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?
2026-04-03 06:49:19.1775198959
End = Aut $\Rightarrow$ algebraic?
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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).