End = Aut $\Rightarrow$ algebraic?

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?

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).
• 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 $\mathbb{R/Q}$? Oct 13, 2019 at 0:49
• 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. Oct 13, 2019 at 13:58