# Homomorphism between finitely generated algebra

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 ?

-
A small comment for fun: an algebraically closed field is necessarily infinite, so you don't need to specify "infinite alg closed". –  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 $c$.

-
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. –  Arsenaler Jan 28 '13 at 15:39
I still think there's some value for you in my post. –  Jesko Hüttenhain Jan 28 '13 at 15:42
Will the result change if $B,C$ are normal? –  Arsenaler Jan 28 '13 at 16:46

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

-
But there is one: $B\to C$ by $(a,b)\mapsto a$. You could even $(a,b)\mapsto b$ if you like. –  Jesko Hüttenhain Jan 28 '13 at 15:25
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.. –  Berci Jan 28 '13 at 15:33
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 $\mathbb R$. –  Jesko Hüttenhain Jan 28 '13 at 15:37
I have edited my question second time. Please check it again. I am sorry if it waste your time. –  Arsenaler Jan 28 '13 at 15:40
@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.. –  Berci Jan 28 '13 at 15:45