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I'm reading the article "Integration in Finite Terms" by Maxwell Rosenlicht and I have a problem with one step in a proof. Rosenlicht wants to prove the following: If $F$ is a differential field of characteristic zero and $K$ an algebraic extension field of $F$, then the derivation on $F$ can be extended to a derivation on $K$ and this extension is unique. After proving uniquess, Rosenlicht then continues as follows: "We now show that such a [differential field] structure on $K$ exists. Using the usual field-theoretic arguments, we may assume that $K$ is a finite extension of $F$, so that we can write $K=F(x)$, for a certain $x\in K$."

Not being an expert in the theory of fields, I don't understand which "usual" arguments he's talking about. An algebraic extension isn't necessary finite, so why can we assume this here?

Frunobulax
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Let $K/F$ be an algebraic extension and let $\mathscr F$ denote the set of finite subextensions of $K/F$ that's the set of subfields $E$ of $K$ containing $F$ such that $E/F$ is a finite extension.

For all $E\in\mathscr F$ your text proves the existence of one and only one derivation $d_E:E\to E$ extending that given on $F$. Since $K=\bigcup\mathscr F$, there exists one (and only one) function $d:K\to K$ such that $d|E=d_E$ for all $E\in\mathscr F$ and this is the required derivation.

Fabio Lucchini
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  • Thanks. I briefly thought about an argument along these lines but dismissed it for the wrong reasons. I now understand that it is perfectly valid. – Frunobulax Aug 21 '18 at 08:04
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    Now that I think of it, the proof could have been worded a tad simpler by directly talking about simple (instead of finite) extensions. If we have a derivative for every $F(x)$ for $x\in K\setminus F$, then due to uniqueness we have one for $K$ because $K$ is the union $\bigcup_{x\in K\setminus F} F(x)$. Right? – Frunobulax Aug 21 '18 at 10:54
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    Yes, you can argue with simple extensions directly. – Fabio Lucchini Aug 21 '18 at 13:17
  • But I was a bit too sloppy. If you argue with simple extensions, uniqueness guarantees that the union of all your derivatives is a function, but that doesn't necessarily mean that it is a derivation. To show that it is a derivation, you need to be able to pick two arbitrary elements from $K$ and combine them. But why should there be an $F(x)$ that contains both? That's why the theorem needs characteristic zero (or, as Rosenlicht points out, separability) - at this point you invoke the primitive element theorem. So, the proof isn't actually shorter this way. – Frunobulax Aug 21 '18 at 13:37
  • As you pointed out, you have to invoke the primitive element theorem to show that the function $d$ is indeed a derivation. – Fabio Lucchini Aug 21 '18 at 15:06