In the category of commutative rings, every domain embeds into a field. Is this true in the category of presentably symmetric monoidal stable $\infty$-categories? Here's what I mean by that.

• Let $PrSt^{L,\otimes}$ be the $\infty$-category of symmetric monoidal closed presentable stable $\infty$-categories and left adjoint strong symmetric monoidal functors.

• Say that $0 \neq \mathcal C \in PrSt^{L,\otimes}$ is a field if there are no nontrivial symmetric monoidal stable accessible localizations of $\mathcal C$.

• Say that $0 \neq \mathcal C \in Pr^{L,\otimes}$ is a domain if $X \otimes Y = 0 \Rightarrow X = 0 \text{ or } Y = 0$ for $X,Y \in \mathcal C$.

• Say that $F: \mathcal C \to \mathcal D \in PrSt^{L,\otimes}$ is an embedding if $F$ is conservative. Note that (just as for commutative rings) we have an orthogonal factorization system on $PrSt^{L,\otimes}$ given by the localizations and the embeddings.

So now the question makes sense:

Question: In $PrSt^{L,\otimes}$, does every domain embed into a field?

Notes:

• It's not hard to show that the converse holds -- if $\mathcal C \to \mathcal D$ is an embedding and $\mathcal D$ is a field, then $\mathcal C$ is a domain.

• One can characterize the fields $\mathcal C \in PrSt^{L,\otimes}$ by the following property: for every nonzero $X \in \mathcal C$, there exist objects $Y_i$ and a colimit diagram $I = \varinjlim_i (Y_i \otimes X)$ where $I$ is the monoidal unit of $\mathcal C$.

• There doesn't seem to be a "field of fractions" construction which automatically turns a domain into a field.

• (At least under Vopenka's Principle) Every nontrivial accessible localization $\mathcal C \to L\mathcal C \in PrSt^{L,\otimes}$ admits a further accessible localization $LC \to \mathcal D$ which is maximal, i.e. such that $\mathcal D$ is a field.

• It's possible (under large cardinal hypotheses) that some sort of ultraproduct of all maximal localizations of a domain $\mathcal C$ might be a field into which $\mathcal C$ embeds diagonally, but I'm not sure.