In theorem A3.5 of Ash's book Abstract Algebra: The Basic Graduate Year (page 20 in this pdf), the author set out to prove the following.

Let $\sigma: F \rightarrow L$ be a field monomorphism where $L$ is algebraically closed. If $E/F$ is algebraic, then $\sigma$ extends to a monomorphism $\tau: E \rightarrow L$.

Here is the full statement and proof:

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The proof is essentially an application of Zorn's lemma. However, the author never explicitly explained why it is necessary for $L$ to be algebraically closed.

Here is a counter-example with the condition removed: Stack $E/L/F$ into a chain of extensions where $E$ is algebraically closed and a proper extension of $L$. As a result, we cannot have any monomorphism from $E$ into $L$.

Now, how do I weave this observation into the proof?

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    $\begingroup$ Well, I would look into (A3.4). $\endgroup$ – Kevin Carlson Aug 2 '15 at 23:08

Take a look at A3.4:

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The fact that $L$ is algebraically closed implies that the number of distinct roots of the polynomial $f\in F[x]$ in an algebraic closure of $F$ is equal to the number of distinct roots of $\sigma f\in L[x]$ in $L$.

  • $\begingroup$ I did not entirely understood A3.4 and skipped it. On second look, it seems that we want $L$ to be algebraically closed so that the image of the extension is in $L$. If $\beta_1, \beta_2 \ldots \beta_m$ are the distinct roots of $\sigma f$, then $F(\alpha) \cong (\sigma F)(\beta_i)$ where $i \in \{ \, 1, 2 \ldots m \, \}$. This gives us the $m$ extensions associated with $\alpha \in E$. If $L$ is not algebraically closed, then some of the extensions may not land on $L$. I still need to stew over the details, but the big picture looks right, TY .... $\endgroup$ – Andy Tam Aug 3 '15 at 14:37

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