I'd really appreciate if someone could help me.
The problem is the following:
Let $\psi_1,...,\psi_m \in k[x_1,\dots,x_n]$ and consider the $k$-algebra homomorphism: $$\phi:k[x_1,...,x_n,y_1,...,y_m] \rightarrow k[x_1,...,x_n],\quad y_j \mapsto \psi_j,\quad x_i \mapsto x_i.$$ Then prove that $\ker(\phi)=I:= \langle y_1-\psi_1,...,y_m-\psi_m \rangle$.
My attempt:
$I \subseteq \ker(\phi)$, because $$x\in I \Rightarrow x=\sum _{j=1}^m \lambda_j(y_j-\psi_j)\Rightarrow \phi(x)=\sum _{j=1}^m \lambda_j(\phi(y_j)-\phi(\psi_j))=\sum _{j=1}^m \lambda_j(\psi_j-\psi_j)=0.$$ I consider the following lexicographic order $y_1>y_2>\cdots>y_m>x_1>\cdots>x_n$.
Let's suppose that generators of $I$ form a Gröbner basis with this order.
Then the normal forms $\pmod I$ are polynomials in $k[x_1,...,x_n]$.
Now considering the inclusion $j$ of $k[x_1,...,x_n]$ into $k[x_1,...,x_n,y_,...,y_m]$, I prove that $\phi \circ j=\mathrm{Id}_{k[x_1,...,x_n]}$, so $k[x_1,...,x_n]\cap \ker(\phi)=0$ so $$k[x_1,...,x_n,y_,...,y_m]/I \twoheadrightarrow k[x_1,...,x_n,y_,...,y_m]/\ker(\phi)$$ is both surjective and injective, and then $I=\ker(\phi)$.
How could I prove what I've supposed it's true?
Thank you in advance.
Set $R=k[x_1,\ldots,x_n,y_1,\ldots,y_m]$. Denote the generators of $I$ by $g_i = y_i - \psi_i$. Then, in the lexicographic order you chose, the leading monomial of $g_i$ is $\operatorname{LM}(g_i) = y_i$. Because these monomials are pairwise coprime, the tuple $(g_1, \ldots, g_m)$ is a Grobner basis.