If $A \to B$ is universal with respect to sending $I$ to the unit ideal and $A$ is an integral domain, does $B$ also have to be an integral domain?

Question

Let $R$ be a commutative ring with unity. Let $A$ be a commutative $R$-algebra. Let $I \subseteq A$ be an ideal with $I \neq \{0\}$. Let $\varphi: A \to B$ be a map of $R$-algebras satisfying the following universal property:

1. $\langle \varphi(I) \rangle = \langle 1_B \rangle$

2. If $\psi: A \to C$ is any map with $\langle \psi(I) \rangle = \langle 1_C \rangle$, then there is a unique map $\tilde{\psi}: B \to C$ making the following diagram commute: $$ \begin{array}{ccl} A & \overset{\varphi}{\longrightarrow} & B \\ & {\scriptstyle \psi \!} \searrow & \downarrow {\!\scriptstyle \exists ! \tilde{\psi}} \\ & & C \\ \end{array} $$

For example, if $I = \langle a \rangle$ is a principal ideal, this just means that $A \to B$ is the localization $A \to A[a^{-1}]$.

Question: If we are in the above situation and $A$ is an integral domain, does this imply that $B$ is an integral domain?

Answer 1

We shall prove something even stronger: if $A$ is an integral domain, and $B$ is an $A$-algebra satisfying the required universal property, then $B$ is an $A$-subalgebra of $K = \mathrm{Frac}(A)$, the field of fractions of $A$, via the unique $A$-algebra map $B \to K$.

First, observe that there exist $\alpha_1, \ldots, \alpha_n \in I$ and $\beta_1, \ldots, \beta_n \in B$ such that $\sum \alpha_i\beta_i = 1$ in $B$. Define $f = \alpha_1x_1 + \cdots + \alpha_nx_n - 1$, $C_A = A[x_1, \ldots, x_n]/f$, and $C_K = K[x_1, \ldots, x_n]/f$. Observe that the ideal $I$ extends to the unit ideal in $K, C_A, C_K$, so we have the following commutative diagram: $$\require{AMScd} \begin{CD} B @>{}>> K\\ @VVV @VVV \\ C_A @>{}>> C_K \end{CD}$$ where the maps $C_A \to C_K$ and $K \to C_K$ are the obvious ones. The commutativity of the diagram comes from the universal property of $B$. Our strategy to prove that $B \to K$ is injective will be to show that $B \to C_A$ and $C_A \to C_K$ are both injective. The first part is straightforward: we have a map $C_A \to B$ given by $x_i \mapsto \beta_i$, and the composed map $B \to C_A \to B$ is $\mathrm{id}_B$ is injective. It follows that the map $B \to C_A$ is also injective.

To prove $C_A \to C_K$ is injective, observe that $C_K = C_A \otimes K$. Thus, our task is equivalent to showing that for $g \in A[x_1, \ldots, x_n]$, if there exists non-zero $a \in A$ with $f \mid ag$, then we also have $f \mid g$. Equivalently, if $f \mid g$ in $K[x_1, \ldots, x_n]$, then we also have $f \mid g$ in $A[x_1, \ldots, x_n]$.

Indeed, let $q \in K[x_1, \ldots, x_n]$ be the unique solution to $g = fq$. If $q$ has coefficients not in $A$, there exists a minimal non-negative integer $d$ such that there exist degree $d$ terms in $q$ with coefficients not in $A$. However, since $f$ has coefficients in $A$ and constant term $-1$, this implies that $g = fq$ also has degree $d$ terms with coefficients not in $A$, a contradiction. Thus, $C_A \to C_K$ is injective, completing the proof.

Answer 2

Yes, if we are in the above situation and $A$ is an integral domain, then this implies that $B$ is an integral domain.

Related question: Ring-theoretic characterization of open affines?

We proceed using several lemmas.

Lemma 1: Let $T = \{f \in A \mid A[f^{-1}] \overset{\cong}{\to} B[\varphi(f)^{-1}]\}$ be the set of those $f \in A$ such that the induced map $A[f^{-1}] \to B[\varphi(f)^{-1}]$ is an isomorphism. Then $T = \sqrt{I}$.

Proof of Lemma 1: For this, we will check three claims:
Claim 1: $T$ is a radical ideal
Claim 2: $I \subseteq T$
Claim 3: $T \subseteq \sqrt{I}$
From these three claims, it will follow by Claim 1 and Claim 2 that $\sqrt{I} \subseteq T$, which together with Claim 3 will imply that $\sqrt{I} = T$, finishing the proof.
Proof of Claim 1: If $f \in A$ and $n \in \mathbb{N}$, then inverting $f$ is the same as inverting $f^n$. Hence the map $A[f^{-1}] \to B[\varphi(f)^{-1}]$ is an isomorphism if and only if the map $A[(f^n)^{-1}] \to B[\varphi(f^n)^{-1}]$ is an isomorphism. So if $f^n \in T$, then $f \in T$. Hence $T$ is a radical ideal.
Proof of Claim 2: Take any $f \in I$. Now consider the following diagram: $$ \begin{array}{ccccc} A & \overset{\varphi}{\longrightarrow} & B & \longrightarrow & B[\varphi(f)^{-1}] \\ & {\scriptstyle \iota_f \!} \searrow & \downarrow {\!\scriptstyle \exists !} & \swarrow {\!\scriptstyle \exists !} & \\ & & A[f^{-1}] & & \\ \end{array} $$ Since $f \in I$, we have $\langle \iota_f(I) \rangle = \langle 1_{A[f^{-1}]} \rangle$. Hence by the universal property of $B$, there exists a unique middle vertical arrow making the left triangle commute.
But now the middle vertical arrow sends the element $\varphi(f)$ to the invertible element $f$, so that by the universal property of the localization $B[\varphi(f)^{-1}]$, there exists a unique right vertical arrow making the right triangle commute. This must be an inverse to the natural map $A[f^{-1}] \to B[\varphi(f)^{-1}]$, so that map is an isomorphism.
Proof of Claim 3: Take $f \in T$. Now we have the following commutative square: $$ \begin{array}{ccc} A & \overset{\varphi}{\longrightarrow} & B \\ {\scriptstyle \iota_f \!} \downarrow & & \downarrow \\ A[f^{-1}] & \overset{\cong}{\longrightarrow} & B[\varphi(f)^{-1}] \\ \end{array} $$ Since the bottom map is an isomorphism, we obtain a triangle $$ \begin{array}{ccl} A & \overset{\varphi}{\longrightarrow} & B \\ & {\scriptstyle \iota_f \!} \searrow & \downarrow \\ & & A[f^{-1}] \\ \end{array} $$ But this implies that $\langle \iota_f(I) \rangle = \langle 1_{A[f^{-1}]} \rangle$. Hence $\iota_f(I) \cap \{1, f, f^2, \ldots\} \neq \emptyset$, so there is some $n \in \mathbb{N}$ with $f^n \in I$, so $f \in \sqrt{I}$.

Lemma 2: $\langle \varphi(T) \rangle = \langle 1_B \rangle$

Proof of Lemma 2: $$ \langle \varphi(T) \rangle = \langle \varphi(\sqrt{I}) \rangle \supseteq \langle \varphi(I) \rangle = \langle 1_B \rangle $$

Lemma 3: If $J \subseteq B$ is an ideal, then $J = \langle 1_B \rangle$ if and only if $\sqrt{J} = \langle 1_B \rangle$

Proof of Lemma 3: If $J = \langle 1_B \rangle$, then $\sqrt{J} \supseteq J = \langle 1_B \rangle$.
On the other hand, if $J \neq \langle 1_B \rangle$, then there is some prime ideal $\mathfrak{q} \subseteq B$ with $J \subseteq \mathfrak{q}$. But then also $\sqrt{J} \subseteq \mathfrak{q}$, so $\sqrt{J} \neq \langle 1_B \rangle$.

Lemma 4: If $D \subseteq B$ is any subset and $n \in \mathbb{N}$, then $\langle f \mid f \in D \rangle = \langle 1_B \rangle$ if and only if $\langle f^n \mid f \in D \rangle = \langle 1_B \rangle$

Proof of Lemma 4: We have $\sqrt{\langle f^n \mid f \in D \rangle} = \sqrt{\langle f \mid f \in D \rangle}$. Hence we are done by Lemma 3.

Lemma 5: If $g \in B$ is such that for all $f \in T$, the image of $g$ under $\iota_{\varphi(f)}: B \to B[\varphi(f)^{-1}]$ is $0$, then $g = 0$.

Proof of Lemma 5: Since by Lemma 2, $\langle \varphi(T) \rangle = \langle 1_B \rangle$, there are $f_1, \ldots, f_n \in T$ such that $\langle \varphi(f_1), \ldots, \varphi(f_n) \rangle = \langle 1_B \rangle$. For each $1 \leq i \leq n$, we know that $\iota_{\varphi(f_i)}(g) = 0$ in $B[\varphi(f)^{-1}]$, which means equivalently that there is some $m_i \in \mathbb{N}$ with $\varphi(f_i)^{m_i} g = 0$ in $B$. Let $m = \underset{1 \leq i \leq n}{\operatorname{max}} m_i$. This implies that in particular $\varphi(f_i)^m g = 0$ in $B$. But now by Lemma 4, we know that $\langle \varphi(f_i)^m \mid 1 \leq i \leq n \rangle = \langle 1_B \rangle$. Hence $g = 0$ in $B$.

Lemma 6: If $g \in B$ is nonzero, then there is some $f \in T$ such that the image of $g$ under $\iota_{\varphi(f)}: B \to B[\varphi(f)^{-1}]$ is nonzero.

Proof of Lemma 6: This is just the contraposition of Lemma 5.

Theorem 7: If $A$ is an integral domain, then $B$ is an integral domain

Proof of Theorem 7: By contradiction. Assume that $A$ is an integral domain, but $B$ is not. Then there are $h, k \in B$ with $h, k \neq 0$ and $h k = 0$. By Lemma 6, there are $f, g \in T$ such that $\iota_{\varphi(f)}(h) \neq 0$ in $B[\varphi(f)^{-1}] \cong A[f^{-1}]$ and $\iota_{\varphi(g)}(k) \neq 0$ in $B[\varphi(g)^{-1}] \cong A[g^{-1}]$.

Now, since $A$ is an integral domain and $f \neq 0$ and $g \neq 0$, we also have $f g \neq 0$. Hence the maps $A[f^{-1}] \to A[(f g)^{-1}]$ and $A[g^{-1}] \to A[(f g)^{-1}]$ are injective. In particular, since $\iota_{\varphi(f)}(h) \neq 0$ in $B[\varphi(f)^{-1}] \cong A[f^{-1}]$, we have that $\iota_{\varphi(f g)}(h) \neq 0$ in $B[\varphi(f g)^{-1}] \cong A[(f g)^{-1}]$. Analogously, since $\iota_{\varphi(g)}(k) \neq 0$ in $B[\varphi(g)^{-1}] \cong A[g^{-1}]$, we have that $\iota_{\varphi(f g)}(k) \neq 0$ in $B[\varphi(f g)^{-1}] \cong A[(f g)^{-1}]$. Since $A[(f g)^{-1}]$ is an integral domain, this implies that $\iota_{\varphi(f g)}(h k) = \iota_{\varphi(f g)}(h) \iota_{\varphi(f g)}(k) \neq 0$ in $B[\varphi(f g)^{-1}] \cong A[(f g)^{-1}]$.

On the other hand, we know that $h k = 0$ in $B$, so that $\iota_{\varphi(f g)}(h k) = 0$ in $B[\varphi(f g)^{-1}] \cong A[(f g)^{-1}]$.

This contradiction finishes the proof.