# k rational points and base change

This could be a tricky question but could help me to better understand these very interesting things.

Let $X$ be an algebraic variety over a field $k$ (in the sense of a k-scheme like in Qing Liu), $K$ an extension of $k$ and define the set $X(K)$ to be the set of morphism of $k$-schemes from $Spec K$ to $X$.

Now is well-known and clear that $X(k)$ are precisely the points in $X$ for which $k(x) = k$. Then one can see that there is a bijection from $X(K)$ to $X_K (K)$ (the base change). The problem is that someone told me (and I've also found it in Qing Liu's book) that in general $X(K)$ is not the set point such that $k(x) \subset K$. If I consider $X_K$ as a $K$-scheme then I can use the identification before to get that these points are precisely the one with $k(x) = K$ and corrispond bijectively with $X(K)$.

So there should be an example of a $k$-scheme X where I pick a point $x \in X(K)$ such that $k(x)$ is not contained in $K$ and if I consider the corrispondent point $y$ in $X_K(K)$ via the bijection I'll get a point with $k(y) = K$. Could someone give some hints to find explicitly this example?

Thank you so much

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You should also have in mind the following : if you take $K=\overline{k}$, then there is a bijection between the closed points of $X$ (i.e. those points $x \in X$ such that the extension $k(x)/k$ is algebraic) and the orbits of $X(\overline{k})$ under the action of $\operatorname{Gal}(\overline{k}/k)$. So, you see that there is a natural map $X(\overline{k}) \to X$, but this map is neither injective (two Galois conjugate points have the same image) nor surjective (it misses the non-closed points). – François Brunault May 25 '11 at 15:54
You should do the exercise in Hartshorne that says "describe the underlying topological space of the affine line over the reals, and compare it to the real numbers and the complex numbers". – Kevin Buzzard May 25 '11 at 18:45
@François: thinking with Galois Action makes it all clearer! The key point is then that $X(K)$ are not in general "points" in the variety $X$ but indeed are equivalency classes. So is meaningless thinking of $k(x)$ copared to $k(y)$ with $y\in X_K$. – Srks May 26 '11 at 13:57
"is well-known and clear that $X(k)$ are the points in $X$ for which $k(x)=k$". Well, it's false. See other comments and answers. – ACL May 27 '11 at 17:32
@ACL - "is well-known and clear that $X(k)$ are the points in $X$ for which $k(x)=k$": here we are assuming that $X$ is a algebraic variety over $k$ and thus this is proven in a straitforward way (see for example Qing Liu). – Srks May 30 '11 at 6:54

Let $X$ be a $k$-scheme and $K/k$ be a field extension. Then $X(K)$ can be identified with the pairs $(x,h)$, where $x \in X$ and $h : k(x) \to K$ is a $k$-homomorphism. Remark that $h$ is not uniquely determined and in general it makes no sense at all to ask whether $k(x) \subseteq K$ or not, because these fields don't lie in some common bigger field.
the homomorphism $h$ is the point! Is it a field homomorphism? then It should be injective and then you'll get $k(x) \subset K$. Am I right? – Srks May 25 '11 at 12:55
Yes it is a field homomorphism (it is optained from $\text{Spec}(K) \to X$ by looking at the residue fields) and thus it is injective, but it is important to remember that this is not a canonical embedding. – Martin Brandenburg May 25 '11 at 12:57
Thank you! to summerize: a point of $X(K)$ has not in general the property of $k(x)\subset K$ but there exists a subfield of $K$ which is isomorphic to $k(x)$. Now when passing to $X_K$, one of the corrisponding point $y$ via the bijection $X(K) \simeq X_K(K)$ is such that $k(y)$ is contained in $K$ via a canonical embedding. – Srks May 25 '11 at 13:19