Why do morphisms of schemes locally of finite type preserve closed points? Let $f:X \to Y$ be a morphism of $k$-schemes locally of finite type, and let $x \in X$ be a closed point. Then $\kappa(x)$ is a finite extension of $k$ by the Nullstellensatz. 
I want to conclude that $f(x)$ is a closed point of $Y$ because $\kappa(f(x))$ is also a finite extension of $k$. Why is $\kappa(f(x))$ a finite extension of $k$?
 A: A morphism of schemes is a morphism of locally ringed spaces.
This means that, writing $y=f(x)$,  the induced $k$-morphism on stalks $f^*_x:\mathcal O_{Y,y }\to \mathcal O_{Y,y }$ sends the maximal ideal $\mathfrak m_y \subset \mathcal O_{Y,y }$ to  the maximal ideal $\mathfrak m_x \subset \mathcal O_{X,x }$ and thus induces a morphism of $k$-extensions $\kappa (y)=\mathcal O_{Y,y }/\mathfrak m_y \to \kappa (x)=\mathcal O_{X,x }/\mathfrak m_x $ .
Since the extension $\kappa (x)/k$ is finite-dimensional, so is its subextension $\kappa (y)/k$
A: The question is local on the domain and codomain, so by shrinking $X$ and $Y$ if necessary, we may assume $f : X \to Y$ is a morphism of affine $k$-schemes of finite type. That is, $Y = \operatorname{Spec} B$ for some $k$-algebra $B$ of finite type, and $X = \operatorname{Spec} A$ for some $B$-algebra $A$ of finite type. 
Let $\mathfrak{m}$ be a maximal ideal of $A$. As you say, $A / \mathfrak{m}$ is a finite field extension of $k$. Consider the resulting homomorphism $B \to A \to A / \mathfrak{m}$. It may not be surjective, but at any rate, its image is an integral domain that is finite-dimensional as a $k$-vector space, so the kernel of $B \to A / \mathfrak{m}$ must be a maximal ideal of $B$. Hence, $f : X \to Y$ preserves closed points.
