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Suppose that $B \rightarrow A$ is an extension of rings where $A$ and $B$ are integral $k$-algebras ($\mathrm{char}\,k = 0$) finitely generated over $k$.

It is well known that if $B \rightarrow A$ is an integral extension, then the induced morphism $\mathrm{Spec}\,A \rightarrow \mathrm{Spec}\,B$ is finite and surjective.

I am interested in the opposite case. Suppose that $B \rightarrow A$ is an extension of rings which is algebraically closed. That is, if $a \in A$ and there exists a polynomial $p(t) \in B[t]$ such that $p(a) = 0$, then $a \in B$. Does this look in any way like a fibration? In particular, is it surjective and does it have connected fibres?

An example of an algebraically closed extension is $k[x] \rightarrow k[x,y]$, which is a fibration.

In this case, it is easy to show that if $B \rightarrow A$ is generically finite, then it is an isomorphism. It is equally easy to show that, if the image of $\mathrm{Spec}\,A$ in $\mathrm{Spec}\,B$ is affine, then it is surjective. To conclude, one would need a "Stein factorisation" result for affine varieties.

Any ideas or counterexamples?

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    $\begingroup$ Do you have examples of such extensions where $A$ is not of the form $B[x_1,\cdots, x_n]$? $\endgroup$ Commented Feb 11, 2021 at 16:00
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    $\begingroup$ Does $B = k[t]$ and $A = B[x,y]/(x y - t)$ give a counter example? $\endgroup$ Commented Feb 11, 2021 at 16:13
  • $\begingroup$ @AntoineLabelle Anything that looks like a product. If $C$ is an integral $k$-algebra, then something of the from $B \rightarrow B \otimes_k C = A$ would be algebraically closed. Otherwise $k[t] \rightarrow k[x,y]$ where $t \rightarrow xy$ is algebraically closed. This is a fibration by hyperbolae. Another one is the radial fibration $k[t] \rightarrow k[x,y,1/x]$ where $t \rightarrow y/x$. $\endgroup$ Commented Feb 11, 2021 at 19:21
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    $\begingroup$ I think that Ben's point is that the fibers are not all the same so it's not a fibration, over t=0 it's two lines crossing while everywhere else it's an hyperbola $\endgroup$ Commented Feb 11, 2021 at 19:36
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    $\begingroup$ @BenC I think your counterexample works. Thank you for pointing it out. I still wonder if it is always surjective $\endgroup$ Commented Feb 12, 2021 at 15:11

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Here is a counterexample for surjectivity: $$B=k[x,y], A=B[X,Y]/(xX+yY-1)$$ It is easy to see that the image of the induced morphism is the plane minus the origin.

Now suppose $a\in A$ is a root of a nonzero polynomial with coefficients in $B$. As $\text{Frac}(A)=k(x,y,X)$ is a purely transcendental extension of $\text{Frac}(B)=k(x,y)$, then $a$ must be in $\text{Frac}(B)$. Write $a=\frac{p}{q}$ for $p,q\in B$ relatively prime. If $a\not\in B$ then $q$ is not a unit, so its zero locus has dimension $1$, so it contains a point $(x_0,y_0)$ other than the origin. Then we can choose $X_0,Y_0$ so that $x_0X_0+y_0Y_0=1$, and then $a$ has a pole at $(x_0,y_0,X_0,Y_0)$, which is a contradiction as $a\in A$.

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Let $B \to A$ be an injective map of finite type $k$-algebra domains with the property that for any $a \in A$ satisfying a polynomial relation $p(a) = 0$ for $p \in B[X]$ we have $a \in B$.

(1) the fibers of $\mathrm{Spec}(A) \to \mathrm{Spec}(B)$ need not be connected. For example, $$ \mathrm{Spec}(k[t][x,y,z]/(xy - t, x(x - y) - 1)) \to \mathrm{Spec}(k[t])$$ This is a family of hyperbolas over $\mathbb{A}^1$ degnerating at $t = 0$ with the plane $x = y$ removed and thus the fiber over $t = 0$ is the coordinate axes minus the origin which is disconnected.

(2) the map $\mathrm{Spec}(A) \to \mathrm{Spec}(B)$ need not be surjective. Take the equation for a in $\mathbb{A}^3$ which both contain the line $x = y = 0$. For example, $f = zx + zy + xy + y$. Let $B = k[x,y]$ and $A = k[x,y,z]_{f}$. Consider, $$ \mathrm{Spec}(k[x,y,z]_{fg}) \to \mathrm{Spec}(k[x,y]) $$ The fiber over $x = y = 0$ is empty.

In both cases, the fraction field is a purely transcendental extension giving counterexamples.

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