Assume we have a finite field $F$ with $p^k$ elements. We start with a single non-zero element $x \in F$, and in one operation we can get the sum or product of any two elements we already have. We need to get to $1$ in the minimal number of operations. Let $C_x$ be the minimal number of operations required for $x$. Alternative definition: $C_x$ is minimal arithmetic circuit complexity for function $f$ with property $f(x) = 1$.
What is maximum value of $C_x$ across all $x$ (lets denote it $C$)?
If we are restricted to multiplication only, then the minimal number of operations for $x$ is $\lambda(\operatorname{ord}_{F^\times}(x))$ where $\lambda(n) = \min_m \ell(mn)$ and $\ell(m)$ is the length of shortest addition chain for $m$. As the set over which minimum is taken is the smallest when $x$ is a generator in $F^\times$, the maximum number of operations needed is $\lambda(|F| - 1)$, and there exists a circuit with complexity that works for all $x$ (lets call such circuits "blind").
Using only multiplication is not always optimal: for example, in $\mathbb Z_5$, we can get from $3$ to $1$ in one operation ($1 \equiv 3 + 3\pmod 5$), but we need two multiplications. However, I was not able to find any example where we can save more than one operation using addition, or where addition allows improvement over multiplication-only for all elements.
Is it possible to do better if addition is available - ie is it possible that $C < \lambda(|F| - 1)$? Is there blind circuit with less than $\lambda(|F| - 1)$ complexity? Is there a blind circuit with complexity $C$?
This question is heavily inspired by Finding the 1 of a finite field. The difference is that in my question we know the field structure, but can't request new elements. A lower bound of $C$ operations for this question implies boundary of at least $C/4$ for the original one (as Alice, when we ask for new elements, can just use elements that are reachable by at most two operations). Also, if for this question there is a blind optimal circuit, then the answer for the original question is not greater than for this one.
