2
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ $K$ (equivalently: $R$) should be of characteristic $0$ for $\varphi$ to exist. $\endgroup$
    – user158047
    Sep 3, 2019 at 11:36
  • $\begingroup$ 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 $\mathbf Q\otimes_\mathbf Z R=K$. $\endgroup$
    – reuns
    Sep 3, 2019 at 15:10
  • $\begingroup$ 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. $\endgroup$
    – rae306
    Oct 7, 2023 at 9:26

0

This site is temporarily in read-only mode and not accepting new answers.

Browse other questions tagged .