I consider a variety over a field $k$, i.e. an integral separated scheme $X$ of finite type over $k$.

One knows by the Nullstellensatz that any closed point on $X$ is a $\bar k-$ rational point (where $\bar k$ denotes the algebraic closure of $k$)as its residue field is finite over $k$.

I know wonder what one can say about the relation between the closedness of a point and its residue field. E.g. it wont hold that any $\bar k-$rational point is closed, but can one say something similar? Or how can one characterize the closed points?

And does the situation change if one assumes the variety furthermore as complete over $k$?

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    $\begingroup$ Dear Descartes, it doesn't really make sense to say that a closed point on $X$ "is" a $\bar k$- rational point . If you want, I'll write a more detailed answer later , but I must run now! $\endgroup$ Commented Sep 10, 2011 at 15:59
  • $\begingroup$ Every rational point is closed... $\endgroup$
    – Matt
    Commented Sep 10, 2011 at 16:25
  • $\begingroup$ Why not, Georges? I thought a $ \bar k$-rational point (which is a morphism of $Spec(\bar k)$ to $X$) is equivalent to giving a point with residue field contained in $\bar k$. $\endgroup$
    – Descartes
    Commented Sep 10, 2011 at 17:08
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    $\begingroup$ A $\bar{k}$-rational point of $X$ (by definition a morphism of $k$-schemes $\mathrm{Spec}(\bar{k})\rightarrow X$) is equivalent to the data of a point $x$ of $X$ and a $k$-monomorphism $k(x)\rightarrow\bar{k}$. $\endgroup$ Commented Sep 10, 2011 at 21:25
  • $\begingroup$ @Keenan. A perfect description: nothing to add or subtract! $\endgroup$ Commented Sep 10, 2011 at 21:29

2 Answers 2


The closed points of a finite type $k$-scheme are precisely the points with residue extension $k(x)/k$ algebraic (equivalently finite). The residue field of a closed point is a domain that is finitely generated as a $k$-algebra, also a field, hence a finite extension of $k$ by (a form of) the Nullstellensatz. For the other implication, assume $X=\mathrm{Spec}(A)$ with $A$ a finitely generated $k$-algebra. If $\mathfrak{p}\in X$ is such that $k(\mathfrak{p})/k$ is algebraic, then $k(\mathfrak{p})$ is integral over the domain $A/\mathfrak{p}$ which is therefore a field, i.e., $\mathfrak{p}$ is maximal.

  • $\begingroup$ Perfect, thanks a lot for the explanation, Keenan! $\endgroup$
    – Descartes
    Commented Sep 10, 2011 at 18:10

A scheme (over a field $k$, say) really has two sorts of points and much confusion arises from the fact that they are not distinguished linguistically. For clarity's sake I'll call them (just here and now!) physical and functorial points.

The physical points They are elements of the underlying set $|X|$. Such an $x\in |X|$ has a residual field $\kappa (x)$ which is an extension $k \to \kappa (x) $. If that extension is an isomorphism, we say that $x$ is rational or $k$-rational.

The functorial points They are $k$ -morphisms from some $k$-scheme $Y$ to $X$. You are interested in the special case where $Y$ corresponds to a fixed algebraic closure $k\to \bar k$. In that special case, a $\bar k$-point $f:Spec (\bar k) \to X$ of $X$ certainly has an image $x=f(\ast)\in X$.
However the crucial point is that this image does not determine $f$ at all. You also have to give yourself a $k$-algebra morphism $\kappa (x) \to \bar k$ in order to define $f$.
So the same $x$ can correspond to billions of $\bar k$-points, say $7$ billion.

An example Consider $k=\mathbb Q$ and $X=Spec( \mathbb Q[T]/\langle T^{7,000,000,000}-2\rangle)=Spec(K)=\lbrace x\rbrace $.
Although $X$ has only one physical point, namely $x$, there are 7 billion different $\bar {\mathbb Q}$- points in $X$ .

[They correspond -via the affine scheme/ring dictionary- to the $\mathbb Q$-algebra morphisms $K \to \bar {\mathbb Q}$, which in turn are uniquely determined by the choice of a 7,000,000,000-th root of 2 in $\bar {\mathbb Q}$]

The case of varieties In the case of a variety $X$, the closed physical points are exactly the images of the $\bar k$-points of $X$ (see Keenan's answer). Completeness of $X$ is irrelevant.

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    $\begingroup$ This is a very nice answer Georges (as your answers tend to be)! $\endgroup$ Commented Sep 10, 2011 at 20:45
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    $\begingroup$ Dear Keenan, considering you have written an answer yourself, your comment is a really gracious gesture: bravo and thank you. $\endgroup$ Commented Sep 10, 2011 at 21:18
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    $\begingroup$ One should mention that for a $k$-variety, $|X| = X(\bar{k})/Gal(\bar{k}/k)$: closed points correspond to Galois orbits of $\bar{k}$-points as the Galois group acts transitively on the embeddings $\kappa(x) \hookrightarrow \bar{k}$. In the case $X = \mathbb{A}^1$, this is a generalization of the fact that the Galois orbit of $\alpha \in \bar{k} = \mathbb{A}^1(\bar{k})$ consists of the roots of its minimal polynomial $P$ over $k$, i.e. corresponds to the maximal ideal $(P)$ of $k[X]$. $\endgroup$
    – AFK
    Commented Sep 10, 2011 at 23:22
  • $\begingroup$ Dear user8882, thanks for your judicious comment. Just a detail: I would use the notation $Aut(\bar k/k)$ instead of $Gal(\bar k /k)$, since $\bar k /k$ is not Galois unless $k$ is perfect. $\endgroup$ Commented Sep 10, 2011 at 23:59
  • $\begingroup$ I love the clarity of your distinction, Georges. And your example! $\endgroup$
    – Descartes
    Commented Sep 11, 2011 at 8:14

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