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! 



Zariski closures of one parameter additive maps in positive characteristic
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.


