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At the end of the proof of Proposition 4.3.30 In Liu's book we have the following situation: $X$ is an algebraic variety over a field $k$, $x\in X$ is a regular closed point of $X$ with $k'=k(x)$ is a separable extension of the base field $k$.

If I understand well I then have to show that the points $x'\in X_{k'}=X\times_k\operatorname{Spec}(k')$ lying over $x$ are $k'$-rationnals.

I've seen somewhere that $x'$ is closed, but $k'$ is not algebraically closed in general I can't conclude that $x'$ is rationnal...

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Take $X=\mathbf{A}_k^1$. Let $k^\prime$ be any finite separable extension of $k$, and choose a primitive element, i.e., a $k$-algebra isomorpjhism $k^\prime\simeq k[T]/(f(T))$ for some irreducible monic $f(T)\in k[T]$. We have $X\times_kk^\prime=\mathbf{A}_{k^\prime}^1$, and the points lying over $x=(f(T))\in\in X$ are those prime ideals $(g(T))\in k^\prime[T]$ such that $(g(T))\cap k[T]=(f(T))$. In particular one must have $f(T)\in g(T)k^\prime[T]$, so we have an injection $k^\prime=k[T]/(f(T))\hookrightarrow k^\prime[T]/(g(T))=k^\prime(x^\prime)$ (for $x^\prime=(g(T))$). This is just the induced map of residue fields for $x^\prime\mapsto x:\mathbf{A}_{k^\prime}^1\to\mathbf{A}_k^1$. But in order for this map to be an isomorphism, we must have $\deg(g(T))=1$, and there's no reason this has to be the case.

As an example, consider $k=\mathbf{Q}$, $f(T)=X^3-2$. Then $k^\prime=k(\alpha)$ is a non-Galois extension of $k$ of degree $3$ ($\alpha$ a root of $X^3-2$). Over $k^\prime$, we have $f(T)=(X-\alpha)g(T)$ where $g(T)$ is an irreducible quadratic in $k^\prime[T]$. Now, we have $f(T)\in(g(T))\cap k[T]$, and since $(f(T))$ is maximal and $(g(T))\cap k[T]$ must be a proper ideal, $x^\prime=(g(T))\in\mathbf{A}_{k^\prime}^1$ lies over $x=(f(T))\in \mathbf{A}_k^1$. But the residue field of $x^\prime$ is a degree $6$ extension of $\mathbf{Q}$, and in particular is not equal to $k^\prime$.

EDIT: Here's a (perhaps) simpler way of looking at this example. Let $k$ be any field, $f(T)\in k[T]$ an irreducible separable polynomial of degree $n>1$ such that $k^\prime=k[T]/(f(T))$ is not Galois, i.e., $f(T)$ does not split. Then, in $k^\prime[T]$, we can write $f(T)=f_1(T)\cdots f_r(T)$ for irreducibles $f_i$ in $k^\prime[T]$, where $2\leq r<n$ because $f$ has a root in $k^\prime$ but doesn't split over $k^\prime$. In particular, some $f_i$, say $f_1$ after re-ordering, has degree bigger than $1$. Consider $X=\mathrm{Spec}(k^\prime)$, a smooth $k$-scheme with a unique ($k^\prime$-rational) point $x$. The base change $X_{k^\prime}$ is the spectrum of $k^\prime\otimes_kk^\prime=k^\prime[T]/(f(T))=\prod_{i=1}^r k^\prime[T]/(f_i(T))$, a product of finite separable extensions of $k^\prime$, one of which (namely $k^\prime[T]/(f_1(T))$) is not equal to $k^\prime$. This factor gives a point $x^\prime\in X_{k^\prime}$ with residue field not equal to $k^\prime$, and of course, all points of $X_{k^\prime}$ lie over the unique point of $X=\mathrm{Spec}(k^\prime)$.

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  • $\begingroup$ The proposition of Liu is extactly so: Let $X$ be an algebraic variety over $k$, and let $x\in X$ be a closed point. Let us suppose that $X$ is smooth at $x$; then $X$ is regular at $x$. Moreover, the converse is true if $k(x)$ is separable over $k$. My question is about the converse. He first treats the case of rational points and then the general case by telling that the points of $X_{k'}$ over $x$ are rationnal (and still regular closed).... So I don't se where is hidden $\deg(g(T))=1$? $\endgroup$ – Macadam May 20 '14 at 20:38
  • $\begingroup$ Dear @Gabriel, Liu's proof is flawed, as my example shows. The result is true, see stacks.math.columbia.edu/tag/00TV. $\endgroup$ – Keenan Kidwell May 20 '14 at 21:08

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