Zariski closures of one parameter additive maps in positive characteristic

Question

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!

Answer 1

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.

Answer 2

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.