maximal ideal not containing some expression

Question

Does there exists a domain $R$ with fraction field $K$, and $x \in K \setminus R$, such that for any maximal ideal $\mathfrak{m} \leq R[x]$, there exists $a \in R$ such that $x-a \in \mathfrak{m}$.

Answer 1

I try to build a counterexample. I hope you are familiar with commutative algebra, and Hilbert's Nullstellensatz.

Call $R = \mathbb{C}[T]$, $x = T^{-1}$. Then $R[x] = \mathbb{C}[T, T^{-1}] \cong \mathbb{C}[T, Y] / (TY-1)$.

Fix a maximal ideal $\mathfrak{m}$ of $\mathbb{C}[T, Y] / (TY-1)$. It has the form $$\mathfrak{m}=(T-\alpha, Y-\beta)/(TY-1)$$ where $\alpha, \beta \in \mathbb{C}$ satisfy $\alpha \beta =1$. Call $a = \alpha^{-2}T \in R$.

Then $x-a = Y - \alpha^{-2}T$. If you evaluate this polynomial at the point $(\alpha, \beta)$ you get by construction $$\beta - \alpha^{-2}\alpha = 0$$

So $x-a \in \mathfrak{m}$.

What I did: I found a domain $R$, $x \in K \setminus R$ such that for any maximal ideal $\mathfrak{m} \leq R[x]$ there exists $a \in R$ such that $x-a \in \mathfrak{m}$. So this is a counterexample.