How to show that $ \left(k[\textbf{x}]/I\right)\otimes_k\left(k[\textbf{y}]/J\right)\cong k[\textbf{x},\textbf{y}]/(I^e+J^e)$?

Question

Consider the polynomial rings $k[\textbf{x}]:=k[x_1,...,x_m]$ and $k[\textbf{y}]:=k[y_1,...,y_n]$ where $m,n\in\mathbb{Z}_{>0}$ and let $I\subset k[\textbf{x}]$ and $J\subset k[\textbf{y}]$ be radical ideals. Then if we denote $k[\textbf{x},\textbf{y}]:=k[x_1,...,x_m,y_1,...,y_n]$, we have natural injections $k[\textbf{x}]\to k[\textbf{x},\textbf{y}]$ and $k[\textbf{y}]\to k[\textbf{x},\textbf{y}]$. Denote by $I^e$ resp. $J^e$ the extension of $I$ resp. $J$ along these injections. Then how can one show that there is an isomorphism of $k$-algebras $$ \left(k[\textbf{x}]/I\right)\otimes_k\left(k[\textbf{y}]/J\right)\cong k[\textbf{x},\textbf{y}]/(I^e+J^e)? $$ I am able to construct a morphism of $k$-algebras $$ \left(k[\textbf{x}]/I\right)\otimes_k\left(k[\textbf{y}]/J\right)\to k[\textbf{x},\textbf{y}]/(I^e+J^e)\\ \overline{x_i}\otimes 1\mapsto \overline{x_i},\quad 1\otimes\overline{y_j}\mapsto\overline{y_j} $$ by using the universal properties involved, where the overline denotes the respective class of the element underneath. Also, by the universal property of the polynomial ring we have a $k$-algebra homomorphism $$ k[\textbf{x},\textbf{y}]\to\left(k[\textbf{x}]/I\right)\otimes_k\left(k[\textbf{y}]/J\right)\\ x_i\mapsto \overline{x_i}\otimes 1,\quad y_j\mapsto 1\otimes\overline{y_j}. $$ But how can one show that the kernel of this homomorphism is given by $I^e+J^e$? That the latter is included in the former is straightforward, but I have no idea for the reverse inclusion.