The question is: What is the algebraic machinery/reasoning behind the following intuition? Given a reduced curve over some field $k$ and a regular function on each of its components such that those functions are zero at every intersection point. Then we should be able to find a globally regular function (via extension by zero) that restrics to each of the given ones on the components.
Put more concretely: Let $X = \operatorname{Spec}(R)$ be a reduced scheme of dimension one over some field $k$. Let $P_1,\ldots,P_m$ be its minimal prime ideals corresponding to the irreducible components $X_i = \operatorname{Spec}(R/P_i)$ of $X$.
Now consider that we have given functions on $f_i + P_i \in R/P_i$ on each of the $X_i$ such that all of them are zero on all intersection points, i.e. for every maximal ideal $P$ of $R$ containing more than one of the $P_i$, we have $(f_i + P_i) \in P(R/P_i)$, that is $f_i \in P$. Then we should be able to find a preimage of $(f_i+P_i)_{i=1,\ldots,m}$ under the canonical injective map $$R \to R/P_1 \oplus \ldots \oplus R/P_m$$ providing $f \in R$ with $f + P_i = f_i + P_i$.
Is this correct and if yes, then how do you prove it?
Many thanks in advance!
After some thought and the help of a friend much smarter than myself, we finally came up with a counterexample. Consider a field $k$ and the ring $R=k[x,y]$ with the ideals $P_1=(x)$, $P_2=(y)$, and $P_3=(x-y)$. We will identify $R/P_1 = k[y]$, $R/P_2=k[x]$ and $R/P_3=k[z]$, where in mapping $R\to R/P_i$ is given by $f\mapsto f(0,y)$, $f\mapsto f(x,0)$ and $f\mapsto f(z,z)$, respectively. Now consider the functions $$ (0, 0, z) \in k[x]\times k[y]\times k[z] $$ All of these functions are zero on their only intersection point, but it is easy to see that they can not have a common preimage in $k[x,y]$. Indeed, assume that $f(x,y)$ is such a preimage. Since $f(x,0)=0$, we know that every monomial in $f$ must contain the factor $x$. Since $f(0,y)=0$, we also know that every monomial in $f$ must contain the factor $y$. In particular, every monomial in $f$ has degree at least $2$. Hence, it is impossible to have $f(z,z)=z$.