It's well known that if $E/k$ is an algebraic extension, then $End(E/k) = Aut(E/k)$. But, what about the other implication?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
Counterexample:
The field of real numbers $\mathbb{R}$ is not algebraic over the rationals $\mathbb{Q}$, but the $\mathbb{Q}$-algebra $\mathbb{R}$ has only the identity endomorphism (hence, a fortiori, only the identity automorphism).
answered Oct 13, 2019 at 0:00
-
$\begingroup$ Perhaps Im confused by notation or something else is implied here, but you are comparing the endomorphisms with the automorphisms ON the set $\mathbb{R/Q}$ - the set which is the subject of discussion - so where are you getting the trivial group containing only the identity and how is that relevant to a discussion on $\mathbb{R/Q}$? $\endgroup$ Oct 13, 2019 at 0:49
-
-
$\begingroup$ Hmm. Are endomorphisms covered there? $\endgroup$ Oct 13, 2019 at 5:26
-
$\begingroup$ Indeed, the notation "$E/k$" is ambiguous since it could also refer to a quotient vector space. Of course, a nonzero vector space always has the zero endomorphism as an endomorphism that is not an automorphism. I have fixed the answer to no longer use the ambiguous notation. $\endgroup$ Oct 13, 2019 at 13:58