Let $K$ be an algebraically closed field. Then there are no non-trivial algebraic field extensions of $K$.
I can understand that if the field extension is of the form $K[x]/\langle p(x)\rangle$, where $p(x)\in K[x]$ is an irreducible polynomial, then as $K$ is algebraically closed, every polynomial in $K[x]$ splits. Hence, $p(x)$ has to be of degree $1$. This causes $K[x]/\langle p(x)\rangle$ to be equal to $K$.
However, how does the field being algebraically closed make the following field extension trivial: $K[x_1,x_2,\dots,x_n]/M$, where $M\subset K[x_1,x_2,\dots,x_n]$ is a maximal ideal?