injectivity($n=1$)

Lemma 16   $\phi:A_n(k)\to A_n(k)$ be a $k$-homomorphism. Then $trans.deg(\phi(\xi_1^p),\dots,\phi(\xi_n^p))$ is equal to $n$.

[proof] We may assume that $k$ is algebraically closed. The field $k(\xi_1,\dots,\xi_n)$ has $n$ linearly independent $k$-derivations $\{\operatorname{ad}(\eta_i)\}_{i=1}^n$. Since $k(\xi_1,\dots,\xi_n)$ is separable over $k$, its transcendent degree is equal to the number of linear independent derivations [3, Theorem 4.4.2.]. Thus we conclude that transcendent degree of $k(\xi_1,\dots,\xi_n)$ is no less than (hence is equal to) $n$. That means, $\xi_1,\dots,\xi_n$ are algebraically independent over $k$.

Lemma 17 ($n=1$)   Any $k$-algebra endomorphism $\phi:A_1(k) \to A_1(k)$ is injective.

[proof] We may assume that the base field $k$ is algebraically closed. $\phi(A_1(k))$ has no zero-divisor except for 0. Thus

$$\phi(A_1(k))\otimes_{\phi(Z(A_1(k))} Q(\phi(Z(A_1(k)))$$

is a skew field which is of finite rank over $Q(\phi(Z(A_1(k))))$. If the transcendent degree of the field $K$ is $1$, then it contradicts with Tsen's theorem. Thus $\phi(\xi^p),\phi(\eta^p)$ are algebraically independent over $k$.

$\qedsymbol$

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