Showing that morphisms of $k$-schemes that agree topologically are equal.

Question

Let $k$ be a field, $X$,$Y$ schemes of locally finite type over $k$. Let $X$ be reduced, $k$ algebraically closed. I want to show:

Two morphisms $f,g: X \rightarrow Y$ over $k$ are equal if and only if the underlying maps of topological spaces agree.

Also I need an example why this fails if the field $k$ is not algebraically closed.

Now, the "only if" part is clear, so I only have to prove the other direction. I`ve got some theorems I think one can work with, such as if $f,g$ agree on a dense, closed subset, then they are already equal. I cant seem to get in a position to use them yet though. I think one can try to reduce this problem to the affine case, but I am not really firm with working with schemes, and the whole algebraic geometry does not really suit me.

This bears resemblance to another question asked, namely Morphisms of $k$-schemes who agree on $\overline{k}$-points. I was not able to construct the proof I need from this though, because the accepted answer uses seperatedness of $X$, which is not given or needed.

Any help would be appreciated!

Answer 1

Let $X=Y=\textrm{Spec}(\mathbb{Z}_p[x])$, where $p\in\mathbb{P}$, and let $\varphi,\psi:X\to Y$ the morphisms of schemes such that \begin{gather} \varphi^{\sharp}:f(x)\in\mathbb{Z}_p[x]\to0\in\mathbb{Z}_p[x],\\ \psi^{\sharp}:f(x)\in\mathbb{Z}_p[x]\to f(x^p)-f(x)\in\mathbb{Z}_p[x]; \end{gather} as maps of topological spaces $\varphi=\psi$ but as morphisms of (affine) schemes $\left(\varphi,\varphi^{\sharp}\right)\neq\left(\psi,\psi^{\sharp}\right)$.

Answer 2

Since Y is locally of finite type and it is sufficient to show, that the morphisms agree locally we can reduce to $\ f,g: Spec(B)\rightarrow Spec(A)$ where A and B are finitely generated k-Algebras and B is reduced.

Now since $Spec(A)= Spec(k[X_1,...X_n]/ \mathfrak{a})$ we have a closed immersion into $A_k^n$. Closed Immersions are Monos, so w.l.o.g. we can assume $Y=A_k^n$. The later is the fiberproduct of n $A_k^1$. By the universal property of the fiberproduct we can finaly reduce to $Y=A_k^1$.

So we have reduced the problem to morphisms $f,g: Spec(B)\rightarrow A_k^1$ i. e. morphisms of k-Algebras $f',g':k[X]\rightarrow B =k[X_1,...X_m]/ \mathfrak{b}$ for some radical ideal $\mathfrak{b}$, where $f'^{-1}(\mathfrak{p})=g'^{-1}(\mathfrak{p})$ for every prime Ideal $\mathfrak{p} \subset B$.

Suppose $f'\neq g'$. That implies $f'(X)\neq g'(X)$. Let $p +\mathfrak{b}=f'(X)$ and $ q +\mathfrak{b}=g'(X)$. We get $p-q\notin \mathfrak{b}$ and since $\mathfrak{b}$ is a radical ideal and $k[X_1,....,X_n]$ jacobson we get some maximal ideal $\mathfrak{m} \subset k[X_1,....,X_n]$ which contains $\mathfrak{b}$ and not $p-q$, , i.e. $f'(X)-g'(X)\notin \mathfrak{m}/\mathfrak{b}.$

Extending the Morphisms to $k[X_1,...,X_m]/\mathfrak{m}$ we get morphisms with kernel $f'^{-1}(\mathfrak{m})=g'^{-1}(\mathfrak{m})$. Since $k$ is algebraically closed $k[X_1,...,X_m]/\mathfrak{m}\cong k$.

Let $\alpha$ be the Element in $k$ that corresponds to $\pi\circ f'(X)$, then $\pi\circ f'(X-\alpha)=0 \in $ and therefore by our assumption on $f',g':$ $\pi\circ g'(X-\alpha)=0$ so $\pi(f'(X-\alpha)-g'(X-\alpha))=\pi(f'(X)-g'(X))=0$ . Which is a contradiction.

As to the counterexample, just try $f;g: \Bbb{R} [X]\rightarrow\Bbb{C}\space X\mapsto i\space ;\space X\mapsto-i.$ They obviously have the same kernel, and since 0 is the only prime in $\Bbb{C}$ that means the respective topological maps agree.