Normal transcendental extension

According to Wikipedia, normal extension are assumed to be algebraic. But one of the definitions

$K/k$ is normal if any $k$-embedding $\sigma : K \rightarrow \Omega$ of $K$ into a fixed algebraic closure $\Omega$ of $K$ does not leave $K$ i.e. $\sigma(K) = K$

is applies perfectly to any field. What happens to transcendental extension $K/k$ that satisfies this property? Are they all essentially algebraically closed?

EDIT: Thanks to @Joanpemo, I realize that I made a mistake in the definition of normal extension: $\Omega$ is supposed to be algebraic closure of the smaller field $k$, not of $K$. Nevertheless, the question is still applicable: We can still ask for transcendental extension $K/k$ satisfying my wrong definition.

• I doubt that a transcendental extension can be embedded into an algebraic one ... – DonAntonio Apr 12 '16 at 23:02
• @Joanpemo Since $\Omega$ is algebraic closure of $K$, it should be possible to embed $K$ into $\Omega$. For instance, take $k = \mathbb{Q}$, $K = \mathbb{R}$ and say $\Omega = \mathbb{C}$ as algebraic closure of $\mathbb{R}$. But of course, in $\mathbb{C}$, we can "move" $\mathbb{R}$ while fixing $\mathbb{Q}$. – An Hoa Apr 12 '16 at 23:04
• As far as I know an algebraic closure of a field is an algebraic extension which is algebraically closed. Perhaps you're confusing with field algebraically closed? – DonAntonio Apr 12 '16 at 23:07
• @Joanpemo Well, $\Omega/K$ is algebraic by definition of algebraic closure but $K/k$ might well be transcendental as in my previous comment. I edited the question to reflect this. – An Hoa Apr 12 '16 at 23:13
• Oh, now I see! You have a mistake in your definition of "normal": $\;K/k\;$ is normal if every embedding $\;K\to\Omega\;,\;\;\Omega$ an alg. closure of the field $\;k\;$ , not $\;K\;$...and which contains $\;K\;$ . So definitely $\;K/k\;$ cannot be transcendental – DonAntonio Apr 12 '16 at 23:20