Finding the kernel of a given map

Question

Let $\mathbb K$ be an algebraic closed field and let $f,g\in\mathbb K[X,Y]$ be two polynomials so that $V_{\mathbb K}(f)$ and $V_{\mathbb K}(g)$ don't have common irreductible components and $(0,0)\in V_{\mathbb K}(f)\cap V_{\mathbb K}(g)$. Let $$ \left<f,g\right>=I_1\cap\dots\cap I_r $$ be the primary decomposition of the ideal so that $V_{\mathbb K}(I_1)=(0,0)$. We have the following map: $$ \frac{\mathbb K[X,Y]}{\left<f,g\right>}\overset{\varphi}{\longrightarrow}\frac{\mathbb K[[X]][Y]}{\left<f,g\right>_*} $$ $$ h+\left<f,g\right>\mapsto h+\left<f,g\right>_* $$ where $\mathbb K[[X]]$ denotes the formal power series ring and $\left< f,g\right>_*$ denotes the ideal in $\mathbb K[[X]][Y]$ generated by $f,g$. I'd like to prove that: $$ \ker\varphi=\frac{I_1}{\left<f,g\right>} $$ Note: I've already proved the inclusion $\frac{I_1}{\left<f,g\right>}\subseteq \ker\varphi$. In fact, if we write the resultant of $f,g$ with respect to $Y$, as $(0,0)$ is a common root: $$ \mathrm{res}_Y(f,g)=X^su(X)\in\left<f,g\right> $$ with $u(0)\neq0$. In particular, $u(X)\in I_2\cap\dots\cap I_r$. So, let $h\in I_1$. It is clear that $hu\in\left<f,g\right>$: \begin{multline} 0+\left<f,g\right>_*=\varphi(0+\left<f,g\right>)=\varphi(hu+\left<f,g\right>)=\varphi(h+\left<f,g\right>)\varphi(u+\left<f,g\right>)=\\ =(h+\left<f,g\right>_*)(u+\left<f,g\right>_*)=hu+\left<f,g\right>_*\;\Rightarrow\;hu\in\left<f,g\right>_* \end{multline} and, as $u(X)$ is a unit in $\mathbb K[[X]]$, we have that $h\in\left<f,g\right>_*$, proving the inclusion. For the other inclusion I've thought about using that $I_1$ is a primary ideal, but I couldn't find a good solution.