Krdytkyu

Let $X,Y$ be $S$-schemes, let $xin X$ and $yin Y$ be points over $sin S$ and let $mathcal O_{Y,y}tomathcal O_{X,x}$ be a local $mathcal O_{S,s}$-homomorphism. If $Y$ is locally of finite presentation at $y$, then this homomorphism extends to an $S$-morphism $Uto Y$ for some neighbourhood $U$ of $x$.

Here a nice proof is given. In brief, everything is reduced to the following algebraic task:

Given $R$-algebras $A,B$ (with $A$ being of finite presentation over $R$) with prime ideals $mathfrak psubset A$ and $mathfrak qsubset B$ that lie over the same prime ideal $mathfrak rsubset R$ and given a local homomorphism $A_{mathfrak p}to B_mathfrak{q}$ over $R_{mathfrak r}$ (hence over $R$ as well), we need to find $bin Bsetminus mathfrak q$ such that the composition $Ato A_{mathfrak p}to B_{mathfrak q}$ factors over $B_bto B_{mathfrak q}$.

In the linked proof, an appropriate $bin B$ is constructed such that the composition $R[T_1,dots,T_n]twoheadrightarrow Ato B_{mathfrak q}$ factors over $B_bto B_{mathfrak q}$ in such a way that the resulting map $R[T_1,dots,T_n]to B_b$ factors through $A$. Here the fact that there are only finitely many relations comes into play. However, my initial proof circumvents this finiteness condition and I don't seem to find a logical error:

In my proof, I simply write $A=R[a_1,dots,a_n]$ (i.e. I only assume $A$ to be of finite type over $R$) and I observe that the images $frac{b_i}{s_i}$ of the $a_i$ along the map $Ato B_{mathfrak q}$ all lie in $B_{b}$ for $b=prod_i s_i$. Hence the map factors over $B_bto B_{mathfrak q}$, regardless of any finiteness condition with regard to the relations between the $a_i$.

Can anybody point out any error in my proof? Otherwise I don't see why the condition of $Y$ being locally of finite presentation over $S$ at $y$ is needed.