Let $A,B,C$ are the coordinate ring of three affine varieties in affine space over an algebraically closed field $k$, $B$ and $C$ are normal, and $i: A\longrightarrow B$ is an injective $k$-algebra homomorphism.

If there is another injective $k$-algebra homomorphism $\varphi$ from $A$ to $C$, can we deduce that there exists a homomorphism $\psi: B\longrightarrow C$ such that $\varphi=\psi\circ i$ ?

If such homomorphism exists, is it unique ?

  • $\begingroup$ A small comment for fun: an algebraically closed field is necessarily infinite, so you don't need to specify "infinite alg closed". $\endgroup$ – Bogdan Jan 28 '13 at 15:44

If $B$ is finitely generated over $k$, then it is certainly finitely generated over $A$, so we have $B=A[x_1,\ldots,x_r]/I$ for some ideal $I$. Furthermore, $C$ becomes an $A$-algebra via the morphism $\varphi$. Now, we get a morphism $\psi_c:A[x_1,\ldots,x_r]\to C$ of $A$-algebras for every $c=(c_1,\ldots,c_r)\in C^r$ by mapping $x_i\mapsto c_i$. If we want a morphism of $A$-algebras $B\to C$, we need to make sure that $I\subseteq\ker(\psi_c)$. Let us assume that $I\ne(1)$, because otherwise we'd have $B=0$, which would be weird.

Let us first assume $A=k$. If $k$ were algebraically closed, we'd know that there is a common zero $a=(a_1,\ldots,a_r)\in Z(I)$ and we could choose $c_i=\varphi(a_i)$. Indeed, it seems that infinite is not enough:

Let $A=C=\mathbb R$ and $B=\mathbb C$. The problem is that you have no algebra morphism $\mathbb C\to \mathbb R$. It would imply that you can map $i$ to some element $\psi(i)\in\mathbb R$ which satisfies $\psi(i)^2=\psi(i^2)=\psi(-1)=-1$.

However, in the case where $k$ is algebraically closed and $A=k$, we can do it. Now, that does not offer much hope: Even if $k$ is algebraically closed, $A$ is not necessarily an algebraically closed extension field.

Edit: Note that we have also established that if such a $\psi$ exists, it does not have to be unique. You might have a lot of choices for your $c$.

| cite | improve this answer | |
  • $\begingroup$ Sorry Jesko, I have confusion with some injective homomorphism, and you posted your answer when I was editing my question. Sorry if my question waste your time. $\endgroup$ – Arsenaler Jan 28 '13 at 15:39
  • 1
    $\begingroup$ I still think there's some value for you in my post. $\endgroup$ – Jesko Hüttenhain Jan 28 '13 at 15:42
  • $\begingroup$ Will the result change if $B,C$ are normal? $\endgroup$ – Arsenaler Jan 28 '13 at 16:46

No. Let $A:=k:=\Bbb R$, and $B:=\Bbb R^2$ with coordinatewise multiplication and $C:=\Bbb C$.

| cite | improve this answer | |
  • $\begingroup$ But there is one: $B\to C$ by $(a,b)\mapsto a$. You could even $(a,b)\mapsto b$ if you like. $\endgroup$ – Jesko Hüttenhain Jan 28 '13 at 15:25
  • $\begingroup$ You are talking about algebra-homomorphisms, not? And what would be the inclusion $i:A\to B$? Should be the diagonal... If you prefer, change the roles, $B:=\Bbb C$. From $\Bbb C$ there are not too many algebra homomorphisms.. $\endgroup$ – Berci Jan 28 '13 at 15:33
  • $\begingroup$ For one thing, no, $\psi$ does not have to be injective. Consider $A=C=k$ and $B=k[x]$ with $\psi: x\mapsto 0$. Secondly, the map $\mathbb R^2\to\mathbb C$, $(a,b)\mapsto a$ is an algebra homomorphism. However, there is a better counterexample: $A=C=\mathbb R$ and $B=\mathbb C$. There is no algebra homomorphism from $\mathbb C$ to $\mathbb R$. $\endgroup$ – Jesko Hüttenhain Jan 28 '13 at 15:37
  • $\begingroup$ I have edited my question second time. Please check it again. I am sorry if it waste your time. $\endgroup$ – Arsenaler Jan 28 '13 at 15:40
  • $\begingroup$ @JeskoHüttenhain: yes, I deleted the injectivity argument in my comment and also realized that if $\Bbb C$ is the source, then the reasoning becomes easier.. $\endgroup$ – Berci Jan 28 '13 at 15:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.