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Let $E/F$ be an algebraic extension.

Let $\sigma:F\to E$ be a homomorphism in the category of fields.

Can it be shown that the extension $E/\sigma(F)$ is also algebraic?

drhab
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2 Answers2

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No. For example, take $E = F = \mathbb{C}$. I claim there are maps $\sigma : \mathbb{C} \to \mathbb{C}$ making $\mathbb{C}$ a non-algebraic extension of itself. This is because every algebraically closed field of characteristic $0$ and the same cardinality as $\mathbb{C}$ is isomorphic to $\mathbb{C}$, and in particular the algebraic closure of $\mathbb{C}(t)$ has these properties.

Qiaochu Yuan
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  • Thank you. I hoped for a "yes", but the fact that I did not encounter anything about it made me suspect that it would be a "no" :-(. I will need some time to digest your answer. – drhab Jan 27 '16 at 15:39
  • I have chosen to accept the answer of @Lubin. This because he produced a cut and dried counterexample. – drhab Jan 28 '16 at 11:43
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Another example, based somehow on Qiaochu’s: Take any field $k$, and let $E=F=k(t_1,t_2,t_3,\cdots)$, a purely transcendental extension of $k$ of countable transcendence degree. Then we can set $\sigma:F\to E$ by $\sigma(t_i)=t_{i+1}$, but identity on $k$. Then $E$ is a transcendental extension of $\sigma(F)$, with transcendence basis $\{t_1\}$.

I suppose that the question to ask is what the situation may be if $\sigma$ is a $k$-morphism when $E$ and $F$ are finitely generated fields over $k$.

Lubin
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  • Thank you. A nice evidence that - at least when it concerns mathematics - you are not "depleted of merit" at all! – drhab Jan 27 '16 at 17:40