I'm trying to show the following (not sure if it's true...):

Let $K$ be a field. Let $B$ be a separable $K-$algebra, so we can write $B = \prod B_i$ as a finite product of finite separable field extensions $B_i/K$. Let $F$ be a finite separate field extension of $K$. Assume we have two $K$-linear morphisms: $$ f, g: B \to F.$$ These two morphisms induce morphisms $$f \otimes id, g \otimes id: B \otimes_K \bar{K} \to F \otimes_K \bar{K}$$ where $\bar{K}$ is an algebraic closure of $K$. Then we obtain morphisms $$ Spec(f \otimes id), Spec(g \otimes id): Spec(F \otimes_K \bar{K}) \to Spec(B \otimes_K \bar{K})$$ in the usual way. Is it true that if $Spec(f \otimes id)$ and $Spec(g \otimes id)$ agree on one element then $f$ and $g$ have to be equal?

  • $\begingroup$ What definition of separable algebra are you using? In the standard definition, $B_i$ need not be fields. $\endgroup$
    – Mohan
    Dec 24 '19 at 0:53
  • $\begingroup$ @Mohan I'm using the one in Lenstra's Galois theory of schemes (Def 1.2 and then Theorem 2.7). So we can just assume that the $B_i$ are fields. $\endgroup$
    – M. Wang
    Dec 24 '19 at 2:44
  • $\begingroup$ @reuns But in this case $F$ would be $K^3$, which is not a field, right? $\endgroup$
    – M. Wang
    Dec 24 '19 at 3:34
  • $\begingroup$ @reuns Oh yes sorry, I want it to be a $K$-linear morphism. So do you think the statement is wrong and I should look for a counterexample? $\endgroup$
    – M. Wang
    Dec 24 '19 at 4:24
  • $L = \overline{K}$, $B$ is a unital finite dimensional $K$-algebra, $F/K$ is a finite separable extension.

    Given a $K$-algebra homomorphism $f : B\to F$ which gives $f_2:B\otimes L\to F\otimes L$ then $\ker(f)$ is a maximal ideal and $\ker(f_2) = \ker(f)\otimes L$,

    given another homomorphism $g : B\to F$, $g_2:B\otimes L\to F\otimes L$ and a prime ideal $p$ of $F\otimes L$

    $f_2^{-1}(p)=g_2^{-1}(p)$ means that $f_2^{-1}(p)$ contains $\ker(f)\otimes L+\ker(g)\otimes L=(\ker(f),\ker(g))\otimes L$, if $\ker(f)\ne \ker(g)$ then they are comaximal so that $(\ker(f),\ker(g))\otimes L=1\otimes L$, a contradiction since $f_2^{-1}(p)$ is a prime ideal.

  • Thus $\ker(f)=\ker(g)$ and we can replace $B$ by $B/\ker(f)$ and assume $B$ is a subfield of $F$, $f$ is the inclusion $B\to F$ and $g$ is a possibly different $K$-embedding $B\to F$.

    Since $F/K$ is separable then $B=K[x]/(u(x)),F=K[x,y]/(u(x),v(x,y))$,

    $g(x) = r(x,y)$ where $r(x,y)$ is one of the roots of $u(T)$ in $F$,

    $F\otimes L=L[x,y]/(u(x),v(x,y))$,

    $f_2$ is the inclusion $L[x]/(u(x))\to L[x,y]/(u(x),v(x,y))$ and $g_2$ is the map $L[x]/(u(x))\to L[x,y]/(u(x),v(x,y))$ sending $x$ to $r(x,y)$

  • The prime ideals of $L[x,y]/(u(x),v(x,y))$ are of the form $p=(x-a,y-b)$ with $a,b\in L$ such that $u(a)=0,v(a,b)=0$.

    $f_2^{-1}(p)$ is generated by $f_2^{-1}(x-a) = (x-a)$.

    $g_2^{-1}(p)$ is generated by $g_2^{-1}(r(x,y)-r(a,b)) = (x- r(a,b))$.

    $f_2^{-1}(p)=g_2^{-1}(p)$ means that $a=r(a,b)$ ie. $r(x,y)=x$ and $f=g$

  • $\begingroup$ Thank you very much! $\endgroup$
    – M. Wang
    Jan 3 '20 at 1:01

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