# Does there always exist a linear shift of a given polynomial such that all coefficients are nonzero?

Let $$K$$ be a field and let $$f(x)=a_0+a_1x+\cdots +a_nx^n\in K[x]$$ be such that $$a_n\neq 0$$. Does there always exists some $$\alpha\in K$$ such that the coefficients of $$f(x-\alpha)$$ are nonzero in every degree $$0\leq i\leq n$$?

For example, if $$K$$ does not have characteristic $$p$$ then the polynomial $$x^n$$ satisfies the above, since the shift $$(x-1)^n$$ has nonzero coefficients in every degree by the binomial theorem. In characteristic $$p$$, the statement seems like it will not be true in general. For example, already in $$K=\mathbb F_p$$, the polynomial $$f(x)=x^p-x$$ cannot be shifted by any element in $$\mathbb F_p$$ so that it has nonzero coefficients in every degree: if $$a\in \mathbb F_p$$ then $$(x-a)^p-(x-a)=x^p-a^p-x+a=x^p-x.$$

Note: I began thinking about this question after a colleague asked why we only consider the lower convex hull in the definition of the $$p$$-adic Newton Polygon. My initial explanation was that "most" polynomials will have at least one coefficient equal to 0 (thus, valuation $$=\infty$$) hence the full convex hull doesn't make much sense, in that the "sides" of the Newton Polygon in such cases are just vertical lines extending up to infinity. But if we're able to shift any polynomial so that all of its coefficients are nonzero, then the full convex hull of points would make sense.

As you note, if $$\operatorname{char}k=p$$ then this fails by virtue of the polynomial $$X^p-X$$.
If $$\operatorname{char}{k}=0$$ then all coefficients of $$f$$ are nonzero if and only if $$f^{(i)}(0)\neq0$$ for all $$0\leq i\leq n$$, where $$f^{(i)}$$ denotes the $$i$$-th derivative of $$f$$. Of course if $$f_{\alpha}:=f(X-\alpha)$$ then $$f_{\alpha}^{(i)}(0)=f^{(i)}(-\alpha),$$ so it suffices to to show that there exists $$\alpha\in k$$ such that $$f^{(i)}(-\alpha)\neq0$$ for all $$i$$. Because the $$f^{(i)}$$ together have only finitely many zeroes (unless $$f=0$$) and $$k$$ is infinite, such an $$\alpha$$ exists.
If $$\operatorname{char}k=0$$ then it is true except for $$f=0$$; expanding $$f(X-\alpha)$$ shows that $$\begin{eqnarray*} f(X-\alpha) &=&\sum_{i=0}^na_i(X-\alpha)^i =\sum_{i=0}^na_i\left(\sum_{j=0}^i\binom{i}{j}(-\alpha)^{i-j}X^j\right)\\ &=&\sum_{i=0}^n\left(\sum_{j=i}^n\binom{j}{i}a_j(-\alpha)^{j-i}\right)X^i, \end{eqnarray*}$$ where the coefficients (I hope I got them right...) $$p_i(a_1,\ldots,a_n,\alpha):=\sum_{j=i}^n\binom{j}{i}a_j(-\alpha)^{j-i},$$ are polynomials in $$a_1,\ldots,a_n,\alpha$$ with coefficients in $$k$$. Note that none of the polynomials $$p_i(a_1,\ldots,a_n,X):=\sum_{j=i}^n\binom{j}{i}a_j(-X)^{j-i},$$ is identically zero because $$a_n\neq0$$ (unless $$f=0$$). Hence for each of them there are only finitely many values of $$\alpha$$ for which $$p_i(a_1,\ldots,a_n,\alpha)=0$$. Because $$k$$ is infinite there are infinitely many values of $$\alpha$$ for which all coefficients are nonzero.