Invertible elements in a discrete valuation ring

Question

I was starting to read 'Local Fields : Jean-Pierre Serre', and in the first section, it says that "the invertible elements of $A$ (a discrete valuation ring) are those elements that do not belong to $m(A)$ (here $m(A)$ is the unique non-zero prime ideal of $A$)".

I do not understand why it is being said that all elements outside $m(A)$ are invertible. Even if we are talking about the field of fractions, the whole $ A\setminus\{0\}$ should be the field of fractions because $A$ being a discrete valuation ring is an integral domain.

$\textbf{Definition}$ : A ring $A$ is called a discrete valuation ring if it is a principal ideal domain that has a unique non-zero prime ideal $m(A)$.

Answer 1

The property of $A$ being a local ring is enough. Let $M$ be its unique maximal ideal. We show that $A^\times = A \setminus M$.

Let $a\in A$ and $(a)$ the principal ideal generated by $a$. We use the easy fact that $a\in A^\times \iff (a) = A$.

• If $a\in M$, then $(a) \subseteq M \neq A$ and hence $a\notin A^\times$.
• If $a\notin M$, then $(a) \nsubseteq M$. Now by the maximality of $M$ we have $(a) = A$ and thus $a\in A^\times$.

Answer 2

For sake of contradiction, assume there is $a\in A/m(A)$ which is not invertible. Then there exists a maximal ideal $I$ of $A$ which contains $a$. Since $A$ is local, $I=m(A)$. Thus $a\in m(A)$, which is absurd.