Every morphism of ringed spaces between algebraic schemes is local

Fix an algebraic closed field $k$. Given a finitely generated $k$-algebra $A$ we can considered the ringed space $$X=(\text{Max}(A),\mathcal{O}_{\text{Max}(A)})$$ where $\text{Max}(A)\subset \text{Spec}(A)$ is the set of maximal ideals of $A$ with the induced topology from $\text{Spec}(A)$ and $\mathcal{O}_{\text{Max}(A)}:=i^{-1}\mathcal{O}_{\text{Spec}(A)}$, this sheaf is easy to compute because $i:\text{Max}(A)\rightarrow \text{Spec}(A)$ is a quasi-homeomorphism, i.e, it induce a bijection between the open subsets. Lets call such a $X$ an affine algebraic scheme.

In general an algebraic scheme is a ringed space with a finite cover of open subsets where each is locally isomorphic to an affine algebraic scheme.

I want to know if the followings fact is true

Every morphism of ringed spaces over $k$ between algebraic schemes is a morphism of locally ringed spaces over $k$.

I ask this because in this notes (in french) the author works with this category but considering as morphisms the morphisms of ringed spaces instead of the morphisms of locally ringed spaces.

As a remark one can prove that this category (with morphisms as morphisms of locally ringed spaces over $k$) is equivalent to the category of schemes of finite type over $k$ and also the full subcategory of reduced algebraic schemes is equivalent to the category of "abstract algebraic varieties" in the context of classical algebraic geometry. So this construction lies in the middle between the classic and the modern approach to algebraic geometry.

• That is true. For every finitely generated $k$-algebra $A$, resp. $B$, for every maximal ideal $\mathfrak{m}\subset A$, resp. $\mathfrak{n}\subset B$, the following compositions are isomorphisms, $k\to A_{\mathfrak{m}} \to A_{\mathfrak{m}}/\mathfrak{m}\cdot A_{\mathfrak{m}}$, resp. $k\to B_{\mathfrak{n}} \to B_{\mathfrak{n}}/\mathfrak{n}\cdot B_{\mathfrak{n}}$. Thus, the composition $A_{\mathfrak{m}}\xrightarrow{\phi} B_{\mathfrak{n}} \to B_{\mathfrak{n}}/\mathfrak{n}\cdot B_{\mathfrak{n}}$ is surjective. So the kernel of the composition is a maximal ideal, namely the unique maximal ideal. – Jason Starr Nov 25 '17 at 21:09
• In my previous comment, the map $\phi$ is an arbitrary $k$-algebra homomorphism from $A_{\mathfrak{m}}$ to $B_{\mathfrak{n}}$. The point is that $\phi^{-1}(\mathfrak{n}\cdot B_{\mathfrak{n}})$ equals $\mathfrak{m}\cdot A_{\mathfrak{m}}$. So, the ring homomorphism $\phi$ is a local ring homomorphism. – Jason Starr Nov 25 '17 at 21:45