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?
Best Answer
A field extension $F\supseteq K$ is called algebraic if every element of $F$ satisfies a polynomial with coefficients in $K$. In particular, for any $\alpha\in F$, there is then a polynomial $p(x)\in K[x]$ of minimal degree such that $p(\alpha)=0$, and then $p(x)$ is irreducible (since if $p(x)=f(x)g(x)$, then $f(\alpha)=0$ or $g(\alpha)=0$ so by minimality of $p$ one of them must be a constant). So if $K$ is algebraically closed, every element of $F$ satisfies a linear polynomial, which means every element of $F$ is actually in $K$. So any algebraic extension of an algebraically closed field is trivial.
However, there is no reason to believe a priori that an extension of the form $K[x_1,\dots,x_n]/M$ is algebraic. The fact that any such extension is algebraic (and hence trivial if $K$ is algebraically closed) is a hard theorem, known as the Nullstellensatz.