Let $k$ be a field and $K=k(x)=\{f/g\mid f, g\in k[x], g\ne 0\}$ the field of rational functions. Let $\mathcal{O}$ be the ring $$\mathcal{O}=\{\frac{f}{g}\in K\mid \deg f\ge\deg g \}\cup\{0\}$$ Then $\mathcal{O}$ is a valuation ring since for any $f/g$ either $\deg f\ge \deg g$ or $\deg g\ge\deg f$. Thus the natural homomorphism $\nu$ defined by \begin{align*} &\nu: K^\times \to\mathbb{Z}\cong\frac{K^\times}{\mathcal{O}^\times}\\ &\nu: \frac{f}{g} \mapsto \deg f-\deg g\end{align*} is a discrete valuation. Let $P=\{a\in\mathcal{O}\mid \nu(a)\gt 0\}$ the non-units of $\mathcal{O}$, I expect this to be the unique maximal ideal of $\mathcal{O}$. However since $x, x+1\in P$ this shows that $1\in P$ if it's an ideal, while $\nu(1)=0$ hence $1\notin P$. What was wrong in the above argument?
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Update: I understand the problem: $\mathcal{O}$ is even not a ring since it is not closed under addtion: $$\frac{1-x}{x}+1=\frac{1}{x}\notin \mathcal{O}$$