Y by
I have the following exercise:
Let
be an algebraic extension of
, and let
be an algebraic closure of
containing
. Prove that if
is an embedding such that
for all
, then
extends to an automorphism of
.
My attempt:
Theorem (*): Suppose that
is an algebraic extension of the field
,
is an algebraically closed field, and
is an embedding. Then, there exists an embedding
that extends
. Moreover, if
is an algebraic closure of
and
is an algebraic extension of
, then
is an isomorphism.
Back to our main problem:
Since
and
is an algebraic extension of
, it follows that
is an algebraic extension of
. Assume that there exists an embedding
such that
for all
. By Theorem (*), there exists an embedding
that extends
. Since
is algebraically closed,
is also an algebraically closed field.
Furthermore, because
for all
and
is an extension of
, we have

This implies that
is an algebraic extension of
. We conclude that
, meaning that
is an automorphism. (Finished!!)
Let choose
be the field of algebraic numbers,
. Consider the embedding
defined by
Then, according to the exercise above,
extends to an isomorphism
How should we interpret
?
Let










My attempt:
Theorem (*): Suppose that











Back to our main problem:
Since












Furthermore, because





This implies that




Let choose






