Proof. Let $R$ be a valuation domain and $K$ its field of fractions. We shall show that the set of all non-units of $R$ is the only maximal ideal of $R$

Let $a$ and $b$ first be such elements of $R$ that $a-b$ is a unit of $R$ we may suppose that $ab \neq 0$ , since otherwise one of $a$ and $b$ is instantly stated to be a unit. Because $R$ is a valuation domain in $K$ therefore e.g. $\frac{a}{b}\in R$ Because now $\frac{a-b}{b} = 1-\frac{a}{b}$ , and $(a-b)^{-1}$ , belong to $R$ so does also the product $\frac{a-b}{b}\cdot(a-b)^{-1} = \frac{1}{b}$ i.e. $b$ is a unit of $R$ We can conclude that the difference $a-b$ must be a non-unit whenever $a$ and $b$ are non-units.

Let $a$ and $b$ then be such elements of $R$ that $ab$ is its unit, i.e. $a^{-1}b^{-1}\in R$ Now we see that $$a^{-1} = b\cdot a^{-1}b^{-1}\in R,\,\,\,b^{-1} = a\cdot a^{-1}b^{-1}\in R ,$$ and consequently $a$ and $b$ both are units. So we conclude that the product $ab$ must be a non-unit whenever $a$ is an element of $R$ and $b$ is a non-unit.

Thus the non-units form an ideal $\mathfrak{m}$ Suppose now that there is another ideal $\mathfrak{n}$ of $R$ such that $\mathfrak{m}\subset\mathfrak{n}\subseteq R$ Since $\mathfrak{m}$ contains all non-units, we can take a unit $\varepsilon$ in $\mathfrak{n}$ Thus also the product $\varepsilon^{-1}\varepsilon$ i.e. 1, belongs to $\mathfrak{n}$ or $R\subseteq\mathfrak{n}$ So we see that $\mathfrak{m}$ is a maximal ideal. On the other hand, any maximal ideal of $R$ contains no units and hence is contained in $\mathfrak{m}$ therefore $\mathfrak{m}$ is the only maximal ideal.