# Isomorphisms of a field into algebraic closure

Let $F$ be any field and $E$ an extension of $F$ with $[E:F]=n$. So $E$ is an algebraic extension and so we can assume that $E$ is inside the algebraic closure $\overline{F}$ of $F$.

Then there are $n$ isomorphisms of $E$ into $\overline{F}$ which fix $F$ elementwise (is this right, independent of characteristic of $F$?). Let these isomoephisms be $\sigma_1,\cdots,\sigma_n$.

Q.1 Suppose $x\in E$ is such that $\sigma_i(x)=x$ for $i=1,2,\cdots,n$. Then should $x$ be in $F$?

If $E$ is Galois extension then answer is yes, but I am considering a variation of that.

Q.2 The isomorphisms $\sigma_1,\cdots,\sigma_n$ not necessarily form a group since $E$ may not be Galois extension. Is the Dedekind's lemma valid, i.e. is it true that $\sigma_1,\cdots,\sigma_n:E\rightarrow \overline{F}$ are independent in the sense that $a_1\sigma_1 + \cdots + a_n\sigma_n=0$ on $E$ ($a_i\in F$) implies $a_1=\cdots=a_n=0$?

I saw Dedekind's theorem in the context where $\sigma_i$'s form a group, but here the situation is general than it.

## 1 Answer

There are $n$ embeddings of $E$ into $\overline F$ fixing $F$ if and only if $E/F$ is a separable extension. If $E/F$ is inseparable, there are fewer.

If $x\in E$, $x\notin F$ and $x$ is separable over $F$ then there are $\sigma_j(x)\ne\sigma_k(x)$. This is because the minimum polynomial of $x$ over $F$ has more than one root in $\overline F$.

Dededkind's lemma works in full generality; there is no need to assume that the $\sigma_j$ form a group.