This is a problem II 2.15b from Hartshorne.
If $f:X\rightarrow Y$ is a morphism of schemes over $k$ and if $p\in X$ with a residue field $k$, then $f(p)$ also has a residue field $k$.
My attempt:
I know that there is an induced $k$-map between $k(f(p))$ and $k(p)=k$, but I don't believe that it implies that $k(f(p))=k$. For me $k$-map means that there is a commutative triangle, i.e. $g:k\rightarrow k(f(p))\rightarrow k$ and $h:k\rightarrow k$ s.t. $g=h$, but unless $h$ is an isomorphism I don't see how it follows that $k(f(p))=k$. I'm probably missing something very simple.
EDIT: As @Roland suggests, there is an abuse of notation in the sense that actually we have to assume that $k\rightarrow k(p)=k$ is be an isomorphism.
Is there a counterexample to the statement without taking into account the abuse of notation? I'm still suspicious about the remark: is there any source which claims that there is such an abuse of notation? I'm not sure that I saw this in Hartshorne.
Any map from $k(f(p))\rightarrow k$ must have a kernel, and that kernel must be an ideal, and in a field, the only ideal is $(0)$. Therefore the kernel is $(0)$, so the map is injective.
It's also surjective, because the identity $k(p)\rightarrow k(p)$ factors through it.
That makes it an isomorphism.