Let $A$ be an integral Noetherian local ring of dimension 1 which is not normal and with residue field a finite field $\mathbb{F}_q$ (so typically some local ring of a singular curve on a finite field). Denote $B$ its integral closure in its fraction field $K$, and $A^{sh}$ the strict henselization. I want to understand $A^{sh}\otimes_ A K$. This is $\prod k(\frak{p}_i)$ where $\frak{p}_i$ goes through the minimal prime ideals of $A^{sh}$.
To each minimal prime ideal $\frak{p}$ of $A^{sh}$ corresponds a unique minimal prime ideal $\frak{q}$ of the normalization of $A^{sh}$, which is $C:=B\otimes_A A^{sh}$ by Tag 0CBM, and we have $k(\mathfrak{p})=k(\mathfrak{q})$ so we are reduced to determining the residue field of minimal primes of $C$.
Now we use Tag 0C25 : minimal primes of $C$ contain a unique maximal ideal and vice versa. Moreover, each maximal ideal of $C$ is above a maximal ideal of $B$, and for each maximal ideal $\mathfrak{m}$ of $B$ the maximal primes of $C$ above it are in bijection with $\mathrm{Gal}(k(\mathfrak{m})/\mathbb{F}_q)$.
For $\mathfrak{m}$ a maximal ideal of $B$, denote $K_{\mathfrak{m}}^{sh}$ the fraction field of $(B_\mathfrak{m})^{sh}$. I am led to conjecture the following : $$A^{sh}\otimes_ A K = \prod_{\mathfrak{m}~\text{maximal ideal of}~B} \prod_{\mathrm{Gal}(k(\mathfrak{m})/\mathbb{F}_q)} K_{\mathfrak{m}}^{sh}$$ Since $B_\frak{m}$ is a DVR, there is an action of $\mathrm{Gal}(k(\mathfrak{m}))$ on $K_{\mathfrak{m}}^{sh}$, and in the above formula there should be a natural action of $\mathrm{Gal}(\mathbb{F}_q)$ acting by permutation.
Does this hold ? Is there some reference where I can find a thorough treatment of that sort of question ?