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I have a problem with proposition 3.2.15 of Algebraic geometry and arithmetic curves of Qing Liu: Let $X$ be an integral algebraic variety over $k$. If $X$ is geometrically reduced then $K(X)$ is a finite separable extension of a purely transcendental extension $k(T_1, \ldots , T_d)$.

First we proove that $K(X)$ is a finite extension of a purely transcendental extension $k(X_1,\ldots,X_r)$: I'm ok with that. Let $L=k(X_1,\ldots,X_r)$ and $L_s$ the separable closure of $L$ is $K(X)$.

Then we proove that for all $b\in K(X)\setminus L_s$ such that $b^p\in L_s$, $L_s[b]$ is a finite separable extension of an (other) purely transcental extension $k(Y_1,\ldots,Y_q)$: I'm ok with that.

My problem and question come with the conclusion: "This imply the proposition by decomposing $K(X)/L_s$ into a sequence of purely inseparable extensions of degree $p = \operatorname{char}(k)$." I don't understand that.

I see that $K(X)=L_s[b_1,\ldots,b_m]$ so $L_s[b]$ would be a first step for induction? But for that I need that all the $b_i$ verifiy $b_i^p\in L_s$ and I don't see how to make that (but I know that $K(X)$ is purely inseparable over $L_s$ but it doesn't help me.)

Then I would need to continue with $L_s[b_1,b_2]$ but how to construct a purely transcendal extension $k(Z_1,\ldots,Z_q)$ over that $L_s[b_1,b_2]$ is finite separable?

M Turgeon
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Gabriel Soranzo
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  • Maybe should I ask my question in mathoverflow, but I thought it is for research thema... isn't? – Gabriel Soranzo May 15 '13 at 19:01
  • No, you shouldn't ask this question on MO, they will close it right away. I will make some changes to your title and tags to make clear to other MSE users what you are looking for. Maybe this will help. If you are not satisfied, feel free to go back to the original version. – M Turgeon May 22 '13 at 17:33
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    You know that $L_s[b_1]$ is separable over $k(Y_1,\dots, Y_q)$. So $b_1$ belongs to the separable closure $L_{s,1}$ of $k(Y_1, \dots, Y_q)$ in $K(X)$. Now restart again with $k(Y_1, \dots, Y_q)\subseteq L_{s,1}$ and $b_2$ which has degree $1$ or $p$ over $L_{s,1}$... Hope this helps. Otherwise email to the author. –  May 26 '13 at 19:36

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It's been 11 years but I think I can answer that. As you said, $$K(X)/L_s \text{ is purely inseparable and }K(X)=L_s[b_1,...,b_m].$$ By the definition of purely inseparable, for each $b_i$ there is a power $b_i^{p^n_i}$ such that $b_i^{p^n_i}\in L_s.$

For simplicity let $b$ denote $b_1$ and $n_1=n$ so that $b^{p^n}\in L_s.$ Then for the element $b^{p^{n-1}}$ you have that: $$b^{p^{n-1}}\not\in L_s\text{ and }(b^{p^{n-1}})^p\in L_s.$$

But then you get that $L_s[b^{p^{n-1}}]$ is a finite separable extension of a purely transcendental extension, which we can denote $L'$. And now we start over: If $K(X)/L'$ is separable we're finished. Otherwise take the separable closure $L'_s$ of $L'$ and repeat. Notice that in the new extension, ie $L'_s$, the element $b^{p^{n-1}}$ is now inside.

Since $[K(X):L'_s]\leq [K(X):L_s]<\infty$, in a finite number of steps you will have the following: An extension $\Lambda$ which is finite separable over some purely transcendental extension and $K(X)=\Lambda[b']$ with $(b')^p\in \Lambda$. And therefore again by what you have written, $\Lambda[b']$ will be finite separable over some purely inseparable extension and we're done.

user128787
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