Definition and some theorems on separable extensions

Question

Theorem 4.3. Let $E$ be a finite extension of $k$. Then $E$ is separable over $k$ if and only if each element of $E$ is separable over $k$.

Definition: Let $E$ be an arbitrary algebraic extension of $k$. We define $E$ to be separable over $k$ if every finitely generated subextension is separable over $k$, i.e. if every extension $k(\alpha_{1},\dots,\alpha_{n})$ with $\alpha_1,\dots,\alpha_n\in E$ is separable over $k$.

Theorem 4.4. Let $E$ be an algebraic extension of $k$, generated by a family of elements $\{\alpha_i\}_{i\in I}$. If each $\alpha_i$ is separable over $k$ then $E$ is separable over $k$.

Proof: Every element of $E$ lies in some finitely generated subfield $k(\alpha_{i_1},\dots,\alpha_{i_n}),$ and as we remarked above, each such subfield is separable over $k$. Hence every element of $E$ is separable over $k$ by Theorem 4.3, and this concludes the proof.

This is the excerpt from Lang's book and in my opinion the proof of Theorem 4.4 is weird and possibly wrong.

My approach was the following: We have to show that $E$ is separable over $k$. By definition we have to show that each finitely generated subextension is separable over $k$. Take such subextension and call it WLOG $k(\alpha_1,\dots,\alpha_n)$. This is finite extension of $k$ and here we can apply Theorem 4.3. But inorder to get desired result we have to show that each element of this subfield is separable over $k$. But can anyone show how to prove this last step?

Would be very grateful for help!

EDIT: Let $E$ be an algebraic extension of $k $, possibly infinite extension. If $E$ is separable over $k$. Could we claim that any element $\alpha\in E$ is separable over $k$?

We see that in the case of finite extension this is true (Theorem 4.3.) What about infinite extension?

Answer 1

Your approach is same as Lang's proof. You have taken the subfield $k(\alpha_1,\dots,\alpha_n)$. Note that every element of $k(\alpha_1,\dots,\alpha_n)$ is in $E$ and every element of $E $ is separable over $k $ by assumption. In particular, every element of $k(\alpha_1,\dots,\alpha_n)$ is separable over $k $.Also note that $k(\alpha_1,\dots,\alpha_n)$ is a finite extension since it is generated by finitely many algebraic elements(you may need to prove this). Now we can use Theorem 4.3 to conclude that $k(\alpha_1,\dots,\alpha_n)$ is separable over $k$, which in turn implies $E $ is separable over $k $.

Your second claim is true. If $E$ is an arbitrary extension of $k $ which is separable over $k $, then $k(\alpha_1,\dots,\alpha_n)$ is separable over $k$(defintion) . In particular, if you take any $\alpha\in E $, consider the subfield $k(\alpha)$ which is separable over $k$. Note that $k(\alpha) $ is finite since $\alpha$ is algebraic. Now by Theorem $4.3$, we can conclude that $\alpha$ is separable over $k$.

Answer 2

The key here is that we are going from an extension generated by finitely many separatble elements to an extension generated by a possibly infinite collection of separable elements, the $\alpha_i$ with $i\in I$.

So, say that $E$ is generated by the $\alpha_i$. If $a\in E$, then there is a finite subcollection of $\alpha_i$s, say $\alpha_{i_1},\ldots,\alpha_{i_n}$, such that $a$ is in fact in $k(\alpha_{i_1},\ldots,\alpha_{i_n})$. Now, since we have already proven that a finite extension given by finitely many separable elements has the property that every element in the extension is separable, it follows that $a$ is separable over $k$ (when viewed as an element of the finite, finitely generated by separable elements extension $k(\alpha_{i_1},\ldots,\alpha_{i_n})$ over $k$), and so is separable, period.

Thus, every element of $E$ is separable over $k$. Therefore, $E$ is separable over $k$.

What you seem to have missed is that the $\alpha_{i_j}$ in the proof are not arbitrary elements, they are elements taken from the family of generators of the extension that are already known to be each separable.