# Spectrum, tensors and algebraic closure

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?

• What definition of separable algebra are you using? In the standard definition, $B_i$ need not be fields. Dec 24 '19 at 0:53
• @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. Dec 24 '19 at 2:44
• @reuns But in this case $F$ would be $K^3$, which is not a field, right? Dec 24 '19 at 3:34
• @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? 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 $$f=g$$

• Thank you very much! Jan 3 '20 at 1:01