3
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ Since $M$ is maximal,$\frac {K[x_1,...,X_n]}{M} \cong K$ (Hilbert Nullstellensatz). $\endgroup$ Jan 18, 2016 at 7:17
  • 6
    $\begingroup$ The title, the first line, and the last line are all asking different questions. The first has a not hard answer, the second is false, and the last has a hard answer. Which one are you trying to ask? $\endgroup$ Jan 18, 2016 at 7:18
  • $\begingroup$ I'm trying to understand why algebraically closed fields cannot have non-trivial algebraic field extensions. In other words, if $K$ is an algberically closed field, why is $K[x_1,x_2,\dots,x_n]/M$ isomorphic to $K$, for every maximal ideal $M\subset K[x_1,x_2,\dots,x_n]$? $\endgroup$
    – user67803
    Jan 18, 2016 at 7:25

1 Answer 1

7
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.