I have a question about correctness of following statement claimed [here][1] in $\boxed{2} \ $:

>Let $k$ arbitrary field, let $f : X \longrightarrow Y$ be a finite dominant morphism between finite type $k$-schemes. Assume that $Y$ is integral and that there is $y \in Y$ such that $f^{-1}(y)$ is scheme-theoretically equal to $q$ reduced points (say $x_1, \ldots, x_q$). Then, we can find an open neighborhood of say $x_1 \in U \subset X$ where $X$ is reduced.

It suffice to consider the **local version** given by a finite monomorphism  $A \longrightarrow B$ of finite-type $k$-algebras with $A=A_{\mathfrak{m}}$ local integral domain ( especially reduced) with unique max ideal $\mathfrak{m}$ corresponding to $y$, furthermore $B=B_{\mathfrak{m}}$ ( ie $B$ localized with respect $\mathfrak{m}$) and  $B/ (\mathfrak{m} \cdot B)=B\otimes_A A/ \mathfrak{m}= B\otimes_A k(y)=k^q$ by assumption. The question becomes why is $B$ reduced?

So philosophically, which preferably most mild conditions should  the map $A\to B$ satisfy such that $B$ "inherits" reducedness from $A$?

It seems that the quoted proof contains a gap I not know how to repair. The strategy in the linked proof is to take basis generators $f_1, \ldots, f_q$  of $B/ (\mathfrak{m} \cdot B)=k^q$ as a $k$ vector space and $e_1, \ldots, e_q$ be some lifts in $B$ of the $f_i$'s.  The $e_i$ induce a morphism of $A_{\mathfrak{m}}$-modules of finite type:

$$ \Phi : A^q \longrightarrow B $$

Using finiteness of $B$ as $A$ module and Nakayama lemma, this map is surjective. 

**(Maybe) Key problem:** But there is a potential gap: it is not clear to me why under the above assumptions the inclusion

$$\mathfrak{m}.\operatorname{Ker}(\Phi) \supset \operatorname{Ker}(\Phi).$$

should hold? If we having this, Nakayama applied again would tell us that that the kernel $\operatorname{Ker}(\Phi)$ is zero and we win. But why this inclusion above holds? If we eg would additionally assume that $B$ is flat over $A$, then it follows from [tag/00HL][2] , but note that flatness wasn't assumed.

Even thought I haven't a conterexample, note that to show the claim above, it would be even sufficient to show an even weaker statement that the kernel $\operatorname{Ker}(\Phi)$ is a radical ideal, but I also not see why thats holds here, only that it is obviously contained in $(\mathfrak{m})^q$.

So finally **the questions** become  
**(I)** if the local version of the statement is true **without any additional assumpions** (eg like flatness),  

then ( if (I) has positive answer ) **(II)** if the given proof can be "repaired relativelly elementary" (eg deduce $\operatorname{Ker}(\Phi)$ is zero or radical), 
  
and **(III)** if (I) is wrong, are there preferably relative mild additional assumpions on $A \to B$ (milder than flatness, compare with comments on the potential gap in the given proof above) turning the statement in a true one, ie that $B$ inherits reducedness from $A$?


  [1]: https://mathoverflow.net/q/453652
  [2]: https://stacks.math.columbia.edu/tag/00HL