I was thinking about that:
Suppose we have two polynomial rings over an arbitrary field $K$, $R=K[x_1,...,x_n]$ and $S=K[y_1,...,y_c]$, and $T=[x_1,...,x_n,y_1,...,y_c]$.
Now let $I$ be an ideal in $R$ and $g=(g_1,...,g_c)$ a list of polynomials in $R$.
Consider the map $r:T\longrightarrow R/I$ which maps $x_i\mapsto\bar{x_i}$ and $y_j\mapsto\bar{g_j}$.
I think $\ker(r)$ is $IT+(y_1-g_1,...,y_c-g_c)$, but I can't prove the inclusion "$\subseteq$".
Any suggestion please?
I add some details to be more precise:
1) The bar over an element means the class in the quotient.
2) Indices $j$ and $i$ runs over $1,...,c$ and $1,...,n$, respectively.
3) $IT$ is the extension of the ideal $I$ in the ring $T$ (which contains $R$).
I have tried to decompose the map $r$ as the composition of $\pi$, the projection in the quotient $R/I$, and the map $r{'}$ which maps $x_i\mapsto x_i$ and $y_i\mapsto g_i$, so I get $F\in\ker(\pi\circ r_1)$ iff $r_1(F)\in I$, but I don't know how to prove this fact.
The problem is that I don't know how to prove $r_1(F) \in I$ iff $F$ is $0$ or $F\in IT$. I know that $\ker(r_1)$ is $(y_1-g_1,...,y_c-g_c)$.
If you are given an element in the kernel you can try to use the Euclidean algorithm to split of multiples of $y_j-g_j$ in order to reduce to the case where your element actually lies in $R$.