I have a question about correctness of following statement claimed here 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 ( esp'ly reduced) with unique max ideal $\mathfrak{m}$ corresponding to $y$, and $B=B_{\mathfrak{m}}$ ( 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 $A$-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? Having this, clearly 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 , 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), ie the "gap" isn't a gap at all

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 $A$?

I have a question about correctness of following statement claimed here 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 ( esp'ly reduced) with unique max ideal $\mathfrak{m}$ corresponding to $y$, and $B=B_{\mathfrak{m}}$ ( 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 $A$-modules of finite type:

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

Since $B$ is finite $A$-module and by 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? Having this, again Nakayama tells 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 , 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), ie the "gap" isn't a gap at all

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 $A$?

I have a question about correctness of following statement claimed here 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 , 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 $A$?

I have a question about correctness of following statement claimed here 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 ( esp'ly reduced) with unique max ideal $\mathfrak{m}$ corresponding to $y$, and $B=B_{\mathfrak{m}}$ ( 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 $A$-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? Having this, clearly 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 , 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), ie the "gap" isn't a gap at all

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 $A$?