Prove that dimension of $A$ is zero iff it is a field, where $A$ is an integral domain.

Question

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.

Answer 1

This means $(0)$ is the only prime ideal, since $A$ is a domain. As the set of non-invertible elements is the union of all prime/maximal ideals, it implies all non-zero elements are invertible.

Answer 2

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...)