I'm having a hard time understanding the theorem in the title, more specifically the proof of the related fact that the image of a dominant morphism contains a dense open set of it's closure. (My sources are Liu "Algebraic Geometry and Arithmetic Curves" pg. 98, Hartshorne "Algebraic Geometry" pg. 94 or Humphreys "Linear Algebraic Groups" pg. 32)

I would like to understand this theorem from a naive stand point of polynomial maps and algebraic sets (i.e. zero sets of polynomials in some affine space), however I need to work over an imperfect field. Upon attempting to understand the commutative algebra (seems to rely on only Noether normalization, localization, and integrality) behind the statement in the initial line, I am unable to see where things fail (if they do) when working over an imperfect field.

The following example confuses me. Suppose $K$ is an imperfect field of characteristic $2$, and $f: K \to K$ is given by $f(k) = k^2$. The Zariski closure of the image is then $K$ and this map is dominant. However, the image is a subgroup of the algebraic group $(K,+)$, and so if it is constructible, then it must be Zariski closed (Humphreys pg. 54). However, the imperfectness of $K$ ensures us that the image is not surjective, contradicting the fact that the Zariski closure of the image is all of $K$.

Apologies if I am missing something obvious, as I am not an algebraic geometer by trade!

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    $\begingroup$ I'm definitely not an algebraic geometer by trade, either, but in this kind of concrete approach (not using scheme language) it's usual to start over an algebraically closed field and then see what happens to results over smaller fields of definition for a variety. Bringing imperfect fields into the picture complicates the classical methods and motivates a transition to schemes. In older books by Borel, Springer, or me, most of the work is done first over an algebraic closure to avoid the fine points about fields of definition (or other rings). $\endgroup$ Nov 11, 2011 at 21:08
  • $\begingroup$ Haha, yeah, I've noticed! It's been difficult for a novice like me to discern exactly what results require a perfect, or algebraically closed field, versus what doesn't. I must be mistaken but it seems to me that the algebra on page 32 leading up to the theorem on the same page does not require any special assumptions on the field. $\endgroup$
    – Confused
    Nov 11, 2011 at 22:47

1 Answer 1


As Jim Humphreys has pointed out in the comments, you have to either work over an algebraically closed field, or use scheme language. It is clear that over an algebraically closed field your example does not make any problems. So let us look at your example from the point of view of schemes.

What you are considering is the morphism $f:\mathbb{A}^1_K\to\mathbb{A}^1_K$ of $K$-schemes which is obtained as follows: recall that $\mathbb{A}^1_K=\operatorname{Spec} K[x]$, hence to give a morphism as above is the same as to give a homomorphism of $K$-algebras, and our homomorphism is given by $x\mapsto x^2$. Then, the morphism $f$ is not only dominant but surjective as a morphism of schemes.

How does one see this? $\mathbb{A}_K^1$ is the set of prime ideals in $K[x]$. There are two types of these: the prime ideal $(0)$, which defines the generic point, and the maximal ideals. Every maximal ideal is generated by a unique monic irreducible polynomial $f$. Hence if $K$ is not algebraically closed, there are strictly more points in $\mathbb{A}_K^1$ than the generic point and those corresponding to elements of $K$. By this classification of prime ideals it is now an easy exercise to deduce surjectivity of $f$.

However, as you rightly remarked, for $K$ imperfect of characteristic two, the induced map on $K$-rational points is not surjective and the image is "weird" in the sense that it is not of the form $S(K)$ for a subscheme $S\subseteq\mathbb{A}^1_K$. But your argument is wrong. The result from Humphreys that you quote again has to be understood as a statement about schemes, and only in the case of an algebraically closed base field can it be interpreted in the "naive" way as a statement about $K$-rational points.

One correct argument is this: if a subset $S\subseteq\mathbb{A}^1_K$ is constructible (i.e., a subscheme), then it is either finite or the complement of a finite subset. This is because the closed proper subsets of $\mathbb{A}_K^1$ are precisely the finite sets of closed points. But if $K$ is imperfect of characteristic two, than the subset $f(K)\subset K$ is infinite and has infinite complement.

By the way, there is a much easier example: take $f:\mathbb{A}^1_K\to\mathbb{A}^1_K$ as above, but with $K=\mathbb{R}$. Then the image of the induced map $\mathbb{R}\to\mathbb{R}$ is the set of nonnegative reals, clearly not "constructible", by the same reason.

Finally, here's the correct version of Chevalley's theorem:

Theorem (EGA IV, 1.8.4.) Let $f:X\to Y$ be a finitely presented morphism of schemes (any morphism between varieties over a field is of this type). Then the image of any constructible subset of $X$ under $f$ is a constructible subset of $Y$.

If $K$ is an algebraically closed field and $X$ and $Y$ are varieties over $K$ and $f$ is a morphism of $K$-varieties, then $X(K)$ can be identified with the set of closed (!) points on $X$, and you obtain the "naive" version.

  • $\begingroup$ Thank you for your comment. A crucial difference between the real case and the imperfect case for the above map though is that in the imperfect case the image is a subgroup like I mentioned, therefore Chevalley's theorem actually fails, at least in the naive sense of understanding it. As I mentioned the associated commutative algebra though doesn't seem to rely on anything to do with separability or the like and so it is still quite confusing to me why it doesn't work. It seems that I need to learn more about maps of $K$-rational points versus maps of the underlying schemes! $\endgroup$
    – Confused
    Nov 11, 2011 at 22:35
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    $\begingroup$ You are right, the associated commutative algebra, i.e. the statements about prime ideals in rings work over any field. But that precisely means you have to translate this into a statement about spectra, not about $K$-rational points. Perhaps this remark may help you: Hilbert's Nullstellensatz is not true over a field which is not algebraically closed. $\endgroup$ Nov 12, 2011 at 10:51
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    $\begingroup$ I edited my last comment into the answer, so I delete it. $\endgroup$ Nov 12, 2011 at 11:07
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    $\begingroup$ The statement in EGA is actually for locally constructible sets. Note also that EGA's definition of constructibility is not the standard one (a finite union of locally closed sets) unless the scheme is noetherian. $\endgroup$ Nov 12, 2011 at 15:22
  • $\begingroup$ @Akhil: that's right. Of course, in the case of varieties over fields, all those things are equivalent. $\endgroup$ Nov 13, 2011 at 18:12

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