# When does the extension of scalars to $\mathbf Q$ give the field of fractions?

Let $$R$$ be an integral domain and let $$K$$ be its field of fractions. Consider the ring $$T=\mathbf Q\otimes_\mathbf Z R$$. We have a homomorphism $$\varphi\colon T\longrightarrow K$$ $$a\otimes r\longmapsto ar\qquad a\in \mathbf Q,r\in R.$$ Under what conditions on $$R$$ is $$\varphi$$ (a) injective, (b) surjective? When is $$T$$ a field? When do $$T$$ and $$K$$ coincide?

• $K$ (equivalently: $R$) should be of characteristic $0$ for $\varphi$ to exist.
– user158047
Sep 3, 2019 at 11:36
• Any subring of the algebraic integers works. Next, is there a subring $R$ of $\mathbf Q(x)$ such that $\mathbf Q\otimes_\mathbf Z R=\mathbf Q(x)$ ? (for each $b \in \mathbf Q(x)$ add $nb$ to $R$ with $n$ chosen such that all the integers remain non inversible) Same question with $\mathbf Q(x_1,\ldots,x_m)$. If so for any field $K$ of characteristic $0$ (with countable transcendental basis over $\mathbf Q$ ?) there is a subring $R$ of $K$ such that $\mathbf Q\otimes_\mathbf Z R=K$. Sep 3, 2019 at 15:10
• If $K$ is an algebraic extension of $\mathbf{Q}$ and $R$ a subring of $K$ such that $K$ is the field of fractions of $R$, this is true. Oct 7, 2023 at 9:26