Derivations and algebraic extensions in characteristic zero

Question

Let $k\subseteq L$ be fields of characteristic zero. Show that the following conditions are equivalent:

(1) $L$ is algebraic over $k$.

(2) For every derivation $d$ of $k$ there exists a unique derivation $D$ of $L$ such that $D_{|k} = d$.

(3) If $\delta$ is a $k$-derivation of $L$, then $\delta = 0$.

Answer 1

$(1)\implies (2)$: Assume $L/k$ is algebraic and let $d\colon k\to k$ be a derivation. Let $\alpha\in L$ and $f(X)=\sum_{j=0}^n a_j X^j\in k[X]$ its minimal polynomial. Then for any derivation $D\colon L\to L$ with $D|_k=d$ it is necessary that $$\begin{align}0&=D(0)\\&=D(f(\alpha))\\&=D\left(\sum_{j=0}^n a_j\alpha ^j\right)\\&=\sum_{j=0}^n D(a_j\alpha^j)\\ &=\sum_{j=0}^n (d(a_j)\alpha^j+ja_j\alpha^{j-1}D(\alpha))\\ &=\sum_{j=0}^n d(a_j)\alpha^j + D(\alpha)\sum_{j=0}^{n-1}(j+1)a_{j+1}\alpha^j\\ &=d(f)(\alpha)+D(\alpha)f'(\alpha)\end{align}$$ As $\deg f'=\deg f-1$ (we are using characteristic zero here!), we must have $f'(\alpha)\ne 0$ and conclude that $$\tag1D(\alpha)=-\frac{d(f)(\alpha)}{f'(\alpha)}.$$ This shows uniqueness of $D$.

On the other hand, for fixed $\alpha$ consider the $k$-linear map $\delta\colon k[X]\to L$, $g\mapsto d(g)(\alpha)+g'(\alpha)D(\alpha)$ (with $D(\alpha)$ given by $(1)$). We have $\delta(a)=d(a)$ for all $a\in k$, and $\delta(f)=0$, and $$\tag2\begin{align}\delta(gh)&=d(gh)(\alpha)+(gh)'(\alpha)D(\alpha)\\&=d(g)(\alpha)h(\alpha)+g(\alpha)d(h)(\alpha)+(g'(\alpha)h(\alpha)+g(\alpha)h'(\alpha))D(\alpha)\\&=\delta(g)h(\alpha)+\delta(h)g(\alpha)\end{align}$$ so that $\delta$ is zero for all multiples of $f$, which means that it factors over $k(\alpha)$, giving us a linear map $\tilde \delta\colon k(\alpha)\to L$ with $\tilde\delta|_k=d$ and that is a derivation because of $(2)$. By the usual Zorn's lemma argument, we find that there exists a derivation on $L$ that extends $d$.

$(2)\implies (3)$: $\delta$ and the trivial derivation on $L$ are both extensions of the trivial derivation on $k$. By uniqueness, we conclude $\delta=0$.

$(3)\implies (1)$: Let $K=\bar k\cap L$ and assume $K\ne L$, say $\alpha\in L\setminus K$ is transcendental. Define $\delta\colon K(\alpha)\to L$ by $\delta(\frac{f(\alpha)}{g(\alpha)})=\frac{f'(\alpha)g(\alpha)-f(\alpha)g'(\alpha)}{g(\alpha)^2}$, note that this can again be extended to all of $L$ and id zero on $k$. From $(3)$ and $\delta(\alpha)=1$, arrive at a contradiction.