Relation between integral domains with isomorphic fraction fields

Question

If $A$ and $B$ are integral domains and they have isomorphic fraction fields, is there any relation between $A$ and $B$?

The question arises from the particular case in which $A$ is a finite type $B$-algebra. For this case I worked the following out, but I am not very comfortable with my argument, specially when switching between ring and algebra morphisms (I don't have much background on commutative algebra).

Let $A$ and $B$ be integral domains such that $A$ is a (non-zero) finite type $B$-algebra and $A_{(0)} \cong_{Ring} B_{(0)}$. Then $A_{(0)} \cong_{B\text{-algebra}} B_{(0)}$.

The $B$-algebra morphism \begin{equation} B \to B[X_{1},...,X_{n}] \twoheadrightarrow A=\frac{B[X_{1},...,X_{n}]}{I} \hookrightarrow A_{(0)}=B_{(0)} \end{equation} sends \begin{equation} 1 \mapsto 1 \mapsto 1+I \mapsto 1 \end{equation} and therefore, since it is a $B$-algebra morphism (this is the part that I couldn't quite work out without leaving the category of rings) \begin{equation} b \mapsto b \mapsto b+I \mapsto b \end{equation} From injectivity of the last arrow we deduce

1. For all $b\in B\setminus \{ 0\}$ we have $b\notin I$, as $b\neq 0$ in $B_{(0)}$.
2. For all $i$, if $X_{i}\notin I$ then $X_{i} + I \mapsto \frac{p_{i}}{q_{i}} \neq 0$ in $B_{(0)}$. Thus $q_{i}X_{i}-p_{i}\in I$.

Therefore $A=B[\frac{p_{1}}{q_{1}},...,\frac{p_{n}}{q_{n}}]=B_q$ where $q$ is the product of the $q_{i}$.

So $A \cong_{B\text{-algebra}} B_{q}$ and in particular $A \cong_{Ring} B_{q}$.

Is this proof correct? Can I change between categories like that?

I've never seen anyone doing it, so that suggests that I can't. And further questions:

Is there a proof of this using only the category of rings?

I feel like there should be a nice easy diagram to prove it using just the epimorphisms involved here, but I could't find it.

Back to the more general question, can we weaken the assumptions somehow? What is the most general statement we can get in this situation?

I've tried Atiyah-Macdonald, Matsumura and Google. The closest thing I could find was this question on StackExchange, but it doesn't really help my case.

Answer 1

Let $A$ and $B$ be integral domains. Saying that $B$ is a $A$-algebra is equivalent to giving a ring morphism $\varphi \colon A \to B$. I assume that you consider the $A$-algebra structures on the fraction fields Frac$(A)$ and Frac$(B)$ induced by $A \hookrightarrow \text{Frac} (A)$ and $A \stackrel{\varphi}{\to} B \hookrightarrow \text{Frac} (B)$.

An morphism $\theta \colon \text{Frac} (A) \to \text{Frac} (B)$ of $A$-algebras with respect to the $A$-algebra structure described above is an ring morphism such that the following diagram commutes:

Now I claim the following:

An injective $A$-algebra morphism $\theta \colon \text{Frac} (A) \to \text{Frac} (B)$ exists if and only if $\varphi$ is injective. Moreover, this morphism $\theta$ is unique, if it exists.

I will first give a proof of the claim and then try to explain where your proof fails.

Proof: ``$\Rightarrow$'': Suppose there is an monomorphism $\theta \colon \text{Frac} (A) \to \text{Frac} (B)$ of $A$-algebras, i. e. the diagram $(\ast)$ commutes. Now $A \hookrightarrow \text{Frac} (A) \stackrel{\theta}{\to} \text{Frac} (B)$ is injective, so by commutativity, the map $A \stackrel{\varphi}{\to} B \hookrightarrow \text{Frac} (B)$ is also injective. Hence, $\varphi$ is injective.

``$\Leftarrow$'': Conversely, suppose that $\varphi$ is injective. In this case $A \stackrel{\varphi}{\to} B \hookrightarrow \text{Frac} (B)$ is injective, so by the universal property of the fraction field there is a unique $\theta \colon \text{Frac} (A) \to \text{Frac} (B)$ such that the following diagram commutes:

Since $\theta$ is a morphism of fields, it is automatically injective.

Unfortunately, the claim can not be extended to surjectivity in a naive way:

• Take $\varphi$ to be the inclusion $\mathbb{Z}[\sqrt{-3}] \hookrightarrow \mathbb{Z} [\frac{1 + \sqrt{-3}}{2}]$, then the identity $\text{id}$ is an $\mathbb{Z}[\sqrt{-3}]$-algebra isomorphism of the (common) fraction field $\mathbb{Q}(\sqrt{3})$ with respect to $\varphi$. However, $\varphi$ is not surjective.

• Take $\varphi$ to be the inclusion $k [T^2] \hookrightarrow k [T]$, where $k$ is a field. The fraction fields are isomorphic as rings via $$k[T] \to k [T^2], \quad T \mapsto T^2. $$ But since the inclusion $k (T^2) \hookrightarrow k(T)$ is the unique $k[T^2]$-algebra morphism with respect to $\varphi$, there cannot be an $k[T^2]$-algebra isomorphism $k(T^2) \to k(T)$.

The second example above is also a counterexample for your claim. I think the problem in your proof is the step $$ B [\tfrac{p_1}{q_1}, ..., \tfrac{p_n}{q_n}] = B_q, \quad \text{ using } q = \prod_{i = 1}^n q_i, $$ paralleling the situation of subrings of $\mathbb{Q}$ (see here). The proof there uses the fact that $\mathbb{Z}$ is a unique factorisation domain and fails, if you work over a ring without this property:

Let $A = \mathbb{Z}[\sqrt{-7}]$ and $\mathcal{O} = \mathbb{Z}[\frac{1 + \sqrt{-7}}{2}]$. We have $\text{Frac} (A) = \text{Frac} (\mathcal{O}) = \mathbb{Q}( \sqrt{-7})$, but $\mathcal{O}$ cannot be a localisation of $A$, since $A^\times = \mathcal{O}^\times = \{ \pm 1\}$.