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Suppose we are in characteristic $p$, and that the field, $K$, that we are working over is imperfect. We have a map $\Theta: K \to K^{\delta}$ where each coordinate function $\Theta_i$ is an additive function. Suppose further that at least one of the $\Theta_i$ has non-zero derivative.

Is the Zariski closure of the image a smooth algebraic set in $K^{\delta}$?

I believe that essentially automatically the transcendental dimension of it's function field is one, as $\dim K = \dim \ker \Theta + \dim \overline{\Theta(K)^z}$.

I want to believe that the tangent space has dimension one also! Apologies if this question is trivially true or false, I'm by no means very familiar with algebraic geometry and am learning on the fly. The lack of the map being automatically separable, and understanding the set only as a parametrization (i.e. not knowing about the functions whose zero set defines it) is confusing me.

Any suggestions towards understanding these tangent spaces would be helpful!

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What is $\delta$? Is it a positive integer? – S. Carnahan Oct 28 '11 at 21:00
Yep, sorry for any confusion. – Confused Oct 28 '11 at 21:09
What do you mean by additive function? If it is a function in the set-theoretic sense that is additive, then Ramsey gave you a counterexample. If it's an additive polynomial ($\sum a_ix^{p^i}$) then the image is a curve. – Felipe Voloch Oct 28 '11 at 22:50
That's interesting. But I'm not sure I totally get what this other notion of "additive" is in the context of this question, Felipe. I do understand the sense in which the polynomial you write is additive, but in what sense could that be a (component) function $K\to K$ (presuming $K$ to be roughly as in my answer)? Oh - do you have in mind your $a_i$ living in the larger field $K$ and the map being $f(t)\mapsto \sum a_i(f(t))^{p^i}$? What about other imperfect $K$? Is there a notion of an additive function $K\to K$ that encapsulates your examples (and excludes those as in my example)? – Ramsey Oct 29 '11 at 2:26
@Ramsey. I have no idea what the appropriately named "Confused" wants. But if $K$ is an arbitrary field, the only reasonable maps from $K$ to $K$ are polynomials. You obviously have on the back of your mind the example of a function field, hence your "t", in general there is no "t" or "larger field", only one field. You could also take derivatives and get additive maps $f \mapsto f'$ (a field is imperfect iff it has non trivial derivations, right?). But we should not be trying to guess what the question should be. It's Confused's duty to come here and clarify. – Felipe Voloch Oct 29 '11 at 8:47
up vote 2 down vote accepted

The image of a map $\Theta(x) = (\Theta_1(x),\ldots,\Theta_{\delta}(x))$ where the $\Theta_i$ are polynomials is always an algebraic variety of dimension at most one, and of dimension one if one of these polynomials is non constant. This follows from general facts. Now, there is no tangent space to the image, what you can talk about is the tangent space to a point $\Theta(a)$ on this set. The tangent space of the parametrization is, of course, $(\Theta_1'(a),\ldots,\Theta_{\delta}'(a))$, but there can be multiple branches if $\Theta$ is not injective. However if $\Theta = \Phi \circ f$, where $f$ is a polynomial in one variable, we have to analyze $\Phi$ instead.

In the special case of additive polynomials, the derivative is constant and, if you assume that one of the components is non-zero, then the derivative is never zero. Hence, the image is a smooth curve if and only if, $\Theta = \Phi \circ f$ where $\Phi$ is injective. Now this is just linear algebra $\Theta$ fails to be injective in the common solutions of $\Theta_i(x-y)=0$ and, putting $f$ as the common factor of the $\Theta_i$ we are done.

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Thanks again for your help. What characterization of tangent space are you using? My entire problem seems to revolve around knowing very little about the coordinate ring of the image variety, and the tangent space characterizations I know require knowledge of the ideal of functions which vanish on the image in some form or another. (which I don't really have) I'm still unsure how to rule out that the tangent space at say the identity doesn't have dimension greater than 1. Is there a source that you can recommend in addition to your comment? – Confused Oct 29 '11 at 20:34
Just use the implicit function theorem. – Felipe Voloch Oct 29 '11 at 21:50
Or better, the tangent space of $\Phi({\mathbb A}^1)$ at $\Phi(a)$ is the image of the tangent space of ${\mathbb A}^1$ at $a$ by $d\Phi_a$ or something. Calculus! – Felipe Voloch Oct 29 '11 at 22:01
This last comment is exactly the issue that I'm worried about, that the $d\Theta$ may not be surjective. Like you say, certainly the tangent space at the identity of the image contains a vector $(\Theta_1'(0), \ldots, \Theta_{\delta}'(0))$, however how does one know that that is everything? As far as I know, this is related to the separability of $\Theta$, which seems to me to be impossible to check directly. – Confused Oct 29 '11 at 22:31
I need to think about an exact statement of the inverse function theorem in this setting, probably it's obvious, but would it imply that these algebraically defined tangent spaces are isomorphic? – Confused Oct 29 '11 at 22:32

I'd like to understand the motivation for this question to see if I'm barking up the wrong tree here, but I think it's false (the later one-dimensional verbiage, that is) as stated.

Let $k$ be an algebraically closed field of of characteristic $p$ and let $K$ be the function field in the single variable $t$ over $k$. So $K$ is an imperfect field of characteristic $p$.

Consider the map $K\to K^2$ defined by $f(t)\mapsto (f(t),f(t+1))$. It's component functions are additive.

Certainly, this map isn't surjective. On the other hand, take a polynomial $P(X,Y)\in K[X,Y]$ vanishing on its image. Clearing denominators from $K$, we may regard $P$ as a polynomial $P(t,X,Y)\in k[t,X,Y]$ that has the property that $P(t,f(t),f(t+1))$ is the zero rational function in $t$ for all $f(t)\in K$. But, given any triple $(a,b,c)\in k^3$ it is easy to find (even a linear polynomial) $f(t)$ such that $f(a)=b$ and $f(a+1)=c$. It follows that $P(t,X,Y)$ is the zero polynomial (since $k$ is algebraically closed).

Thus there is non-zero $P$ vanishing on the image, which means that the Zariski closure of the image is all of $K^2$.

This seems to have nothing to do with the characteristic. The argument (if non-bogus) works fine for any algebraically closed $k$. I think the issue is that "additive" is an odd condition here from the point of view of algebraic geometry.

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Thanks for taking the time to think of my question, however my poor wording seems to be a problem. More explicitly, the map $\Theta$ is a homomorphism $\Theta: K \to K^{\delta}$ $x \mapsto (\sum_{j=0}^n a_{1j} x^{p^{j}}, \ldots, \sum_{j=0}^n a_{\delta j} x^{p^{j}})$ with the $a_{ij} \in K$. The derivative condition amounts to at least one of the $a_{i0} \neq 0$. I am thinking of this as a one-parameter subgroup of $K^{\delta}$, and upon taking the Zariski closure of the image, I'm having a hard time understanding the tangent space of this Zariski closed set. – Confused Oct 29 '11 at 14:17
Gotcha. Just additive polynomials. This would have been a natural interpretation indeed - somehow my mind jumped to the far more general notion. – Ramsey Oct 29 '11 at 15:33

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