Field of fractions and integral closure of $\mathbb{Q} + x\mathbb{R}[x]$

Question

Let $R=\mathbb{Q}+ x\mathbb{R}[x]$.

What is its field of fractions and the integral closure in its field of fractions?

Answer 1

I will give a brief sketch, perhaps you can fill in the details: The field of fractions is $\mathbb R(x)$.

The integral closure of $R$ is $K+x\mathbb R[x]$, where $K=\overline{\mathbb Q}\cap\mathbb R$ is the field of real algebraic numbers. Indeed, clearly all elements of $K+x\mathbb R[x]$ are in the integral closure. Conversely, suppose $f\in\mathbb R(x)$ is in the integral closure of $R$. In fact, since $\mathbb R[x]$ is integrally closed, we must have $f\in\mathbb R[x]$. But now $f$ satisfies some monic polynomial equation $$f^n+a_{n-1}f^{n-1}+\cdots+a_0=0$$ where $a_i\in R$. Substituting $x=0$ (i.e., applying the surjection $ev_0\colon\mathbb R[x]\to\mathbb R$) gives: $$f(0)^n+a_{n-1}(0)f(0)^{n-1}+\cdots+a_0(0)=0.$$ By definition all $a_i(0)\in\mathbb Q$, so $f(0)\in K$, as desired.