ideals are maximal ideals."
In a domain $\{0\}$ is prime, and in a zero-dimensional domain it is also maximal.
So, there are no other ideals besides $\{0\}$ and $A$. Do you know what this implies about $A$? (There are already more than one question on the site answering this, if you formulate the correct question...)
