When is "base changing morphisms of schemes" surjective?

Question

Suppose $X, Y$ are $S$-schemes and $S'\to S$ is a morphism. Every $S$-morphism $f:X\to Y$ gives rise to a $S'$-morphism $f':X'\to Y'$ (where $f'$, $X'$, $Y'$ are the base changes).

Under which assumptions is $Mor_S(X,Y)\to Mor_{S'}(X',Y')$ surjective? Is it possible to specify properties of $X$ and/or $Y$ and/or $S'\to S$ which guarantee surjectivity?

(Sorry for the vague question. I don't have a particular setting in mind, any example is appreciated.)

Answer 1

Here is an example showing that in general surjectivity does not hold. Take: $$S=\operatorname {Spec}\mathbb R,\quad S'=\operatorname {Spec}\mathbb C, \quad X=\mathbb A^1_\mathbb R=\operatorname {Spec}\mathbb R[t],\quad Y=\operatorname {Spec}(\frac {\mathbb R[x,y]}{\langle x^2+y^2+1\rangle})$$

Then $Mor_S(X,Y)=\emptyset$ because at the level of rings there is no $\mathbb R-$ algebra morphism $$\frac {\mathbb R[x,y]}{\langle x^2+y^2+1\rangle} \to \mathbb R[t]$$

However $X'=\mathbb A^1_\mathbb C=\operatorname {Spec}\mathbb C[t],\quad Y'=\operatorname {Spec}(\frac {\mathbb C[x,y]}{\langle x^2+y^2+1\rangle})$ and now there are many $S'$-morphisms $X'\to Y'$ corresponding to the many $\mathbb C$-algebra morphisms $$ \frac {\mathbb C[x,y]}{\langle x^2+y^2+1\rangle} \to \mathbb C[t]$$

Conclusion
The natural map $$Mor_S(X,Y)=\emptyset \to Mor_{S'}(X',Y')\neq \emptyset$$ cannot be surjective

Answer 2

If one modifies the question slightly then one can create a bevy of examples:

• Let $G$ and $H$ be reductive groups over a field $k$. Let $L_1$ and $L_2$ be algebraically closed extensions of $k$. Then, the map $\mathrm{Hom}(G_{L_1},G_{L_1})\to\mathrm{Hom}(G_{L_2},H_{L_2})$. For example, taking $H=\mathrm{GL}_n$ basically says that representations into some fixed $n$-dimensional space don't change as you increase field. This is essentially because representations of split groups are determined by data (the theory of highest weights) that doesn't change as the field changes. In particular, the key property that determines all of this in some sense is the fact that $\mathrm{Hom}(\mathbb{G}_m,\mathbb{G}_m)=\mathbb{Z}$ regardless of what field you're over.
• The same is true for abelian varieties (e.g. see Theorem 3.19 of this for a novel proof).
• If $X$ is an etale scheme over $k$ then the map $X(L_1)\to X(L_2)$ is a bijection for any extension of separably closed fields $L_2/L_1$. In some sense, this is what cuases the previous example to work.

To give you a huge list of non-examples note that one instance of the question you're asking is whether or not for $X/\mathbb{Q}$ the map $X(\mathbb{Q})\to X(\mathbb{C})$ is an isomorphism (it's already injective).

This happens (For $X/\mathbb{Q}$ finite type), I suspect, essentially if and only if $X/\mathbb{Q}$ is finite and $X_\mathrm{red}$ is finitely many copies of $\mathrm{Spec}(\mathbb{Q})$. A rough argument (I haven't though it deeply through) might be the following. Notice that $X(\mathbb{Q})$ is always countable--cover $X$ with finitely many affines and note that for an affine the result follows easily from Noether normalization. If $n=\dim X>0$ then affine locally we have a surjection $X\to \mathbb{A}^n$ which shows that $X(\mathbb{C})$ is uncountable. So, $\dim X=0$. Thus, since $X$ is of finite type this implies that $\mathcal{O}(X)$ is a finite product of Artin local rings. Note that $X(\mathbb{Q})$ and $X(\mathbb{C})$ are agnostic to replacing $X$ by $X_\mathrm{red}$. So we may as well assume that $\mathcal{O}(X)$ is a finite product of fields. It's then easy to see that the conclusion happens if and only if all those fields are $\mathbb{Q}$.